Free Math Worksheets

- Write the accumulation function as an integral from the fixed lower limit to \(x\). - Find an antiderivative of the polynomial. - Subtract the antiderivative value at the lower limit. - Check that the resulting formula equals zero at \(x=1\).

1. An antiderivative is \(F(t)=\frac{1}{4}t^4-t^3+t^2\). 2. By the Fundamental Theorem of Calculus, \(I_1(x)=F(x)-F(1)\). 3. Since \(F(1)=\frac{1}{4}-1+1=\frac{1}{4}\), \(I_1(x)=\frac{1}{4}x^4-x^3+x^2-\frac{1}{4}\).

\(I_1(x)=\frac{1}{4}x^4-x^3+x^2-\frac{1}{4}\)

- Use the definition of an accumulation function. - Account for the coefficient in the exponent when integrating \(e^{kt}\). - Recall that \(e^0=1\). - Subtract the entire antiderivative value at the lower limit.

1. An antiderivative of \(e^{0.5t}+2\) is \(F(t)=2e^{0.5t}+2t\). 2. Therefore, \(I_0(x)=F(x)-F(0)\). 3. Since \(F(0)=2\), \(I_0(x)=2e^{0.5x}+2x-2\).

\(I_0(x)=2e^{0.5x}+2x-2\)

- Use the definition of an accumulation function with lower limit \(2\). - Find an antiderivative of the polynomial. - Evaluate the antiderivative at the fixed lower limit. - Check that your formula equals zero at \(x=2\).

1. An antiderivative is \(F(t)=2t^3-2t^2+5t\). 2. Therefore, \(I_2(x)=F(x)-F(2)\). 3. Since \(F(2)=16-8+10=18\), \(I_2(x)=2x^3-2x^2+5x-18\).

\(I_2(x)=2x^3-2x^2+5x-18\)

- Use an integral with variable upper limit and fixed lower limit \(0\). - Account for the coefficient in the exponent of \(e^{2t}\). - Evaluate the antiderivative at \(t=0\). - Check that the final formula gives \(I_0(0)=0\).

1. An antiderivative is \(F(t)=\frac{1}{2}e^{2t}-e^t\). 2. Therefore, \(I_0(x)=F(x)-F(0)\). 3. Since \(F(0)=\frac{1}{2}-1=-\frac{1}{2}\), \(I_0(x)=\frac{1}{2}e^{2x}-e^x+\frac{1}{2}\).

\(I_0(x)=\frac{1}{2}e^{2x}-e^x+\frac{1}{2}\)

- Account for the factor \(2\) in the exponent. - Use \(e^0=1\). - Take natural logarithms after isolating the exponential. - Simplify with logarithm properties.

1. An antiderivative is \(F(x)=\frac{1}{2}e^{2x}\). 2. Apply the bounds: \(\frac{1}{2}e^{2k}-\frac{1}{2}=4\). 3. Therefore, \(e^{2k}=9\). Taking natural logarithms gives \(2k=\ln(9)\), so \(k=\frac{1}{2}\ln(9)=\ln(3)\).

\(k=\ln(3)\)

- Express the area as an accumulation function of \(k\). - Use the power rule. - Set the area formula equal to \(72\). - Take the cube root.

1. Since \(f(x)\ge 0\), the area is \(A(k)=\int_0^k x^2\,dx\). 2. Evaluate: \(A(k)=\left[\frac{x^3}{3}\right]_0^k=\frac{k^3}{3}\). 3. Set \(\frac{k^3}{3}=72\). Then \(k^3=216\), so \(k=6\).

\(k=6\)

- Express the area as a definite integral. - Apply the power rule. - Set the area expression equal to \(16\). - Use the positive fourth root.

1. Since the function is nonnegative on \([0, k]\), \(A(k)=\int_0^k\frac{1}{4}x^3\,dx\) \(=\left[\frac{x^4}{16}\right]_0^k=\frac{k^4}{16}\). 2. Set \(\frac{k^4}{16}=16\). Then \(k^4=256\), so \(k=4\) because \(k>0\).

1) \(A(k)=\frac{k^4}{16}\) 2) \(k=4\)

- Rewrite the integrand using a negative exponent. - Evaluate the accumulation as a function of \(b\). - Set it equal to \(0.8\). - Solve the reciprocal equation.

1. The function is positive, so the area is \(\int_1^b x^{-2}\,dx\). 2. An antiderivative is \(-\frac{1}{x}\). Thus, \(\left[-\frac{1}{x}\right]_1^b=1-\frac{1}{b}\). 3. Set \(1-\frac{1}{b}=0.8\). Then \(\frac{1}{b}=0.2\), so \(b=5\).

\(b=5\)

- Rewrite the reciprocal power with a negative exponent. - Evaluate the integral as a function of \(k\). - Solve the equation containing \(k\) in a denominator. - Apply the condition \(k>1\).

1. Rewrite \(f(x)=4x^{-3}\). An antiderivative is \(F(x)=-\frac{2}{x^2}\). 2. Apply the bounds: \(2-\frac{2}{k^2}=1.5\). 3. Thus, \(\frac{2}{k^2}=0.5\), so \(k^2=4\). Since \(k>1\), \(k=2\).

\(k=2\)

- Account for the factor \(2\) in the exponent. - Evaluate upper bound minus lower bound. - Isolate the exponential containing \(k\). - Use the one-to-one property of the exponential function.

1. An antiderivative is \(F(t)=\frac{1}{2}e^{2t}\). 2. Apply the bounds: \(\frac{1}{2}e^2-\frac{1}{2}e^{2k}=\frac{1}{2}(e^2-e)\). 3. Simplify: \(e^{2k}=e\). Therefore, \(2k=1\), so \(k=\frac{1}{2}\).

\(k=\frac{1}{2}\)

- Evaluate a general accumulation function at its lower limit. - Identify a necessary property of any accumulation function. - Check whether \(H\) has a real zero.

1. Every accumulation function of the form \(I_a(x)=\int_{a}^{x}h(t)\,dt\) satisfies \(I_a(a)=0\). 2. For every real \(x\), \(1+x^2>0\), so \(H(x)=\frac{1}{1+x^2}>0\). 3. Therefore, \(H\) has no zero. No lower limit \(a\) can satisfy \(H(a)=0\), so \(H\) cannot be an accumulation function.

No. Every accumulation function has a zero at its lower limit, but \(H(x)>0\) for every real \(x\).

- Recall an antiderivative of \(\frac{1}{t}\) for \(t>0\). - Substitute the coordinates of the given point. - Simplify \(\ln(e^2)\). - Undo the natural logarithm with the exponential function.

1. Since \(t>0\), an antiderivative of \(\frac{1}{t}\) is \(\ln t\). 2. Therefore, \(I_a(x)=\ln x-\ln a\). 3. Use the point \((e^2, 1)\): \(\ln(e^2)-\ln a=1\). 4. Since \(\ln(e^2)=2\), \(2-\ln a=1\), so \(\ln a=1\). Therefore, \(a=e\).

\(a=e\)

- Interpret \(I_u(x)\) as a definite integral with a variable upper limit. - Find an antiderivative of each power term. - Apply the Fundamental Theorem of Calculus at the upper and lower limits. - Be careful when subtracting the value at the lower limit.

1. An antiderivative of \(f(t)=1.5t^2-4t\) is \(F(t)=0.5t^3-2t^2\). 2. By the Fundamental Theorem of Calculus, \(I_u(x)=F(x)-F(u)\). 3. Substitute the two limits: \(I_u(x)=(0.5x^3-2x^2)-(0.5u^3-2u^2)\). 4. Therefore, \(I_u(x)=0.5x^3-2x^2-0.5u^3+2u^2\).

\(I_u(x)=0.5x^3-2x^2-0.5u^3+2u^2\)

- Rewrite the fraction using a negative exponent. - Apply the power rule for antiderivatives. - Evaluate the antiderivative at both limits. - Check that your result equals zero when \(x=2\).

1. Rewrite the integrand as \(f(t)=4t^{-2}\). 2. An antiderivative is \(F(t)=-4t^{-1}=-\frac{4}{t}\). 3. By the Fundamental Theorem of Calculus, \(I_2(x)=F(x)-F(2)=-\frac{4}{x}-(-2)\). 4. Thus, \(I_2(x)=2-\frac{4}{x}\), for \(x>0\).

\(I_2(x)=2-\frac{4}{x}\), for \(x>0\)

- Write the condition as an integral equation. - Recall an antiderivative of \(\frac{1}{x}\). - Use \(\ln 1=0\). - Undo the natural logarithm with the exponential function.

1. Evaluate the accumulation: \(\int_1^k\frac{3}{x}\,dx=[3\ln x]_1^k=3\ln k\), since \(k>1\). 2. Set the result equal to \(6\): \(3\ln k=6\), so \(\ln k=2\). 3. Apply the exponential function: \(k=e^2\), which satisfies \(k>1\).

\(k=e^2\)

- Write the integral as a function of the upper bound. - Factor the resulting polynomial equation. - Exclude \(b=0\) because \(b>0\). - Interpret zero accumulation as balanced signed areas.

1. Define the accumulation function \(A(b)=\int_0^b(x^2-4x+3)\,dx\). An antiderivative gives \(A(b)=\frac{b^3}{3}-2b^2+3b\). 2. Set \(A(b)=0\): \(b\left(\frac{b^2}{3}-2b+3\right)=0\). 3. Since \(b>0\), solve \(b^2-6b+9=0\), or \((b-3)^2=0\). Therefore, \(b=3\).

\(b=3\)

- Rewrite the reciprocal square using a negative exponent. - Evaluate the antiderivative at the variable upper bound. - Clear the denominator to obtain a quadratic equation. - Apply the condition \(b>1\).

1. An antiderivative is \(F(x)=x+\frac{1}{x}\). 2. Apply the bounds: \(b+\frac{1}{b}-2=0.5\), so \(b+\frac{1}{b}=2.5\). 3. Multiply by \(b\): \(b^2-2.5b+1=0\). The solutions are \(b=2\) and \(b=0.5\). 4. The condition \(b>1\) leaves \(b=2\).

\(b=2\)

- Use the Fundamental Theorem of Calculus. - When finding the accumulation function, evaluate an antiderivative at \(x\) and at the fixed lower limit. - Differentiate the formula you obtain and compare it with \(f(x)\).

1. An antiderivative of \(f\) is \(F(x)=0.5x^3-3x^2\). 2. For a), \(\int_{0}^{4}f(x)\,dx=F(4)-F(0)=32-48=-16\). 3. For b), \(I_0(x)=F(x)-F(0)=0.5x^3-3x^2\). 4. Differentiate: \(\frac{d}{dx}I_0(x)=1.5x^2-6x=f(x)\). Thus, the derivative of the accumulation function equals the original integrand.

a) \(-16\) b) \(I_0(x)=0.5x^3-3x^2\), and \(\frac{d}{dx}I_0(x)=f(x)\).

- Recall the derivative of an accumulation function. - Horizontal tangents occur where the derivative is zero. - Factor \(x^3-4x\) to find its zeros. - You can use the theorem directly in part b) rather than differentiating the expanded formula.

1. An antiderivative of \(g(t)=t^3-4t\) is \(\Phi(t)=\frac{1}{4}t^4-2t^2\). 2. Therefore, \(G_{-2}(x)=\Phi(x)-\Phi(-2)=\frac{1}{4}x^4-2x^2+4\). 3. By the Fundamental Theorem of Calculus, \(\frac{d}{dx}G_{-2}(x)=g(x)\). Horizontal tangents occur where \(g(x)=0\). 4. Solve: \(x^3-4x=x(x^2-4)=x(x-2)(x+2)=0\). Thus, \(x=-2,0,2\).

a) \(G_{-2}(x)=\frac{1}{4}x^4-2x^2+4\) b) \(x=-2,0,2\)

- Evaluate each integral as a function of \(z\). - Set the accumulation equal to the given value. - Solve the resulting polynomial equation. - Keep only positive solutions.

1. For a), \(\int_0^z(3x^2+6x)\,dx=z^3+3z^2\). Set \(z^3+3z^2=54\). The positive solution is \(z=3\). 2. For b), \(\int_0^z(4x+1)\,dx=2z^2+z\). Set \(2z^2+z-10=0\). Factoring gives \((2z+5)(z-2)=0\), so the positive solution is \(z=2\).

a) \(z=3\) b) \(z=2\)

- Treat the integral as an accumulation function of \(z\). - Apply the lower bound carefully. - Factor the equation in part a). - Use the one-to-one property of the exponential function in part b).

1. For a), an antiderivative is \(F(x)=\frac{x^3}{3}-x\). Applying the bounds gives \(\frac{z^3}{3}-z+\frac{2}{3}=\frac{2}{3}\). Thus, \(z\left(\frac{z^2}{3}-1\right)=0\). The positive solution is \(z=\sqrt{3}\). 2. For b), \(\int_0^z e^x\,dx=e^z-1\). Therefore, \(e^z-1=e^3-1\), so \(e^z=e^3\) and \(z=3\).

a) \(z=\sqrt{3}\) b) \(z=3\)

- Use the relationship between an accumulation function and its derivative. - A derivative value gives the slope of the graph at that input. - Recall the value of \(\ln(e)\).

1. By the Fundamental Theorem of Calculus, \(\frac{dG}{dx}=\sqrt{x}+\ln(x)\). 2. The slope at \(x=e\) is \(\left.\frac{dG}{dx}\right|_{x=e}=\sqrt{e}+\ln(e)\). 3. Since \(\ln(e)=1\), the slope is \(\sqrt{e}+1\).

\(\sqrt{e}+1\)

- Determine the value every accumulation function has at its lower limit. - Translate that condition into a property of the graph. - Check each given function for zeros. - Respect the domain of the logarithmic function.

1. An accumulation function must satisfy \(I_a(a)=0\). Conversely, if an antiderivative has a zero at \(x=a\), then it equals the integral of its derivative from \(a\) to \(x\). 2. \(F(x)=(x-3)^2\), so \(F(3)=0\). Therefore, \(F\) can be an accumulation function with lower limit \(a=3\). 3. Since \(e^x>0\), \(G(x)=e^x+2>2\) for every real \(x\). It has no zero, so it cannot be an accumulation function of this form. 4. \(H(x)=0\) when \(\ln(x)=1\), so \(x=e\). Because \(e\) is in the domain, \(H\) can be an accumulation function with lower limit \(a=e\).

\(F\): Yes, with lower limit \(a=3\). \(G\): No, because it has no zero. \(H\): Yes, with lower limit \(a=e\).

- Include an arbitrary constant when writing all antiderivatives. - An accumulation function equals zero at its lower limit. - Determine when the quadratic \(F_c\) has a real zero. - For part c), solve \(G(a)=0\).

1. For a), integrating gives \(F_c(x)=x^2-4x+c\), where \(c\in\mathbb{R}\). 2. For b), an antiderivative can be an accumulation function exactly when it has at least one real zero. The equation \(x^2-4x+c=0\) has a real solution when its discriminant is nonnegative: \(16-4c\geq0\). Therefore, \(c\leq4\). 3. For c), an accumulation function with lower limit \(a\) must satisfy \(G(a)=0\). Solve \(a^2-4a=a(a-4)=0\), giving \(a=0\) or \(a=4\).

a) \(F_c(x)=x^2-4x+c\), where \(c\in\mathbb{R}\) b) \(c\leq4\) c) \(a=0\) or \(a=4\)

- Consider an integral whose upper and lower limits are equal. - Use the Fundamental Theorem of Calculus to find the derivative. - Relate the sign of the derivative to monotonicity. - A strictly increasing function cannot cross the x-axis more than once.

1. For a), \(I_2(2)=\int_{2}^{2}f(t)\,dt=0\), so \(x=2\) is a zero. 2. For b), the Fundamental Theorem of Calculus gives \(\frac{d}{dx}I_2(x)=\sqrt{x^4+1}\). Since \(x^4+1\geq1\), the derivative is positive for every real \(x\). Therefore, \(I_2\) is strictly increasing on \(\mathbb{R}\). 3. For c), a strictly increasing function can take the value \(0\) at most once. Since \(I_2(2)=0\), \(x=2\) is its only zero.

a) \(x=2\) b) \(\frac{d}{dx}I_2(x)=\sqrt{x^4+1}>0\), so \(I_2\) is strictly increasing. c) \(x=2\) is the only zero.

- Evaluate the accumulation function at its lower limit. - Use the Fundamental Theorem of Calculus to determine its derivative. - Relate the sign of \(g\) to the monotonicity of \(G_a\). - Decide how many times a strictly decreasing function can take the value zero.

1. At the lower limit, \(G_a(a)=\int_{a}^{a}g(t)\,dt=0\), so \(x=a\) is a zero. 2. By the Fundamental Theorem of Calculus, \(\frac{d}{dx}G_a(x)=g(x)\). 3. Since \(g(x)<0\) for every real \(x\), the derivative of \(G_a\) is always negative. Thus, \(G_a\) is strictly decreasing. 4. A strictly decreasing function takes each value at most once. Because \(G_a(a)=0\), there can be no other zero. Therefore, \(G_a\) has exactly one zero.

\(G_a(a)=0\), and \(\frac{d}{dx}G_a(x)=g(x)<0\), so \(G_a\) is strictly decreasing. Therefore, \(x=a\) is its unique zero.

- Evaluate the integral as a function of \(k\). - Recognize the resulting equation as quadratic in \(k^2\). - Substitute \(u=k^2\). - Apply the condition \(k>0\).

1. An antiderivative is \(F(x)=x^4-3x^2\). 2. The equation becomes \(k^4-3k^2=4\), or \(k^4-3k^2-4=0\). 3. Let \(u=k^2\). Then \(u^2-3u-4=(u-4)(u+1)=0\). 4. Since \(k\) is real and positive, \(k^2=4\), so \(k=2\).

\(k=2\)

- Account for the inner factor when integrating. - Isolate the cosine expression. - Use the allowed interval to select the correct angle. - Verify the value in the original integral.

1. An antiderivative is \(F(x)=-\frac{1}{2}\cos(2x)\). 2. Apply the bounds: \(-\frac{1}{2}\cos(2a)+\frac{1}{2}=1\). 3. Thus, \(\cos(2a)=-1\). For \(a\in[0, \pi]\), the unique solution is \(2a=\pi\), so \(a=\frac{\pi}{2}\).

\(a=\frac{\pi}{2}\)

- Equal geometric areas correspond to a signed integral of \(0\). - Evaluate the integral as a function of \(k\). - Factor the resulting equation. - Check that the interval crosses the x-intercept.

1. Equal areas above and below the x-axis mean that the signed integral is zero: \(\int_0^k(9-x^2)\,dx=0\). 2. Evaluate: \(9k-\frac{k^3}{3}=0\). 3. Factor: \(k\left(9-\frac{k^2}{3}\right)=0\). Since \(k>0\), \(k^2=27\), so \(k=3\sqrt{3}\). 4. Because \(3\sqrt{3}>3\), the interval extends past the x-intercept at \(x=3\), so it includes regions on both sides of the x-axis.

\(k=3\sqrt{3}\approx 5.20\)

- Solve the equation \(f(x)=0\). - Use the relationship between an accumulation function and its derivative. - Apply the second derivative test to classify the critical point.

1. For a), solve \(4-e^{0.5x}=0\). Then \(e^{0.5x}=4\), so \(0.5x=\ln(4)\) and \(x=2\ln(4)\). 2. For b), the Fundamental Theorem of Calculus gives \(\frac{d}{dx}I_0(x)=f(x)\). Thus, the only critical point occurs at \(x=2\ln(4)\). 3. Since \(\frac{d^2}{dx^2}I_0(x)=f'(x)=-0.5e^{0.5x}\), this second derivative is negative at the critical point. Therefore, \(I_0\) has a local maximum at \(x=2\ln(4)\).

a) \(x=2\ln(4)\approx2.773\) b) \(I_0\) has a local maximum at \(x=2\ln(4)\).

- Evaluate the antiderivative at both variable bounds. - Expand carefully before subtracting. - Solve the resulting quadratic equation. - Both real solutions are allowed.

1. An antiderivative is \(F(x)=x^2-6x\). 2. Apply the variable bounds: \(F(2k)-F(k)=3k^2-6k\). 3. Set this equal to \(9\): \(3k^2-6k=9\), or \(k^2-2k-3=0\). 4. Factor: \((k-3)(k+1)=0\). Therefore, \(k=3\) or \(k=-1\).

\(k=3\) or \(k=-1\)

- Account for the linear exponent when integrating. - Use \(e^0=1\) at the lower bound. - Cancel matching terms on both sides. - Take natural logarithms after isolating the exponential.

1. An antiderivative is \(F(x)=2e^{x/2}-x\). 2. Apply the bounds: \(2e^{k/2}-k-2=4-k\). 3. The \(-k\) terms cancel, leaving \(2e^{k/2}=6\), so \(e^{k/2}=3\). 4. Take natural logarithms: \(\frac{k}{2}=\ln(3)\), so \(k=2\ln(3)\).

\(k=2\ln(3)\)

- Express the area as an accumulation function of \(b\). - Include the lower-bound value. - Solve the resulting cubic equation. - Check simple values greater than \(1\).

1. The function is positive for \(x\ge 1\), so \(\int_1^b(3x^2+2x)\,dx=78\). 2. An antiderivative is \(G(x)=x^3+x^2\). Therefore, \(b^3+b^2-2=78\), or \(b^3+b^2-80=0\). 3. Factor: \(b^3+b^2-80=(b-4)(b^2+5b+20)\). The quadratic factor has discriminant \(25-80=-55<0\), so it has no real roots. Therefore, the only real solution, and hence the solution with \(b>1\), is \(b=4\).

\(b=4\)

- Evaluate the accumulation using an antiderivative. - Solve the resulting quadratic equation. - Apply the condition \(k>2\). - Interpret a definite integral as signed area.

1. An antiderivative is \(F(x)=\frac{x^2}{2}-3x\). 2. Apply the bounds: \(\frac{k^2}{2}-3k+4=7.5\). This becomes \(k^2-6k-7=0\). 3. Factor: \((k-7)(k+1)=0\). The condition \(k>2\) gives \(k=7\). 4. Geometrically, the signed area from \(x=2\) to \(x=7\) is \(7.5\): the positive area above the x-axis exceeds the negative area below it by \(7.5\) square units.

\(k=7\). The signed area between the graph and the x-axis on \([2, 7]\) is \(7.5\) square units.

- Integrate with respect to \(t\), not \(x\). - Evaluate the accumulation at the variable upper limit. - Test simple integer roots of the cubic. - Check whether the remaining quadratic has real roots.

1. An antiderivative with respect to \(t\) is \(F(t)=t^3-t\). 2. Apply the bounds: \(x^3-x-(1^3-1)=6\), so \(x^3-x-6=0\). 3. Since \(x=2\) is a root, factor: \(x^3-x-6=(x-2)(x^2+2x+3)\). The quadratic has negative discriminant, so the only real solution is \(x=2\).

\(x=2\)

- Use the Fundamental Theorem of Calculus. - Determine whether the denominator can be zero or negative. - Relate a positive derivative to monotonicity. - Consider how changing the lower limit affects two antiderivatives of the same function.

1. For a), the Fundamental Theorem of Calculus gives \(\frac{d}{dx}F_c(x)=\frac{2}{x^2+1}\). 2. For b), \(x^2+1>0\) for every real \(x\), so \(\frac{2}{x^2+1}>0\). Therefore, \(F_c\) is strictly increasing on \(\mathbb{R}\). 3. For c), changing \(c\) changes only the additive constant and therefore shifts the graph vertically. The derivative remains \(\frac{2}{x^2+1}\), so the monotonicity does not change.

a) \(\frac{d}{dx}F_c(x)=\frac{2}{x^2+1}\) b) The derivative is positive for every real \(x\), so \(F_c\) is strictly increasing. c) No. Changing \(c\) shifts the graph vertically but does not change its derivative.

- Use the Fundamental Theorem of Calculus to find the derivative. - Determine the sign of \(-(x^4+3)\). - Recall the value of an accumulation function at its lower limit. - Use monotonicity to compare \(G_2(10)\) with \(G_2(2)\).

1. For a), the Fundamental Theorem of Calculus gives \(\frac{d}{dx}G_a(x)=-(x^4+3)\). 2. Since \(x^4+3>0\) for all real \(x\), the derivative is always negative. Therefore, \(G_a\) is strictly decreasing on \(\mathbb{R}\). 3. For b), \(G_2(2)=0\). Because \(10>2\) and \(G_2\) is strictly decreasing, \(G_2(10)<G_2(2)=0\). Thus, \(G_2(10)\) is negative.

a) \(\frac{d}{dx}G_a(x)=-(x^4+3)<0\), so \(G_a\) is strictly decreasing. b) \(G_2(10)<0\)

- Use the value an accumulation function has at its lower limit. - Factor \(F(x)\) to find all zeros. - Differentiate \(F\) to recover the integrand.

1. An accumulation function must satisfy \(I_a(a)=0\), so the possible values of \(a\) are the zeros of \(F\). 2. Factor: \(F(x)=x(x^2-3x+2)=x(x-1)(x-2)\). Thus, \(a=0,1,2\). 3. By the Fundamental Theorem of Calculus, the integrand is the derivative of \(F\): \(f(x)=3x^2-6x+2\).

\(a\in\{0,1,2\}\), and \(f(x)=3x^2-6x+2\)

- Find an antiderivative of the polynomial. - Evaluate it at \(x\) and at the lower limit \(2\). - Differentiate the resulting accumulation function. - Compare the derivative with the original integrand.

1. An antiderivative is \(F(t)=0.5t^3-2t^2+t\). 2. Therefore, \(I_2(x)=F(x)-F(2)\). Since \(F(2)=4-8+2=-2\), \(I_2(x)=0.5x^3-2x^2+x+2\). 3. Differentiate: \(\frac{d}{dx}I_2(x)=1.5x^2-4x+1=f(x)\).

1. \(I_2(x)=0.5x^3-2x^2+x+2\) 2. \(\frac{d}{dx}I_2(x)=1.5x^2-4x+1=f(x)\)

- Recall an antiderivative of cosine. - Use the exact sine value at \(\frac{\pi}{2}\). - Differentiate the formula you obtain. - An integral over an interval of zero width equals zero.

1. An antiderivative of cosine is sine, so \(I_{\pi/2}(x)=\sin(x)-\sin\left(\frac{\pi}{2}\right)=\sin(x)-1\). 2. Differentiate: \(\frac{d}{dx}I_{\pi/2}(x)=\cos(x)=f(x)\). Therefore, \(I_{\pi/2}\) is an antiderivative of \(f\). 3. An integral with equal limits is zero, so \(I_{\pi/2}\left(\frac{\pi}{2}\right)=0\).

1. \(I_{\pi/2}(x)=\sin(x)-1\) 2. \(\frac{d}{dx}I_{\pi/2}(x)=\cos(x)=f(x)\), so yes. 3. \(I_{\pi/2}\left(\frac{\pi}{2}\right)=0\)

- Differentiate the given accumulation function to recover the integrand. - The accumulation function equals zero at its lower limit. - Test simple integer values when solving the resulting cubic. - Check whether any remaining factor has real zeros.

1. Differentiate the accumulation function: \(f(x)=\frac{d}{dx}(2x^3+x^2-3)=6x^2+2x\). 2. At the lower limit, the accumulation function must equal zero. Thus, \(2a^3+a^2-3=0\). 3. Factor: \(2a^3+a^2-3=(a-1)(2a^2+3a+3)\). The quadratic factor has discriminant \(9-24=-15\), so it has no real zeros. Therefore, \(a=1\).

\(f(x)=6x^2+2x\) and \(a=1\)

- Differentiate the accumulation function to find the integrand. - Use the value an accumulation function has at its lower limit. - Solve the resulting exponential equation.

1. Differentiate the accumulation function: \(f(x)=\frac{d}{dx}(e^{x-1}-1)=e^{x-1}\). 2. At the lower limit, \(I_a(a)=0\). Thus, \(e^{a-1}-1=0\), so \(e^{a-1}=1\). 3. Since \(e^0=1\), \(a-1=0\), giving \(a=1\).

\(f(x)=e^{x-1}\) and \(a=1\)

- Find an antiderivative of the polynomial. - Evaluate at the variable upper limit and the fixed lower limit. - Substitute \(x=3\) carefully. - Differentiate the accumulation function term by term.

1. An antiderivative is \(F(t)=\frac{1}{3}t^3-3t^2+5t\). Since \(F(0)=0\), \(I_0(x)=\frac{1}{3}x^3-3x^2+5x\). 2. Evaluate: \(I_0(3)=9-27+15=-3\). 3. Differentiate: \(\frac{d}{dx}I_0(x)=x^2-6x+5=f(x)\).

1. \(I_0(x)=\frac{1}{3}x^3-3x^2+5x\) 2. \(I_0(3)=-3\) 3. \(\frac{d}{dx}I_0(x)=x^2-6x+5=f(x)\)

- Treat the parameter \(a\) as a fixed number. - Use a different dummy variable inside the integral. - Evaluate an antiderivative at \(x\) and at \(a\). - Terms containing no \(x\) disappear when differentiating with respect to \(x\).

1. An antiderivative is \(F(t)=\frac{1}{16}t^4-\frac{1}{2}t^2\). Therefore, \(I_a(x)=F(x)-F(a)\), so \(I_a(x)=\frac{1}{16}x^4-\frac{1}{2}x^2-\frac{1}{16}a^4+\frac{1}{2}a^2\). 2. The terms involving only \(a\) are constant with respect to \(x\). Thus, \(\frac{d}{dx}I_a(x)=\frac{1}{4}x^3-x=f(x)\).

1. \(I_a(x)=\frac{1}{16}x^4-\frac{1}{2}x^2-\frac{1}{16}a^4+\frac{1}{2}a^2\) 2. \(\frac{d}{dx}I_a(x)=\frac{1}{4}x^3-x=f(x)\)

- Write the accumulation function using an antiderivative. - Substitute the coordinates of the given point. - Rearrange the resulting cubic equation. - Test simple integer roots, then factor completely.

1. Find the accumulation function: \(I_a(x)=[t^3-3t^2]_a^x=x^3-3x^2-(a^3-3a^2)\). 2. Use the point \((3, 4)\): \(I_a(3)=4\), so \(-(a^3-3a^2)=4\). 3. Rearrange and factor: \(a^3-3a^2+4=0\), \((a+1)(a-2)^2=0\). 4. Therefore, the possible lower limits are \(a=-1\) and \(a=2\).

\(a=-1\) or \(a=2\)

- Write an integral for the area from \(0\) to \(k\). - Use the power rule to integrate \(x^{1/4}\). - Isolate \(k^{5/4}\). - Use exponent rules to solve for \(k\).

1. Represent the area with an accumulation integral: \(\int_0^k x^{1/4}\,dx=25.6\). 2. Evaluate the integral: \(\left[\frac{4}{5}x^{5/4}\right]_0^k=\frac{4}{5}k^{5/4}\). 3. Solve the equation: \(\frac{4}{5}k^{5/4}=25.6\), so \(k^{5/4}=32\). 4. Raise both sides to the \(\frac{4}{5}\) power: \(k=32^{4/5}=(2^5)^{4/5}=16\).

\(k=16\)

- Evaluate the integral in terms of \(b\). - Set the resulting expression equal to \(12\). - Solve the quadratic equation. - Apply the condition that \(b\) is positive.

1. Evaluate the accumulation in terms of \(b\): \(\int_0^b(2x+1)\,dx=[x^2+x]_0^b=b^2+b\). 2. Set the result equal to \(12\): \(b^2+b=12\), so \(b^2+b-12=0\). 3. Factor: \((b-3)(b+4)=0\). Thus, \(b=3\) or \(b=-4\). 4. Because \(b\) must be positive, \(b=3\).

\(b=3\)

- Evaluate an antiderivative at \(x\) and at the lower limit. - Every accumulation function equals zero at its lower limit. - Rewrite \(F\) as a square plus a constant to determine whether it has a zero.

1. An antiderivative is \(T(t)=t^4-2t^2\). Thus, \(I_1(x)=T(x)-T(1)=x^4-2x^2+1\). 2. Every accumulation function has a zero at its lower limit. Rewrite \(F(x)=x^4-2x^2+5=(x^2-1)^2+4\). Since \(F(x)\geq4\) for every real \(x\), it has no zero. Therefore, \(F\) cannot be an accumulation function of \(f\).

1. \(I_1(x)=x^4-2x^2+1\) 2. No. Since \(F(x)=(x^2-1)^2+4\geq4\), \(F\) has no zero, but every accumulation function must have a zero at its lower limit.

- Differentiate the given accumulation function. - Use the condition \(I_a(a)=0\). - Solve the resulting quadratic equation. - Compare the definitions of an antiderivative and an accumulation function.

1. Differentiate: \(f(x)=\frac{d}{dx}I_a(x)=x-2\). 2. The lower limit must satisfy \(I_a(a)=0\). Thus, \(\frac{1}{2}a^2-2a-6=0\), or \(a^2-4a-12=0\). Factoring gives \((a-6)(a+2)=0\), so \(a=6\) or \(a=-2\). 3. By the Fundamental Theorem of Calculus, the derivative of an accumulation function is its integrand, so it is an antiderivative. An arbitrary antiderivative is an accumulation function only if it has a zero that can serve as the lower limit.

1. \(f(x)=x-2\) 2. \(a=6\) or \(a=-2\) 3. Every accumulation function differentiates to its integrand. An antiderivative is an accumulation function only when it has at least one zero.

- Recall an antiderivative of \(\frac{1}{t}\). - Use a linear substitution for the integral on the right. - Simplify both definite integrals. - Use the one-to-one property of the natural logarithm.

1. Evaluate the right side: \(2\int_0^1\frac{1}{2t+1}\,dt =2\left[\frac{1}{2}\ln(2t+1)\right]_0^1 =\ln 3\). 2. Evaluate the left side: \(\int_1^x\frac{1}{t}\,dt=[\ln t]_1^x=\ln x\). 3. Set the results equal: \(\ln x=\ln 3\). Since the natural logarithm is one-to-one, \(x=3\).

\(x=3\)

- Find an antiderivative of the entire integrand. - Evaluate carefully at the lower limit \(0\). - Compare the structure of both sides after simplifying.

1. An antiderivative of \(e^x+1\) is \(F(x)=e^x+x\). 2. Apply the bounds: \(\int_0^k(e^x+1)\,dx=e^k+k-1\). 3. Set this equal to the given value: \(e^k+k-1=e^3+2\), so \(e^k+k=e^3+3\). 4. Substitution shows that \(k=3\) satisfies the equation. Because \(e^k+k\) is strictly increasing, this solution is unique.

\(k=3\)

- What value must an accumulation function have at its lower limit? - Use the Fundamental Theorem of Calculus to relate the derivative of \(I_{-2}\) to the integrand. - Find where the integrand equals zero. - Use the sign of the integrand to determine where the accumulation function increases or decreases.

1. Since \(I_{-2}(-2)=0\), the graph must pass through \((-2,0)\). All three curves satisfy this condition. 2. By the Fundamental Theorem of Calculus, \(\frac{d}{dx}I_{-2}(x)=-\frac{1}{2}x+1\). The derivative is zero at \(x=2\), so the graph must have a horizontal tangent there. This eliminates curve \(r\), whose vertex is at \(x=1\). 3. The derivative changes from positive to negative at \(x=2\), so \(I_{-2}\) has a local maximum there. Equivalently, its second derivative is \(-\frac{1}{2}<0\). This eliminates the upward-opening curve \(q\). 4. Therefore, curve \(p\) is the graph of \(I_{-2}\).

The correct curve is \(p\).

- What value must the accumulation function have at its lower limit? - Relate the derivative of the accumulation function to \(f\). - Use the zeros of \(f\) to locate horizontal tangents. - Use the sign of \(f\) to determine where the accumulation function increases or decreases.

1. Since the lower limit is \(2\), the accumulation function satisfies \(I_2(2)=0\). Graphs \(1\) and \(3\) pass through \((2,0)\), while Graph \(2\) does not. 2. By the Fundamental Theorem of Calculus, the derivative of \(I_2\) is \(f\). The graph of \(f\) has zeros at \(x=0\) and \(x=4\), so the accumulation function has horizontal tangents at those values. 3. Because \(f(x)<0\) for \(x<0\), \(f(x)>0\) for \(0<x<4\), and \(f(x)<0\) for \(x>4\), the accumulation function decreases, then increases, then decreases. It has a local minimum at \(x=0\) and a local maximum at \(x=4\). 4. Only Graph \(1\) has these features, so Graph \(1\) represents \(I_2\).

Graph \(1\) represents \(I_2\).

- Interpret each definite integral as signed area between the graph and the x-axis. - Areas below the x-axis contribute negative values. - Reversing the limits changes the sign of an integral. - Break the regions into triangles and trapezoids.

1. a) From \(t=0\) to \(t=1\), the graph forms a triangle above the x-axis with signed area \(\frac{1}{2}\). From \(t=1\) to \(t=2\), it forms a congruent triangle below the x-axis with signed area \(-\frac{1}{2}\). Therefore, \(I_0(2)=0\), so the statement is true. 2. b) For \(0<x\leq1\), the accumulated area is positive. For \(1<x<2\), the negative area accumulated after \(t=1\) is less than \(\frac{1}{2}\), so it does not cancel all of the earlier positive area. Thus, \(I_0(x)>0\) for every \(x\in(0,2)\), and the statement is true. 3. c) The area from \(t=-1\) to \(t=0\) is a trapezoid with bases \(2\) and \(1\) and width \(1\), so its area is \(\frac{2+1}{2}\cdot1=1.5\). Because the limits are reversed, \(I_0(-1)=-1.5\). The statement is false. 4. d) From \(t=-3\) to \(t=-1\), the triangle below the x-axis has signed area \(-1\), and the triangle above the x-axis has signed area \(1\). These areas cancel, so \(\int_{-3}^{-1}f(t)\,dt=0\). Therefore, \(I_0(-3)=I_0(-1)\), and the statement is true.

a) True b) True c) False; \(I_0(-1)=-1.5\). d) True

- What value does an accumulation function have when its upper and lower limits are equal? - Relate the derivative of \(I_{-2}\) to \(f\). - Use the sign of \(f\) to determine whether \(I_{-2}\) increases or decreases. - Examine the graph of \(f\) on \([-2,2]\).

1. a) Every accumulation function satisfies \(I_a(a)=0\). Therefore, \(I_{-2}(-2)=0\), so its graph must pass through \((-2,0)\). Curve \(q\) does not pass through this point, so it can be eliminated. 2. b) By the Fundamental Theorem of Calculus, the derivative of \(I_{-2}\) is \(f\). On \([-2,2]\), \(f(x)\geq0\), and \(f(x)>0\) for \(-2<x<2\). Thus, \(I_{-2}\) increases throughout this interval. 3. Of the remaining curves, \(p\) increases on \([-2,2]\), while \(r\) decreases. Therefore, \(p\) is the graph of \(I_{-2}\).

a) Curve \(q\) can be eliminated because it does not pass through \((-2,0)\). b) Curve \(p\) is correct because \(f(x)\geq0\) on \([-2,2]\), so \(I_{-2}\) must increase there.

- Relate the derivative of the accumulation function to \(f\). - Use the sign of \(f\) to determine where \(I_2\) increases or decreases. - Recall what happens when the limits of a definite integral are reversed. - Determine the sign of the signed area from \(-2\) to \(2\).

1. a) By the Fundamental Theorem of Calculus, the derivative of \(I_2\) is \(f\). For \(x<-2\), \(f(x)>0\), so \(I_2\) increases. For \(-2<x<2\), \(f(x)<0\), so \(I_2\) decreases. Because the derivative changes from positive to negative at \(x=-2\), \(I_2\) has a local maximum there. 2. b) By reversing the limits, \(I_2(-2)=\int_2^{-2}f(t)\,dt=-\int_{-2}^{2}f(t)\,dt\). The graph of \(f\) lies below the x-axis between \(-2\) and \(2\), so \(\int_{-2}^{2}f(t)\,dt<0\). Its opposite is positive. In fact, \(I_2(-2)=\frac{32}{3}\).

a) \(I_2\) increases for \(x<-2\) and decreases on \((-2,2)\), so it has a local maximum at \(x=-2\). b) Reversing the limits changes the sign of the negative signed area: \(I_2(-2)=-\int_{-2}^{2}f(t)\,dt=\frac{32}{3}>0\).

- The sign of \(f\) determines where \(F\) increases or decreases. - Use signed areas to compare values of \(F\). - Areas below the x-axis count negatively. - The derivative of \(F\) is \(f\).

1. The function \(F\) increases where \(f>0\) and decreases where \(f<0\). The main candidates for the absolute maximum are \(x=2\) and the endpoint \(x=6\). 2. From \(x=-2\) to \(x=2\), the two triangular areas each have area \(2\), so \(F(2)=F(-2)+4=4\). 3. The signed area from \(x=2\) to \(x=6\) is \(-2.5\), so \(F(6)=4-2.5=1.5\). Therefore, the absolute maximum is \(F(2)=4\). 4. On \([0,5]\), the signed areas are \(+2\), \(-0.5\), and \(-2\). Thus, \(\int_0^5 f(x)\,dx=2-0.5-2=-0.5\). 5. Since \(F'=f\), \(F'(4)=f(4)=-1\).

a) Absolute maximum at \(x=2\), with value \(4\) b) \(-0.5\) c) \(F'(4)=-1\)

- What value must an accumulation function have at its lower limit? - Subtracting a constant does not change a derivative. - Use the slope of \(F\) to determine the sign of \(f\). - Apply the Fundamental Theorem of Calculus.

1. An accumulation function \(A_a(x)=\int_a^x f(t)\,dt\) satisfies \(A_a(a)=0\). The graph of \(F\) has no zeros, so statement a is false. 2. The function \(H=F-2\) has the same derivative as \(F\), and \(H(2)=0\). Therefore, \(H(x)=\int_2^x f(t)\,dt\), so statement b is true. 3. Since \(F\) is decreasing at \(x=1\), \(f(1)=F'(1)<0\). Statement c is false. 4. By the Fundamental Theorem of Calculus, \(\int_0^2 f(x)\,dx=F(2)-F(0)=2-4=-2\). Statement d is true.

a) False b) True c) False d) True

- A local minimum occurs when the derivative changes from negative to positive. - An accumulation function represents signed area. - Look for symmetry between the two regions. - Use the sign of \(f\) to determine monotonicity.

1. A local minimum of \(F\) occurs where \(f\) changes from negative to positive. The graph shows this at \(x=-1\) and \(x=3\). 2. The positive area on \([-1,1]\) and the negative area on \([1,3]\) have equal magnitude by symmetry. Therefore, \(A_{-1}(3)=0\). 3. Since \(f>0\) on \((-1,1)\), every antiderivative is strictly increasing there. Since \(f<0\) on \((1,3)\), every antiderivative is strictly decreasing there.

a) \(x=-1\) and \(x=3\) b) \(A_{-1}(3)=0\); the equal positive and negative signed areas cancel. c) Increasing on \((-1,1)\) and decreasing on \((1,3)\)

- Interpret \(I_0(x)\) as signed area between the graph and the x-axis. - Areas above the x-axis are positive, and areas below it are negative. - Use the sign changes of \(f\) to locate extrema of \(I_0\). - Break the area through \(x=5\) into triangles and a trapezoid.

1. a) The signed area from \(0\) to \(2\) is \(-2\), so \(I_0(2)=-2\). From \(2\) to \(4\), a positive area of \(2\) is added, giving \(I_0(4)=0\). From \(4\) to \(6\), another positive area of \(2\) is added, so \(I_0(6)=2\). From \(6\) to \(8\), an area of \(2\) is subtracted, giving \(I_0(8)=0\). Together with \(I_0(0)=0\), the zeros are \(x=0\), \(x=4\), and \(x=8\). 2. b) The derivative of \(I_0\) is \(f\). Since \(f\) changes from positive to negative at \(x=6\), \(I_0\) reaches its maximum there. The maximum value is \(I_0(6)=2\). 3. c) Since \(I_0(4)=0\), \(I_0(5)\) is the area under \(f\) from \(4\) to \(5\). This region is a trapezoid with parallel sides \(2\) and \(1\) and width \(1\), so \(I_0(5)=\frac{2+1}{2}\cdot1=1.5\).

a) \(x=0\), \(x=4\), and \(x=8\) b) The maximum value is \(2\), attained at \(x=6\). c) \(I_0(5)=1.5\)

- An accumulation function has value \(0\) at its lower limit. - The slope of \(F\) gives the sign of \(f\). - At a local extremum, the derivative is \(0\). - Use symmetry and the Fundamental Theorem of Calculus.

1. An accumulation function must have a zero at its lower limit. The graph of \(F\) remains above the x-axis, so statement a is false. 2. On \([-2,-1]\), \(F\) is increasing, so \(f=F'>0\). Statement b is true. 3. The graph has a local maximum at \((0,3)\), so \(F(0)=3\) and \(f(0)=F'(0)=0\). Thus, \(F(0)+f(0)=3\), so statement c is false. 4. By the Fundamental Theorem of Calculus, \(\int_{-2}^{2}f(x)\,dx=F(2)-F(-2)\). The graph is symmetric about the y-axis, so these values are equal. Statement d is true.

a) False b) True c) False d) True

- Interpret \(I(x)\) as the net signed area accumulated from \(0\) to \(x\). - Areas above the x-axis are positive, and areas below it are negative. - When integrating to the left of \(0\), account for the reversed limits. - Relate the derivative of \(I\) to \(f\) to locate extrema. - Remember the value of an accumulation function at its lower limit.

1. a) From \(0\) to \(2\), the region is a triangle above the x-axis with base \(2\) and height \(2\), so \(I(2)=\frac{1}{2}\cdot2\cdot2=2\). From \(2\) to \(4\), an equal triangular area lies below the x-axis, so \(I(4)=2-2=0\). From \(-1\) to \(0\), the area above the axis is \(1\), and reversing the limits gives \(I(-1)=-1\). 2. b) The derivative of \(I\) is \(f\). At \(x=-1\), \(f\) changes from negative to positive, so \(I\) has a local minimum. At \(x=2\), \(f\) changes from positive to negative, so \(I\) has a local maximum. 3. c) The lower limit gives \(I(0)=0\), and part a gives \(I(4)=0\). From \(-2\) to \(-1\), the signed area is \(-1\), and from \(-1\) to \(0\), it is \(1\). These cancel, so \(I(-2)=-\int_{-2}^{0}f(t)\,dt=0\). On the intervals between these values, the constant sign of \(f\) makes \(I\) strictly monotonic, so there are no additional zeros. Thus, the zeros are \(x=-2\), \(x=0\), and \(x=4\).

a) \(I(2)=2\), \(I(4)=0\), and \(I(-1)=-1\) b) Local minimum at \(x=-1\); local maximum at \(x=2\) c) \(x=-2\), \(x=0\), and \(x=4\)

- Translate equal areas above and below the x-axis into a condition on signed area. - Find an antiderivative of \(f\). - Evaluate the definite integral from \(1\) to \(k\). - Use the visible factor corresponding to \(k=1\). - Apply the condition \(k>1\).

1. Equal areas above and below the x-axis mean the net signed area is zero: \(\int_1^k(3x^2-12x+9)\,dx=0\). 2. An antiderivative is \(F(x)=x^3-6x^2+9x\). 3. Apply the bounds: \(k^3-6k^2+9k-4=0\). 4. Factor: \(k^3-6k^2+9k-4=(k-1)^2(k-4)\). 5. The condition \(k>1\) eliminates \(k=1\), so \(k=4\).

\(k=4\)

- Identify the geometric shape of the region under the line. - Express the lengths of the two parallel sides using function values. - Apply the area formula for a trapezoid. - Evaluate an antiderivative at the upper and lower limits. - Compare the simplified formulas from the two methods.

1. a) The region under the line from \(1\) to \(x\) is a trapezoid. Its parallel vertical sides have lengths \(f(1)=3\) and \(f(x)\), and the distance between them is \(x-1\). Therefore, \(I_1(x)=\frac{3+f(x)}{2}(x-1)=\frac{1}{2}(3+f(x))(x-1)\). 2. b) Substitute \(f(x)=-x+4\): \(I_1(x)=\frac{1}{2}(7-x)(x-1)=-0.5x^2+4x-3.5\). 3. c) An antiderivative of \(f(t)=-t+4\) is \(F(t)=-0.5t^2+4t\). Thus, \(I_1(x)=F(x)-F(1)=(-0.5x^2+4x)-3.5=-0.5x^2+4x-3.5\). This matches the result from the trapezoid method.

a) The trapezoid has parallel sides of lengths \(3\) and \(f(x)\), separated by a distance of \(x-1\), so its area is \(\frac{1}{2}(3+f(x))(x-1)\). b) \(I_1(x)=-0.5x^2+4x-3.5\) c) Using \(F(t)=-0.5t^2+4t\) gives \(I_1(x)=F(x)-F(1)=-0.5x^2+4x-3.5\), the same result.

- What value must an accumulation function have at its lower limit? - Express a definite integral using two values of the same antiderivative. - Compare an arbitrary antiderivative with the additional zero condition for an accumulation function. - Changing the lower limit shifts an accumulation function vertically.

1. a) An accumulation function satisfies \(J_b(b)=0\). The graph has zeros at \(x=-3\) and \(x=0\). Since \(b\neq0\), \(b=-3\). 2. b) The graph gives \(J_b(0)=0\) and \(J_b(-1)=2\). Therefore, \(\int_b^0g(x)\,dx=J_b(0)=0\), and \(\int_{-1}^0g(x)\,dx=J_b(0)-J_b(-1)=-2\). 3. c) Since the derivative of \(J_b\) is \(g\), the derivative of \(G\) is also \(g\). Thus, \(G\) is an antiderivative of \(g\). The graph shows that \(J_b(x)\geq0\) on \([-4,0]\), so \(G(x)=J_b(x)+1\geq1\). Because \(G\) has no zero in the interval, no point can serve as a lower limit for an accumulation function there. 4. d) Changing the lower limit gives \(J_{-1}(x)=J_b(x)-J_b(-1)=J_b(x)-2\). Its zeros occur where \(J_b(x)=2\). From the graph, this happens at \(x=-4\) and \(x=-1\).

a) \(b=-3\) b) \(\int_b^0g(x)\,dx=0\) and \(\int_{-1}^0g(x)\,dx=-2\) c) \(G\) differentiates to \(g\), but \(G(x)\geq1\) on \([-4,0]\), so it has no zero that could serve as a lower limit. d) \(x=-4\) and \(x=-1\)

- Use \(F^{\prime}=f\) and look for a sign change from negative to positive. - Interpret the definite integral as signed area and divide the region into triangles and trapezoids. - Areas below the x-axis count negatively. - An accumulation function equals \(0\) at its lower limit.

1. Since \(F^{\prime}=f\), a local minimum of \(F\) occurs where \(f\) changes from negative to positive. The graph shows this sign change only at \(x=4\), so that is the only local-minimum x-value in the interior. 2. On \([-3,-2]\), the signed area is a trapezoid with area \(\frac{1+2}{2}\cdot1=\frac32\). On \([-2,0]\), the area is a triangle with area \(\frac12\cdot2\cdot2=2\). On \([0,1]\), the signed area is \(-\frac12\cdot1\cdot1=-\frac12\). Therefore, \(\int_{-3}^{1}f(x)\,dx=\frac32+2-\frac12=3\). 3. From \(-4\) to \(0\), the signed area is \(4\). From \(0\) to \(4\), it is \(-4\), and from \(4\) to \(5\), it is \(\frac12\). Thus, \(A(5)=\frac12\). 4. By definition, \(A(-4)=0\). The accumulation increases to \(4\) at \(x=0\), then decreases to \(0\) at \(x=4\), and is positive afterward. Therefore, the zeros are \(x=-4\) and \(x=4\).

a) \(x=4\) b) \(\int_{-3}^{1} f(x)\,dx=3\) c) \(A(5)=\frac12\); the zeros are \(x=-4\) and \(x=4\).

- Find an antiderivative and evaluate it at \(x\) and \(0\). - Factor the resulting cubic equation. - Interpret the definite integral as signed area. - Use the sign of \(f\) to determine where \(I_0\) decreases or increases. - Compare the positive region with the reflection of the negative region across \(x=2\).

1. a) Evaluate the integral: \(I_0(x)=\int_0^x(0.5t^2-2)\,dt=\frac{1}{6}x^3-2x\). Set this equal to zero: \(x\left(\frac{1}{6}x^2-2\right)=0\). Thus, \(x=0\) or \(x^2=12\), giving \(x=0\) and \(x=\pm\sqrt{12}=\pm2\sqrt{3}\). 2. b) First, \(I_0(0)=0\). On \((0,2)\), the graph of \(f\) is below the x-axis, so \(I_0\) decreases and is negative. On \((2,4)\), \(f(x)>0\), so \(I_0\) increases strictly. Reflect the negative region on \([0,2]\) across the line \(x=2\). The positive region on \([2,4]\) has the same width and its graph lies above that reflected region except at \(x=2\), so its area is larger than the magnitude of the negative area. Therefore, \(I_0(4)>0\). By continuity, \(I_0\) has one zero in \((2,4)\), and strict increase there makes that zero unique. Together with \(x=0\), there are exactly two zeros on \([0,4]\).

a) \(x=-2\sqrt{3}\), \(x=0\), and \(x=2\sqrt{3}\) \((\approx-3.46,0,3.46)\) b) One zero is \(x=0\). The accumulation function is negative on \((0,2]\), then increases strictly and becomes positive before \(x=4\), so it crosses zero exactly once more.

- Interpret \(F(4)\) as an area under the graph of \(f\). - Identify the geometric shape on \([0,4]\). - Relate the derivative of \(F\) to \(f\). - Use the sign of \(f\) to determine the monotonicity of \(F\). - Compare the slope and concavity of the candidates at and around \(x=0\).

1. a) On \([0,4]\), the region under \(f\) is a right triangle with base \(4\) and height \(4\). Therefore, \(F(4)=\frac{1}{2}\cdot4\cdot4=8\). 2. b) Since \(f(x)\geq0\) throughout \([-4,4]\), the accumulation function must increase on the entire interval. Candidate \(2\) does not have this behavior and also does not pass through the origin. 3. By the Fundamental Theorem of Calculus, the derivative of \(F\) at \(0\) is \(f(0)=4\). Candidate \(3\) has a horizontal tangent at the origin, so it cannot represent \(F\). 4. Candidate \(1\) passes through the origin, increases on the full interval, has slope \(4\) at the origin, is concave up for \(x<0\), and is concave down for \(x>0\). Therefore, Candidate \(1\) is the graph of \(F\).

a) \(F(4)=8\) b) Candidate \(1\) is the graph of \(F\).

- The sign of \(f\) gives the sign of \(F^{\prime}\). - Look for zeros of \(f\) where the sign changes. - Use \(F(2)-F(-2)=\int_{-2}^{2}f(x)\,dx\) and estimate the area from the grid.

1. Local extrema of \(F\) occur where \(f=F^{\prime}\) changes sign. At \(x=-2\), \(f\) changes from negative to positive, so \(F\) has a local minimum. At \(x=2\), \(f\) changes from positive to negative, so \(F\) has a local maximum. 2. Since \(f>0\) on \((-2,2)\), \(F\) is strictly increasing on \([-2,2]\). 3. By the Fundamental Theorem of Calculus, \(F(2)=F(-2)+\int_{-2}^{2}f(x)\,dx\). Since \(F(-2)=0\), estimate the positive area under the graph. Counting grid squares gives about \(2.7\). The exact area for the displayed parabola is \(\frac83\approx2.67\), consistent with the estimate.

a) Local minimum at \(x=-2\); local maximum at \(x=2\) b) \([-2,2]\) c) \(F(2)\approx2.7\)

- An accumulation function satisfies \(A(0)=0\). - Zeros and sign changes of \(f\) determine local extrema of \(A\). - Local extrema of \(f\) determine inflection points of \(A\). - Use the symmetry of \(f\) to determine the symmetry of \(A\).

1. The function \(A\) is the antiderivative of \(f\) that satisfies \(A(0)=0\). 2. The zeros of \(f\) are \(x=-2\), \(x=0\), and \(x=2\). The sign of \(f\) changes from negative to positive at \(x=-2\), from positive to negative at \(x=0\), and from negative to positive at \(x=2\). Therefore, \(A\) has local minima at \(x=-2\) and \(x=2\), and a local maximum at \(x=0\). 3. From the graph, the local extrema of \(f\) occur at approximately \(x=-1.2\) and \(x=1.2\). These are the approximate inflection-point x-values of \(A\). 4. The graph of \(f\) is symmetric about the origin, so \(f\) is odd. Therefore, \(A(x)=\int_0^x f(t)\,dt\) is even. The graph of \(A\) is symmetric about the y-axis, is W-shaped, and passes through the local maximum \((0,0)\).

The graph is symmetric about the y-axis and W-shaped. It has a local maximum at \((0,0)\), local minima at \(x=-2\) and \(x=2\), and inflection points at approximately \(x=-1.2\) and \(x=1.2\).

- Symmetry about the y-axis means \(f\) is even. - Write \(F(x)=\int_0^x f(t)\,dt\). - Compare the accumulated areas from \(0\) to \(x\) and from \(0\) to \(-x\).

1. Symmetry about the y-axis means that \(f\) is even: \(f(-x)=f(x)\). 2. The condition \(F(0)=0\) makes \(F\) the accumulation function \(F(x)=\int_0^x f(t)\,dt\). 3. For an even integrand, \(F(-x)=\int_0^{-x}f(t)\,dt=-\int_0^x f(t)\,dt=-F(x)\). 4. Therefore, \(F\) is odd, and its graph is symmetric about the origin.

The graph of \(F\) is symmetric about the origin.

- What value must an accumulation function have at its lower limit? - Use the Fundamental Theorem of Calculus to relate the slope of the graph to \(f\). - Interpret a positive or negative slope as a sign of the integrand. - Use the condition \(a>0\) to choose between the zeros.

1. An accumulation function satisfies \(I_a(a)=0\), so \(a\) must be a zero of the graph. The zeros are \(x=-2\) and \(x=1\). Since \(a>0\), \(a=1\). 2. By the Fundamental Theorem of Calculus, the derivative of \(I_a\) is \(f\). The slope of the graph at \(x=0\) is \(-1\), so \(f(0)=-1\). 3. At \(x=-2\), the graph of \(I_a\) is increasing, so its derivative is positive. Therefore, \(f(-2)>0\).

1. \(a=1\) 2. \(f(0)=-1\) 3. \(f(-2)>0\)

- An accumulation function equals zero at its lower limit. - The derivative of the accumulation function is the original function. - Zeros of \(f\) correspond to horizontal tangents of \(I_0\). - The sign of \(f\) determines where \(I_0\) increases or decreases.

1. Every accumulation function with lower limit \(0\) satisfies \(I_0(0)=0\). Thus, its graph must pass through the origin. This eliminates \(G_2\). 2. By the Fundamental Theorem of Calculus, \(I_0^{\prime}(x)=f(x)\). At \(x=0\), \(I_0^{\prime}(0)=f(0)=2\), so the accumulation graph must have positive slope \(2\) at the origin. 3. Graph \(G_1\) is increasing at the origin, while \(G_3\) is decreasing there. Therefore, \(G_1\) represents \(I_0\). 4. This also agrees with the fact that \(f(x)>0\) on \((-\frac{\pi}{2}, \frac{\pi}{2})\), so \(I_0\) must increase on that interval.

Graph \(G_1\).

- What value must the accumulation function have at the lower limit? - Relate the derivative of \(I_1\) to \(f\). - Use the sign of \(f\) to determine where \(I_1\) increases or decreases. - Check a value of \(I_1\) by calculating a triangular area.

1. The accumulation function must satisfy \(I_1(1)=0\). Candidates \(1\) and \(2\) pass through \((1,0)\), while Candidate \(3\) does not. 2. By the Fundamental Theorem of Calculus, the derivative of \(I_1\) is \(f\). Since \(f(x)>0\) on \((1,3)\), \(I_1\) must increase throughout that interval. Candidate \(1\) increases there, while Candidate \(2\) decreases. 3. As a check, the area under \(f\) from \(1\) to \(3\) is a triangle with base \(2\) and height \(2\), so \(I_1(3)=\frac{1}{2}\cdot2\cdot2=2\). Candidate \(1\) has this value. 4. Therefore, Candidate \(1\) is the graph of \(I_1\).

Candidate \(1\) is the graph of \(I_1\).

- What is the value of an integral with equal limits? - Relate the derivative of \(J\) to \(f\). - Reversing the limits changes the sign of a definite integral. - Use signed areas under the graph.

1. a) Since the limits are equal, \(J(2)=\int_2^2f(t)\,dt=0\). The statement is true. 2. b) The derivative of \(J\) is \(f\). The graph of \(f\) is positive before \(x=6\) and negative after \(x=6\), so \(J\) increases and then decreases. Thus, \(J(6)\) is the maximum value on \([0,8]\). The statement is true. 3. c) Reversing the limits gives \(J(0)=-\int_0^2f(t)\,dt\). The triangular area from \(0\) to \(2\) is \(2\), so \(J(0)=-2\). The statement is false. 4. d) From \(2\) to \(6\), the accumulated positive area is \(6\), so \(J(6)=6\). From \(6\) to \(8\), the negative triangular area is \(-2\), giving \(J(8)=4\). Since \(J\) increases from \(0\) to \(6\) and then decreases only to \(4\), \(J(x)>0\) for every \(x\in(2,8]\). The statement is true.

a) True b) True c) False; \(J(0)=-2\). d) True

- What value must an accumulation function have at its lower limit? - Interpret the integral as a balance of signed areas. - Find the areas of the triangles above and below the x-axis. - Compare the magnitudes of the positive and negative contributions.

1. a) An accumulation function satisfies \(I_a(a)=0\), so \(a\) must be a zero of the accumulation curve. The graph has zeros at \(x=0\) and \(x=4\). Therefore, \(a=0\) or \(a=4\). 2. b) On \([0,2]\), the graph of \(f\) forms a triangle above the x-axis with area \(\frac{1}{2}\cdot2\cdot2=2\). 3. On \([2,5]\), the graph forms a triangle below the x-axis with area magnitude \(\frac{1}{2}\cdot3\cdot3=4.5\). 4. Therefore, \(I_0(5)=2-4.5=-2.5\). The negative area has greater magnitude than the positive area, so the net signed area is negative.

a) \(a=0\) or \(a=4\) b) The positive area is \(2\), and the negative area has magnitude \(4.5\), so \(I_0(5)=2-4.5=-2.5\).

- An accumulation function is \(0\) at its lower limit. - The sign of \(f\) determines whether \(A\) increases or decreases. - A sign-changing zero of \(f\) gives a local extremum of \(A\).

1. Every accumulation function satisfies \(A(1)=0\). Candidate graphs 1 and 3 pass through \((1,0)\), while graph 2 does not. 2. The function \(f\) is positive on \((1,3)\), so \(A\) must be increasing there. Candidate 1 increases on that interval, while candidate 3 decreases. 3. At \(x=3\), \(f\) changes from positive to negative, so \(A\) has a local maximum there. Candidate 1 has that behavior. 4. Therefore, candidate graph 1 represents \(A\).

Candidate graph 1

- Interpret the accumulation function as signed area starting at \(x=1\). - Areas above the x-axis are positive, and areas below it are negative. - After \(x=6\), use the constant value of \(f\) to write a linear expression for \(I_1\). - Use the sign of \(f\) to determine the monotonicity of \(I_1\).

1. a) From \(1\) to \(2\), the positive triangular area is \(1\). From \(2\) to \(4\), the positive rectangular area is \(4\). From \(4\) to \(5\), the positive triangular area is \(1\). Therefore, \(I_1(5)=1+4+1=6\). 2. b) From \(5\) to \(6\), the signed triangular area is \(-1\), so \(I_1(6)=5\). For \(x>6\), \(f(t)=-2\), and \(I_1(x)=I_1(6)+\int_6^x(-2)\,dt=5-2(x-6)=17-2x\). Solving \(17-2x=0\) gives \(x=8.5\). 3. c) Since \(f(x)>0\) for \(1<x<5\), \(I_1\) increases on \([1,5]\). Since \(f(x)<0\) for \(5<x\leq6\), \(I_1\) decreases on \([5,6]\). The sign change from positive to negative at \(x=5\) gives a local maximum, with \(I_1(5)=6\).

a) \(I_1(5)=6\) b) \(x=8.5\) c) \(I_1\) increases on \([1,5]\), decreases on \([5,6]\), and has a local maximum of \(6\) at \(x=5\).

- The slope of \(F\) is \(f\). - Use \(\int_a^bf(x)\,dx=F(b)-F(a)\). - An accumulation function is \(0\) at its lower limit. - Use the increasing or decreasing behavior of \(F\) to determine the sign of \(f\).

1. At \(x=0\), the graph of \(F\) has a local maximum, so \(F^{\prime}(0)=f(0)=0\). Statement a is true. 2. By the Fundamental Theorem of Calculus, \(\int_{-2}^{2}f(x)\,dx=F(2)-F(-2)\). The graph is symmetric about the y-axis, so \(F(2)=F(-2)\). Statement b is true. 3. An accumulation function \(A_a(x)=\int_a^x f(t)\,dt\) must equal \(0\) at \(x=a\). The graph of \(F\) has no zeros, so it cannot be an accumulation function. Statement c is false. 4. At \(x=1\), the graph of \(F\) is decreasing, so \(f(1)=F^{\prime}(1)<0\). Statement d is true.

a) True b) True c) False d) True

- Use the values of the antiderivative at the endpoints. - What value must an accumulation function have at its lower limit? - Relate \(f\) to the slope of its antiderivative \(F\). - Examine whether the graph is increasing or decreasing at \(x=0\).

1. a) By the Fundamental Theorem of Calculus, \(\int_0^2f(x)\,dx=F(2)-F(0)\). The graph shows \(F(2)=0\) and \(F(0)=4\), so the integral equals \(-4\). 2. b) An antiderivative is an accumulation function when it has a zero that can serve as the lower limit. The graph of \(F\) has zeros at \(x=-4\) and \(x=2\). Therefore, either \(a=-4\) or \(a=2\) is possible. 3. c) Since \(f(0)\) equals the derivative of \(F\) at \(0\), its sign is the sign of the slope of \(F\) there. The graph is decreasing at \(x=0\), so \(f(0)<0\).

a) \(\int_0^2f(x)\,dx=-4\) b) \(F\) has zeros, so it can be written as an accumulation function. One possible lower limit is \(a=2\); \(a=-4\) also works. c) \(f(0)<0\) because \(F\) is decreasing at \(x=0\).

- An antiderivative with a zero can be written as an accumulation function with that zero as the lower limit. - The slope of \(F\) is \(f\). - Use \(\int_a^bf(x)\,dx=F(b)-F(a)\). - Horizontal tangents of \(F\) give zeros of \(f\).

1. True. The graph of \(F\) has zeros, including \(x=-1\). Since \(F^{\prime}=f\) and \(F(-1)=0\), the Fundamental Theorem of Calculus gives \(F(x)=\int_{-1}^x f(t)\,dt\). 2. False. At \(x=4\), the graph of \(F\) is decreasing, so \(f(4)=F^{\prime}(4)<0\). 3. True. \(\int_0^3f(x)\,dx=F(3)-F(0)=8-1.25=6.75\). 4. False. Zeros of \(f\) occur at horizontal tangents of \(F\). On \([-2,4]\), the graph of \(F\) has a local minimum at \(x=-1\) and a local maximum at \(x=3\), so \(f\) has two zeros.

a) True b) False c) True d) False

- Does the antiderivative have a zero that could serve as a lower limit? - Interpret \(f(x)\) as the slope of \(F\). - Use endpoint values of the antiderivative to evaluate the definite integral. - A horizontal tangent of \(F\) corresponds to what value of \(f\)?

1. a) An accumulation function must have a zero that can serve as its lower limit. The graph of \(F\) stays above the x-axis and has a minimum value of \(2\), so it has no zero. The statement is false. 2. b) The value \(f(1)\) equals the slope of \(F\) at \(x=1\). The graph is decreasing there, so the slope is negative. The statement is true. 3. c) By the Fundamental Theorem of Calculus, \(\int_0^4f(x)\,dx=F(4)-F(0)\). The graph shows \(F(4)=4\) and \(F(0)=4\), so the integral is \(0\). The statement is true. 4. d) Zeros of \(f\) occur where \(F\) has a horizontal tangent. The graph has a local minimum at \(x=2\), so \(f(2)=0\). The statement is true.

a) False b) True c) True d) True

- Interpret the accumulation function as signed area. - Relate the value of \(f(x)\) to the slope of \(J\). - A local maximum occurs where the derivative changes from positive to negative. - Start with \(J(3)=0\), and look for later cancellation of equal positive and negative signed areas.

1. The value \(J(6)=\int_3^6f(t)\,dt\) is the signed area under the graph from \(3\) to \(6\). Estimating from the grid gives about \(1.4\). The exact value for the displayed function is \(\frac{3\sqrt{2}}{\pi}\approx1.35\). 2. The derivative of \(J\) is \(f\). At \(x=5\), \(f\) changes from positive to negative, so \(J\) has a local maximum there. 3. The graph passes through \((3,0)\). It decreases on \([0,1]\), increases on \([1,5]\), and decreases on \([5,9]\). It has a local minimum at \(x=1\), a local maximum at \(x=5\), and an endpoint minimum at \(x=9\). The signed areas cancel again at \(x=7\), so the x-intercepts are \(x=3\) and \(x=7\). Useful values are \(J(1)=J(9)=-\frac{6}{\pi}\approx-1.91\) and \(J(5)=\frac{6}{\pi}\approx1.91\).

1. \(J(6)\approx1.4\); an estimate near \(1.35\) is reasonable. 2. \(x=5\) 3. The graph decreases on \([0,1]\), increases on \([1,5]\), and decreases on \([5,9]\). Its x-intercepts are \(x=3\) and \(x=7\); it has a local minimum at \(x=1\), a local maximum at \(x=5\), and an endpoint minimum at \(x=9\).

- The derivative of an accumulation function equals its integrand. - A local maximum occurs when the derivative changes from positive to negative. - Interpret the integral as a signed-area balance. - An integral with equal upper and lower limits is zero.

1. By the Fundamental Theorem of Calculus, \(J^{\prime}(x)=f(x)\). Therefore, the zeros of \(J^{\prime}\) are the zeros of \(f\): \(x=1.5\), \(x=4.5\), and \(x=7.5\). 2. A local maximum of \(J\) occurs where \(f\) changes from positive to negative. On \([0,6]\), this happens at \(x=1.5\). 3. Since \(J(1)=\int_1^1f(t)\,dt=0\), compare this with \(J(4)=\int_1^4f(t)\,dt\). 4. The small positive area from \(1\) to \(1.5\) is outweighed by the negative area from \(1.5\) to \(4\). Thus, \(J(4)<0\), so \(J(1)>J(4)\).

a) \(x=1.5\), \(x=4.5\), and \(x=7.5\) b) \(x=1.5\) c) \(J(1)>J(4)\)

- Use two points on the line to find its slope and y-intercept. - Apply the power rule for antiderivatives. - Consider the cases \(x<2\) and \(x>2\) separately. - Reversing the limits changes the sign of a definite integral.

1. a) The line passes through \((0,2)\) and \((2,0)\). Its slope is \(\frac{0-2}{2-0}=-1\), and its y-intercept is \(2\). Thus, \(g(t)=-t+2\). 2. b) Integrate: \(J_2(x)=\int_2^x(-t+2)\,dt=[-0.5t^2+2t]_2^x\) \(=-0.5x^2+2x-2=-0.5(x-2)^2\). 3. c) For \(x>2\), the graph of \(g\) lies below the x-axis, so the signed area from \(2\) to \(x\) is negative. For \(x<2\), the graph lies above the x-axis, but the integral runs from \(2\) to the left, reversing the sign of that positive area. Therefore, the integral is negative in both cases.

a) \(g(t)=-t+2\) b) \(J_2(x)=-0.5x^2+2x-2=-0.5(x-2)^2\) c) For \(x>2\), the accumulated area is below the x-axis. For \(x<2\), the area is above the axis but the limits are reversed. In either case, \(J_2(x)<0\).

- Start with the value of \(G(0)\). - Interpret the integral as signed area between the graph and the x-axis. - Track when positive area can cancel accumulated negative area. - Use the symmetry of \(g\) to relate the positive and negative sides.

1. Since the limits are equal, \(G(0)=0\), so \(x=0\) is one zero. 2. For \(0<x<3\), the graph of \(g\) lies below the x-axis, so \(G\) decreases and remains negative. 3. For \(3<x<5\), \(g\) is positive, so \(G\) increases. The positive area from \(3\) to \(5\) exceeds the earlier negative area, so the balance reaches zero exactly once, at approximately \(x=4.2\). 4. The graph of \(g\) is symmetric about the origin. Therefore, \(G\) is symmetric about the y-axis, giving another zero at approximately \(x=-4.2\). 5. Thus, \(G\) has exactly three zeros on the interval.

Exactly three zeros: \(x=0\) and approximately \(x=\pm4.2\).

- The sign of \(f\) determines the monotonicity of an antiderivative. - A positive-to-negative sign change gives a local maximum. - Compare positive and negative signed areas from \(x=-2\) to \(x=3\).

1. Since \(F^{\prime}=f\), an antiderivative is strictly decreasing where \(f<0\). The sign of the factored cubic is negative on \((-\infty, -2)\) and \((1, 3)\). 2. At \(x=1\), \(f\) changes from positive to negative, so every antiderivative has a local maximum there. 3. Since \(F(-2)=0\), \(F(3)=\int_{-2}^{3}f(x)\,dx\). The positive signed area on \([-2,1]\) is \(\frac{63}{40}\), while the signed area on \([1,3]\) is \(-\frac{8}{15}\). Their sum is \(\frac{25}{24}>0\), so \(F(3)>0\).

a) \((-\infty, -2)\) and \((1, 3)\) b) \(f\) changes from positive to negative at \(x=1\), so \(F\) has a local maximum there. c) \(F(3)>0\)

- Interpret the accumulation function as signed area. - Find the areas of the triangles above and below the x-axis. - Use the sign of \(g\) to determine where \(I_0\) increases or decreases. - The antiderivative of a linear function is quadratic.

1. a) From \(0\) to \(4\), the graph forms a triangle above the x-axis with base \(4\) and height \(2\). Thus, \(I_0(4)=\frac{1}{2}\cdot4\cdot2=4\). 2. From \(4\) to \(8\), the graph forms an equal triangle below the x-axis, contributing \(-4\). Therefore, \(I_0(8)=4-4=0\). 3. b) The derivative of \(I_0\) is \(g\). Since \(g>0\) on \((0,4)\) and \(g<0\) on \((4,8)\), \(I_0\) increases and then decreases. Its maximum occurs at \(x=4\), with value \(4\). 4. c) The graph begins at \((0,0)\), passes through \((2,2)\), reaches its maximum at \((4,4)\), passes through \((6,2)\), and ends at \((8,0)\). Because \(g\) is piecewise linear, \(I_0\) consists of connected parabolic arcs.

a) \(I_0(4)=4\) and \(I_0(8)=0\) b) The maximum value is \(4\), attained at \(x=4\). c) The graph passes through \((0,0)\), \((2,2)\), \((4,4)\), \((6,2)\), and \((8,0)\), using connected parabolic arcs.

- Interpret the integral as signed area. - Zeros and sign changes of \(f\) determine extrema of \(I_0\). - Areas above the x-axis are positive, and areas below are negative. - The value of \(f(x)\) is the slope of \(I_0\) at \(x\).

1. a) The triangular area from \(0\) to \(2\) is \(2\), so \(I_0(2)=2\). From \(2\) to \(4\), an equal triangular area lies below the x-axis, so \(I_0(4)=2-2=0\). 2. b) The derivative of \(I_0\) is \(f\). At \(x=2\), \(f\) changes from positive to negative, so \(I_0\) has a local maximum. At \(x=4\), \(f\) changes from negative to positive, so \(I_0\) has a local minimum. 3. At the left endpoint \(x=0\), \(I_0\) increases immediately to the right, so \(x=0\) is an endpoint minimum. At the right endpoint \(x=5\), \(I_0\) increases as it approaches the endpoint, so \(x=5\) is an endpoint maximum. Also, \(I_0(5)=1\). 4. c) The graph starts at \((0,0)\), rises to \((2,2)\), and then falls to \((4,0)\). Its pieces are parabolic because \(f\) is piecewise linear.

a) \(I_0(2)=2\) and \(I_0(4)=0\) b) Endpoint minimum at \(x=0\); local maximum at \(x=2\); local minimum at \(x=4\); endpoint maximum at \(x=5\) c) The graph runs from \((0,0)\) through \((2,2)\) to \((4,0)\) in connected parabolic arcs.

- Use the Fundamental Theorem of Calculus to build \(I_1\). - Use the given value \(G(2)=20\) to determine the constant. - Find the minimum possible value of \(x^4+\frac{1}{x^4}\) for \(x>0\). - A shifted antiderivative is an accumulation function exactly when it has a zero.

1. For a), an antiderivative is \(x^4+x^{-4}\). Therefore, \(I_1(x)=x^4+\frac{1}{x^4}-2\). 2. For b), write \(G(x)=x^4+\frac{1}{x^4}+C\). The condition \(G(2)=20\) gives \(16+\frac{1}{16}+C=20\), so \(C=\frac{63}{16}\). 3. For \(x>0\), \(x^4+\frac{1}{x^4}\geq2\), with equality at \(x=1\). Hence, \(G(x)\geq2+\frac{63}{16}=\frac{95}{16}>0\). Thus, \(G\) has no zero and cannot be an accumulation function. 4. For c), \(G+k\) can be an accumulation function exactly when it has a zero at some \(a>0\). Its minimum value is \(\frac{95}{16}+k\). A zero exists exactly when this minimum is nonpositive, so \(k\leq-\frac{95}{16}\).

a) \(I_1(x)=x^4+\frac{1}{x^4}-2\) b) \(G(x)=x^4+\frac{1}{x^4}+\frac{63}{16}\), and \(G(x)\geq\frac{95}{16}>0\), so it is not an accumulation function. c) \(k\leq-\frac{95}{16}\)

- Evaluate the accumulation function at its lower limit. - Use the sign of \(g\) to determine where \(J_c\) decreases and increases. - Determine the sign of the minimum at \(x=4\). - Use end behavior, the Intermediate Value Theorem, and monotonicity.

1. For a), solve \(1-\frac{4}{x}=0\), giving \(x_0=4\). 2. For b), \(J_c(c)=0\), so \(x=c\) is one zero. 3. By the Fundamental Theorem of Calculus, \(\frac{d}{dx}J_c(x)=g(x)\). Because \(g(x)<0\) on \([c,4)\), \(J_c\) decreases from \(0\) to a negative minimum at \(x=4\). 4. For \(x>4\), \(g(x)>0\), so \(J_c\) is strictly increasing. Also, \(J_c(x)=x-4\ln(x)-\left(c-4\ln(c)\right)\), which approaches \(\infty\) as \(x\to\infty\). 5. Therefore, the Intermediate Value Theorem gives one zero for \(x>4\). Strict increase on that interval makes it unique. Together with \(x=c\), \(J_c\) has exactly two zeros for \(x\geq c\).

a) \(x_0=4\) b) One zero is \(x=c\). The function decreases to a negative minimum at \(x=4\), then increases strictly to \(\infty\), so it crosses the x-axis exactly once more for \(x>4\).

- Find an antiderivative and evaluate the fixed lower bound. - Recognize the equation as quadratic in \(k^2\). - Include both square roots for each positive value of \(k^2\). - Remember that a variable endpoint may lie to the left of the fixed endpoint.

1. An antiderivative is \(F(x)=\frac{x^4}{4}-\frac{3x^2}{2}\), and \(F(-2)=-2\). 2. The equation becomes \(\frac{k^4}{4}-\frac{3k^2}{2}+2=0.75\). Multiplying by \(4\) gives \(k^4-6k^2+5=0\). 3. Let \(u=k^2\). Then \((u-1)(u-5)=0\). Thus, \(k=\pm1\) or \(k=\pm\sqrt{5}\). 4. Geometrically, the oriented accumulation from \(-2\) to the variable endpoint \(k\) is \(0.75\). For \(k<-2\), the reversed orientation of the definite integral is included.

\(k\in\{-\sqrt{5}, -1, 1, \sqrt{5}\}\)

- Simplify the rational expression into powers of \(x\). - An accumulation function with lower limit \(a\) must equal \(0\) at \(x=a\). - Keep the interval within one connected part of the domain. - Reverse the power and chain rules for \(k\).

1. For a), rewrite \(g(x)=x-4x^{-2}\). One antiderivative is \(G(x)=\frac{1}{2}x^2+\frac{4}{x}\). 2. For b), \(H'(x)=x-4x^{-2}=g(x)\), so \(H\) is an antiderivative on each interval not containing \(0\). An accumulation function with lower limit \(a\) must satisfy \(H(a)=0\). 3. Solve \(\frac{1}{2}a^2+\frac{4}{a}-4.5=0\), which is equivalent to \(a^3-9a+8=0\). Since \(a=1\) is a root, \(H(x)=\int_1^x g(t)\,dt\) on \((0, \infty)\). 4. For c), reverse the power and chain rules: \(K(x)=\frac{5}{8}(2x-7)^4\).

a) \(G(x)=\frac{1}{2}x^2+\frac{4}{x}\) b) Yes. One possible lower limit is \(a=1\), giving \(H(x)=\int_1^x g(t)\,dt\) on \((0, \infty)\). c) \(K(x)=\frac{5}{8}(2x-7)^4\)

- Interpret \(G(x)\) as accumulated signed area. - Zeros and sign changes of \(g\) determine local extrema of \(G\). - Local extrema of \(g\) determine inflection points of \(G\). - Use symmetry to compare the positive and negative areas over one full period.

1. Since \(G(4)=\int_0^4g(t)\,dt\) and \(g>0\) on \((0,4)\), statement a is true. 2. At \(x=4\), \(g=G^{\prime}\) changes from positive to negative, so \(G\) has a local maximum, not a local minimum. Statement b is false. 3. At \(x=2\), \(g\) has a local maximum, so \(g\) changes from increasing to decreasing. Therefore, \(G\) changes concavity and has an inflection point there. Statement c is true. 4. The positive signed area on \([0,4]\) and the negative signed area on \([4,8]\) have equal magnitude by symmetry. Thus, \(G(8)=\int_0^8g(t)\,dt=0\), so statement d is true.

a) True b) False c) True d) True

- The lower limit of an accumulation function is a zero of that function. - Express definite integrals as differences of values of the displayed accumulation function. - Accumulation functions of the same integrand differ by a constant. - Interpret the slope of \(I_a\) as the value of \(f\).

1. a) Since \(I_a(a)=0\), the lower limit must be a zero of the graph. The only zero in \([0,4]\) is \(x=2\), so \(a=2\). 2. b) The graph gives \(\int_2^0f(x)\,dx=I_2(0)=3\). Also, \(\int_0^4f(x)\,dx=I_2(4)-I_2(0)=3-3=0\). 3. c) The accumulation functions are related by \(I_0(x)=I_2(x)-I_2(0)=I_2(x)-3\). Thus, \(I_0(x)=0\) where the displayed graph has value \(3\). This occurs at approximately \(x=0\) and \(x=4\). 4. d) The value \(f(2)\) is the slope of \(I_a\) at \(x=2\). The graph has a local minimum and a horizontal tangent there, so \(f(2)=0\).

a) \(a=2\) b) \(\int_2^0f(x)\,dx=3\) and \(\int_0^4f(x)\,dx=0\) c) \(x\approx0\) and \(x\approx4\) d) \(f(2)=0\)