Free Math Worksheets

- Find an antiderivative of each integrand. - Substitute the upper and lower limits carefully. - Pay special attention when the upper limit is less than the lower limit. - Integrate each term separately.

1. For a), an antiderivative is \(F(x)=\frac{1}{2}x^2+x\). Thus, \(F(4)-F(-2)=12-0=12\). 2. For b), an antiderivative is \(F(t)=2t-0.4t^2\). Thus, \(F(5)-F(0)=10-10=0\). 3. For c), an antiderivative is \(F(u)=0.75u^2-3u\). Thus, \(F(1)-F(4)=-2.25-0=-2.25\).

a) \(12\) b) \(0\) c) \(-2.25\)

- Integrate with respect to \(x\), treating \(a\) as a constant. - Use the power rule. - Apply the upper and lower bounds carefully. - Terms containing a factor of \(0\) vanish at the lower bound in b).

1. For a), an antiderivative is \(F(x)=\frac{ax^4}{4}-ax\). Therefore, \(F(2)-F(1)=2a-\left(-\frac{3a}{4}\right)=\frac{11a}{4}\). 2. For b), an antiderivative is \(G(x)=\frac{a+1}{3}x^3\). Thus, \(G(1)-G(0)=\frac{a+1}{3}\).

a) \(\frac{11a}{4}\) b) \(\frac{a+1}{3}\)

- Rewrite the square-root denominator as a negative power. - Treat \(a\) as a constant. - Track signs carefully at the negative lower bound. - Symmetry can also simplify part a).

1. For a), an antiderivative is \(F(x)=\frac{ax^3}{3}+a^3x\). Evaluating from \(-2\) to \(2\) gives \(\frac{16a}{3}+4a^3\). 2. For b), write the integrand as \(ax^{-1/2}\). An antiderivative is \(G(x)=2a\sqrt{x}\). Therefore, \(G(4)-G(1)=4a-2a=2a\).

a) \(\frac{16a}{3}+4a^3\) b) \(2a\)

- Apply the power rule to find an antiderivative. - Use upper-limit value minus lower-limit value. - Use parentheses when subtracting a negative value. - A constant factor can remain in front while you integrate.

1. An antiderivative of \(f(x)=\frac{1}{2}x^2-2\) is \(F(x)=\frac{1}{6}x^3-2x\). 2. Evaluate at the limits: \(F(4)=\frac{64}{6}-8=\frac{8}{3}\) and \(F(1)=\frac{1}{6}-2=-\frac{11}{6}\). 3. Therefore, \(F(4)-F(1)=\frac{8}{3}-\left(-\frac{11}{6}\right)=\frac{27}{6}=\frac{9}{2}=4.5\).

\(4.5\)

- Apply the power rule term by term. - Remember how an even power affects a negative input. - Keep the result as a fraction until the final step. - Put the entire lower-limit value in parentheses when subtracting.

1. An antiderivative is \(F(x)=\frac{1}{16}x^4+\frac{1}{2}x^2\). 2. Evaluate at the limits: \(F(2)=1+2=3\) and \(F(-1)=\frac{1}{16}+\frac{1}{2}=\frac{9}{16}\). 3. Thus, \(F(2)-F(-1)=3-\frac{9}{16}=\frac{39}{16}=2.4375\).

\(\frac{39}{16}\), or \(2.4375\)

- Treat \(k\) as a constant while integrating. - Use the power rule to find an antiderivative. - Be careful with even powers at the negative lower bound. - Subtract the entire lower-bound expression.

1. An antiderivative is \(F(x)=\frac{x^4}{4}-\frac{kx^2}{2}\). 2. Evaluate the endpoints: \(F(2)=4-2k\) and \(F(-1)=\frac{1}{4}-\frac{k}{2}\). 3. Subtract: \(F(2)-F(-1)=\frac{15}{4}-\frac{3k}{2}\).

\(\frac{15}{4}-\frac{3k}{2}\)

- Rewrite the reciprocal-square term using a negative exponent. - Treat \(a\) as a constant. - Find an antiderivative before applying the bounds. - Subtract the entire lower-bound expression.

1. Rewrite \(\frac{1}{x^2}\) as \(x^{-2}\). An antiderivative is \(F(x)=\frac{ax^3}{3}-\frac{1}{x}\). 2. Evaluate: \(F(3)=9a-\frac{1}{3}\) and \(F(1)=\frac{a}{3}-1\). 3. Subtract: \(F(3)-F(1)=\frac{26a}{3}+\frac{2}{3}\).

\(\frac{26a}{3}+\frac{2}{3}\)

- Pay attention to the order of the limits. - Integrate the polynomial term by term. - Check the value of the antiderivative at \(x=3\) carefully.

1. An antiderivative is \(F(x)=x^4-2x^3+x^2-5x\). 2. The endpoint values are \(F(0)=0\) and \(F(3)=81-54+9-15=21\). 3. Because the integral runs from \(3\) to \(0\), \(F(0)-F(3)=0-21=-21\).

\(-21\)

- How do you evaluate a definite integral when an antiderivative is known? - Write each integral separately using \(F\). - Pay attention to the order of the limits when substituting into \(F\). - What happens to every sign inside brackets preceded by a minus sign?

1. By the Fundamental Theorem of Calculus, \(\int_a^b f(x)\,dx=F(b)-F(a)\). 2. Reversing the limits gives \(\int_b^a f(x)\,dx=F(a)-F(b)\). 3. Therefore, \(-\int_b^a f(x)\,dx=-[F(a)-F(b)]=F(b)-F(a)\), which equals \(\int_a^b f(x)\,dx\).

The Fundamental Theorem of Calculus gives \(\int_a^b f(x)\,dx=F(b)-F(a)\) and \(-\int_b^a f(x)\,dx=-[F(a)-F(b)]=F(b)-F(a)\). Thus the two expressions are equal.

- Recall which antiderivatives produce sine and cosine, including their signs. - Use the exact trigonometric values at \(0\) and \(\frac{\pi}{2}\). - Apply upper-limit value minus lower-limit value.

1. An antiderivative is \(F(x)=-3\cos(x)-2\sin(x)\). 2. At the upper limit, \(F\left(\frac{\pi}{2}\right)=-3\cdot0-2\cdot1=-2\). 3. At the lower limit, \(F(0)=-3\cdot1-2\cdot0=-3\). 4. Therefore, \(F\left(\frac{\pi}{2}\right)-F(0)=-2-(-3)=1\).

\(1\)

- Account for the inner factor when integrating the sine function. - Use the exact cosine values at \(0\) and \(\frac{\pi}{2}\). - Form a linear equation in \(a\). - Solve after evaluating the integral.

1. An antiderivative is \(F(x)=-2a\cos(0.5x)\). 2. Apply the bounds: \([-2a\cos(0.5x)]_0^\pi=-2a\cos\left(\frac{\pi}{2}\right)+2a\cos(0)=2a\). 3. Solve \(2a=10\), giving \(a=5\).

\(a=5\)

- Use the logarithmic antiderivative for \(\frac{1}{x+2}\). - Combine the difference of logarithms. - Rewrite \(16\) as a power of \(2\). - Compare the coefficients of \(\ln(2)\).

1. An antiderivative is \(F_k(x)=k\ln(x+2)\). 2. Apply the bounds: \(\int_0^2\frac{k}{x+2}\,dx=k(\ln(4)-\ln(2))=k\ln(2)\). 3. Since \(\ln(16)=\ln(2^4)=4\ln(2)\), \(k\ln(2)=4\ln(2)\), so \(k=4\).

\(k=4\)

- How does changing the value subtracted from \(x\) affect a graph horizontally? - The derivative of a cumulative distribution function is its density. - What cumulative probability lies at the mean of a symmetric normal distribution? - Use the symmetry of the normal curve.

1. Increasing \(\mu\) shifts the entire cumulative distribution graph to the right without changing its shape. 2. The derivative of the cumulative distribution function is the normal density. At \(x=\mu\), the slope is \(\frac{1}{\sigma\sqrt{2\pi}}\). Therefore, increasing \(\sigma\) decreases the maximum slope and makes the center of the graph less steep. 3. By symmetry, \(\Phi_{\mu,\sigma}(\mu)=0.5\) for every \(\sigma>0\). Thus, all graphs with fixed \(\mu\) pass through \((\mu, 0.5)\).

1. The graph shifts to the right. 2. The maximum slope decreases as \(\sigma\) increases. 3. \((\mu, 0.5)\)

- Apply the power rule to each term. - Use the Fundamental Theorem of Calculus to connect the antiderivative and the definite integral. - Use parentheses and check signs carefully when substituting the negative lower limit.

1. Using the power, sum, and constant multiple rules, an antiderivative is \(F(x)=2x^3-x^2+x\). 2. Evaluate the endpoint values: \(F(2)=16-4+2=14\) and \(F(-1)=-2-1-1=-4\). 3. Therefore, \(F(2)-F(-1)=14-(-4)=18\).

\(18\)

- Integrate the sum term by term. - Keep each constant multiple when applying the power rule. - Differentiate your antiderivative mentally to check that it gives the original integrand.

1. Using the power, sum, and constant multiple rules, an antiderivative is \(F(x)=\frac{1}{32}x^4+\frac{1}{2}x^3-4x\). 2. Evaluate: \(F(4)=8+32-16=24\) and \(F(0)=0\). 3. Therefore, \(F(4)-F(0)=24\).

\(24\). The antiderivative uses the power rule, constant multiple rule, and sum rule.

- Identify the variable named in the differential. - Integrate the polynomial term by term. - Evaluate the antiderivative at both limits. - Use the Fundamental Theorem of Calculus to subtract the endpoint values.

1. Using the power, constant multiple, and sum rules, an antiderivative is \(F(u)=2u^3-2u^2+3u\). 2. Evaluate: \(F(2)=16-8+6=14\) and \(F(1)=2-2+3=3\). 3. By the Fundamental Theorem of Calculus, \(F(2)-F(1)=14-3=11\).

The integral is \(11\). The rules used are the power rule, constant multiple rule, sum rule, and the Fundamental Theorem of Calculus.

- Use parentheses when substituting the negative lower limit. - Apply the power rule to each term. - Use the difference rule for the two terms. - Subtract the lower-limit value from the upper-limit value.

1. Using the power and difference rules, an antiderivative is \(F(v)=\frac{1}{4}v^4-\frac{1}{2}v^2\). 2. Evaluate: \(F(2)=4-2=2\) and \(F(-1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\). 3. By the Fundamental Theorem of Calculus, \(F(2)-F(-1)=2-\left(-\frac{1}{4}\right)=\frac{9}{4}\).

The integral is \(\frac{9}{4}\), or \(2.25\). The rules used are the power rule, difference rule, and the Fundamental Theorem of Calculus.

- Use the power rule to find an antiderivative. - Substitute the upper limit first. - Evaluate the lower-limit value and subtract it.

1. An antiderivative is \(F(t)=t^3-t^2\). 2. Evaluate: \(F(4)=64-16=48\) and \(F(0)=0\). 3. Therefore, \(F(4)-F(0)=48\).

\(48\)

- Integrate the polynomial term by term. - Use parentheses when substituting the negative lower limit. - Recall that an even power of a negative number is positive. - Keep exact fractions until the final step.

1. An antiderivative is \(F(x)=\frac{1}{8}x^4-\frac{3}{2}x^2+x\). 2. Evaluate at the upper limit: \(F(2)=2-6+2=-2\). 3. Evaluate at the lower limit: \(F(-1)=\frac{1}{8}-\frac{3}{2}-1=-\frac{19}{8}\). 4. Therefore, \(F(2)-F(-1)=-2+\frac{19}{8}=\frac{3}{8}\).

\(\frac{3}{8}\), or \(0.375\)

- Treat \(k\) as a constant while integrating. - Use the exact cosine values at \(0\) and \(\pi\). - Set the evaluated integral equal to the given expression. - Solve the resulting linear equation.

1. An antiderivative is \(F(x)=-k\cos(x)+\frac{x^2}{2}\). 2. Apply the bounds: \(F(\pi)-F(0)=2k+\frac{\pi^2}{2}\). 3. Set this equal to the given value: \(2k+\frac{\pi^2}{2}=\frac{\pi^2}{2}+4\). Thus, \(2k=4\) and \(k=2\).

\(k=2\)

- Recall the antiderivative of cosine and use the power rule for \(x\). - Use radians for the trigonometric limit. - Subtract the lower-limit value from the upper-limit value.

1. An antiderivative is \(F(x)=3\sin(x)-x^2\). 2. At the upper limit, \(F\left(\frac{\pi}{2}\right)=3-\frac{\pi^2}{4}\). 3. Since \(F(0)=0\), the integral is \(3-\frac{\pi^2}{4}\).

\(3-\frac{\pi^2}{4}\)

- Treat \(k\) as a constant. - Be careful with the sign when integrating sine. - Use the cosine values at \(0\) and \(\pi\). - Set the evaluated integral equal to \(0\).

1. An antiderivative is \(F(x)=-k\cos(x)-x\). 2. Apply the bounds: \(F(\pi)-F(0)=2k-\pi\). 3. Set the integral equal to \(0\): \(2k-\pi=0\), so \(k=\frac{\pi}{2}\).

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

- Apply the power rule to find antiderivatives. - Integrate sums and differences term by term. - Rewrite the radical as a rational power. - For part c), consider the symmetry of an odd function on a symmetric interval.

1. For a), an antiderivative is \(F(x)=\frac{1}{3}x^3-2x\). Thus, \(F(3)-F(0)=3\). 2. For b), rewrite the integrand as \(x^{-1/2}\). An antiderivative is \(F(x)=2\sqrt{x}\). Thus, \(F(4)-F(1)=4-2=2\). 3. For c), an antiderivative is \(F(x)=\frac{1}{4}x^4\). Thus, \(F(2)-F(-2)=4-4=0\).

a) \(3\) b) \(2\) c) \(0\)

- Use the relationship between a definite integral and an antiderivative. - Apply the power rule term by term. - Use parentheses when substituting the negative lower limit. - Interpret \([F(x)]_a^b\) as \(F(b)-F(a)\).

1. An antiderivative is \(F(x)=x^4-2x^3+2x\). 2. Evaluate: \(F(2)=16-16+4=4\), and \(F(-1)=1+2-2=1\). 3. Therefore, \(F(2)-F(-1)=4-1=3\).

\(3\)

- Apply the power rule to each polynomial term. - Evaluate the antiderivative at both limits. - Subtract the lower-limit value from the upper-limit value.

1. An antiderivative is \(F(x)=x^3-2x^2+x\). 2. Evaluate: \(F(3)=27-18+3=12\), and \(F(1)=1-2+1=0\). 3. Therefore, \(F(3)-F(1)=12\).

\(12\)

- Treat \(a\) as a constant while integrating. - Apply upper bound minus lower bound. - Form an equation in \(a\). - Solve the linear equation.

1. An antiderivative is \(F(x)=\frac{x^3}{3}-ax\). 2. Apply the bounds: \(F(3)-F(1)=\left(9-3a\right)-\left(\frac{1}{3}-a\right)=\frac{26}{3}-2a\). 3. Set this equal to \(\frac{2}{3}\): \(\frac{26}{3}-2a=\frac{2}{3}\), so \(2a=8\) and \(a=4\).

\(a=4\)

- Use the Fundamental Theorem of Calculus with the given antiderivative. - You do not need to find an explicit formula for \(f\). - Recall that \(e^0=1\), and check the sign of the lower-limit value.

1. Since \(F\) is already given as an antiderivative, \(\int_{0}^{2}f(x)\,dx=F(2)-F(0)\). 2. Evaluate: \(F(2)=(2-2)e^2=0\), and \(F(0)=(0-2)e^0=-2\). 3. Therefore, \(F(2)-F(0)=0-(-2)=2\).

\(2\)

- Recall which function has derivative \(\frac{1}{x}\). - Find the antiderivative term by term. - Apply the Fundamental Theorem of Calculus. - Simplify \(\ln(e^k)\) using logarithm properties.

1. On this positive interval, an antiderivative is \(F(x)=5\ln(x)-x\). 2. Evaluate: \(F(e^2)=5\ln(e^2)-e^2=10-e^2\), and \(F(1)=5\ln(1)-1=-1\). 3. Therefore, \(F(e^2)-F(1)=10-e^2-(-1)=11-e^2\).

\(11-e^2\)

- Find an antiderivative for each term. - Account for the coefficient in the exponent when integrating \(e^{kx}\). - Evaluate at the upper and lower limits. - Recall that \(e^0=1\).

1. An antiderivative is \(F(x)=2e^{0.25x}+2x\), because differentiating \(2e^{0.25x}\) gives \(\frac{1}{2}e^{0.25x}\). 2. Evaluate: \(F(4)=2e+8\), and \(F(0)=2\). 3. Therefore, \(F(4)-F(0)=2e+6\).

\(2e+6\)

- Compare the numerator with the derivative of the denominator. - Use the logarithmic antiderivative pattern. - Apply the Fundamental Theorem after finding an antiderivative. - Recall that \(\ln(1)=0\).

1. Since the derivative of \(x^2+1\) is \(2x\), an antiderivative is \(F(x)=\frac{3}{2}\ln(x^2+1)\). 2. Apply the Fundamental Theorem: \(\int_0^2\frac{3x}{x^2+1}\,dx=\left[\frac{3}{2}\ln(x^2+1)\right]_0^2\). 3. Evaluate: \(\frac{3}{2}\ln(5)-\frac{3}{2}\ln(1)=\frac{3}{2}\ln(5)\).

\(\frac{3}{2}\ln(5)\)

- Use \(F(b)-F(a)=\int_a^b f(x)\,dx\). - Break the region into simple geometric shapes. - Areas below the x-axis count negatively.

1. By the Fundamental Theorem of Calculus, \(F(4)-F(1)=\int_1^4 f(x)\,dx\). 2. From \(x=1\) to \(x=3\), the graph forms a triangle above the x-axis with base \(2\) and height \(2\), so its area is \(\frac12\cdot2\cdot2=2\). 3. From \(x=3\) to \(x=4\), the graph forms a triangle below the x-axis with base \(1\) and height \(1\), so its signed area is \(-\frac12\). 4. Therefore, \(F(4)-F(1)=2-\frac12=\frac32\).

\(F(4)-F(1)=\frac32\)

- Use \(F(3)-F(0)=\int_0^3f(x)\,dx\). - Determine the sign of the signed area under \(f\) on \([0,3]\). - Positive values of \(f\) make an antiderivative increase.

1. By the Fundamental Theorem of Calculus, \(F(3)=F(0)+\int_0^3 f(x)\,dx\). 2. The graph of \(f\) is above the x-axis on \([0,3)\) and equals \(0\) at \(x=3\). Therefore, \(\int_0^3 f(x)\,dx>0\). 3. Thus, \(F(3)=1+\int_0^3 f(x)\,dx>1\).

\(F(3)>1\)

- Find an antiderivative for each integrand. - Apply \(F(b)-F(a)\). - For linear or constant functions, a geometric interpretation using rectangles, triangles, or trapezoids may help.

1. For a), an antiderivative is \(F(x)=0.5x\). Thus, \(F(6)-F(-4)=3-(-2)=5\). Geometrically, this is the area of a rectangle with width \(10\) and height \(0.5\). 2. For b), an antiderivative is \(F(x)=3x-\frac{1}{2}x^2\). Thus, \(F(4)-F(0)=12-8=4\). This is a net area because the line crosses the x-axis at \(x=3\). 3. For c), an antiderivative is \(F(t)=t^2+t\). Thus, \(F(3)-F(-2)=12-2=10\).

a) \(5\) b) \(4\) c) \(10\)

- Use the graph of the antiderivative \(g\), not the graph of \(f\). - Read \(g(3)\) and \(g(0)\). - Apply the Fundamental Theorem of Calculus.

1. By the Fundamental Theorem of Calculus, \(\int_0^3f(x)\,dx=g(3)-g(0)\). 2. From the graph, \(g(3)=4\) and \(g(0)=1\). 3. Therefore, \(\int_0^3f(x)\,dx=4-1=3\).

\(3\)

- Read the values of \(g\) at both integration limits. - Use \(\int_a^bf(x)\,dx=F(b)-F(a)\). - Equal endpoint values give an integral of \(0\).

1. By the Fundamental Theorem of Calculus, \(\int_{-2}^{2}f(x)\,dx=g(2)-g(-2)\). 2. The graph shows \(g(2)=0\) and \(g(-2)=0\). 3. Therefore, the integral is \(0-0=0\).

\(0\)

- First evaluate the integral using an antiderivative. - Find where the graph crosses the \(x\)-axis. - Identify which part of the graph lies above the axis and which part lies below it. - Interpret what an integral value of zero means about the two signed-area contributions.

1. An antiderivative of \(f(x)=-\frac{1}{4}x+1\) is \(F(x)=-\frac{1}{8}x^2+x\). Therefore, \(I=F(8)-F(0)=\left(-\frac{1}{8}\cdot 64+8\right)-0=-8+8=0\). 2. The graph crosses the \(x\)-axis at \(x=4\). On \([0,4]\), the graph is above the axis and contributes signed area \(\int_{0}^{4}f(x)\,dx=2\). On \([4,8]\), it is below the axis and contributes signed area \(\int_{4}^{8}f(x)\,dx=-2\). These contributions cancel, so the net signed area is \(0\).

The integral is \(I=0\). The positive area above the \(x\)-axis on \([0,4]\) equals the magnitude of the negative area below the \(x\)-axis on \([4,8]\), so the net signed area is zero.

- Use the Fundamental Theorem of Calculus. - Evaluate the antiderivative at the upper limit and subtract its value at the lower limit. - Recall how portions of a graph below the x-axis contribute to a definite integral. - Consider what a net area of zero means.

1. For a), an antiderivative is \(F(x)=\frac{1}{3}x^3-x^2-3x\). Then \(F(2)=-\frac{22}{3}\) and \(F(-1)=\frac{5}{3}\), so \(F(2)-F(-1)=-\frac{22}{3}-\frac{5}{3}=-9\). 2. For b), an antiderivative is \(G(x)=\frac{1}{8}x^4-2x^2\). Since \(G(4)=0\) and \(G(0)=0\), the integral is \(0\). 3. A net area of \(0\) means that the positive and negative signed areas on \([0,4]\) have equal magnitudes and cancel.

a) \(-9\) b) \(0\). The signed areas above and below the x-axis on \([0,4]\) cancel.

- Treat \(k\) as a constant while integrating. - Find an antiderivative first. - Apply the bounds and form an equation in \(k\). - Solve the resulting linear equation.

1. An antiderivative is \(F(x)=\frac{kx^2}{2}+x\). 2. Apply the Fundamental Theorem: \(F(3)-F(1)=\left(\frac{9k}{2}+3\right)-\left(\frac{k}{2}+1\right)=4k+2\). 3. Set the result equal to \(10\): \(4k+2=10\), so \(k=2\).

\(k=2\)

- Identify the function variable and the dummy variable of integration. - Reversing the limits changes the sign of an integral. - Apply the Fundamental Theorem of Calculus at the variable upper limit.

1. For a), reverse the limits: \(I(x)=-\int_{10}^{x}\cos(z^2)\,dz\). Therefore, \(\frac{dI}{dx}=-\cos(x^2)\). 2. For b), the Fundamental Theorem of Calculus gives \(\frac{dA}{du}=\frac{e^u}{u+3}\). 3. For c), treat \(k\) as a constant. Then \(\frac{dF}{dy}=k\sin(y)\).

a) \(\frac{dI}{dx}=-\cos(x^2)\) b) \(\frac{dA}{du}=\frac{e^u}{u+3}\) c) \(\frac{dF}{dy}=k\sin(y)\)

- First verify the antiderivative. - Check the required order for subtracting endpoint values. - Use parentheses when substituting a negative number. - Review how an odd power affects a negative number.

1. An antiderivative is \(F(x)=2x-x^3\). 2. The endpoint values are \(F(2)=4-8=-4\) and \(F(-1)=-2-(-1)=-1\). 3. Therefore, \(F(2)-F(-1)=-4-(-1)=-3\), so Luke is correct. 4. Maya reversed the order of subtraction and computed \(F(-1)-F(2)\). 5. Tom substituted \(1^3\) instead of \((-1)^3\) at the lower limit, causing a sign error.

Luke is correct. Maya reversed the limits by computing \(F(-1)-F(2)\) instead of \(F(2)-F(-1)\). Tom used \(1^3\) instead of \((-1)^3\) when substituting the lower limit.

- Check each transformation separately. - Recall that \(a^{-n}=\frac{1}{a^n}\). - Pay close attention to both the negative base and the negative exponent. - After evaluating the antiderivative, subtract the lower-limit value from the upper-limit value.

1. Steps 1 through 3 are correct. The integrand is \(4x^{-3}\), and an antiderivative is \(-2x^{-2}\). 2. The first error is in step 4. The student used \((-2)^{-2}=4\), but \((-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}\). 3. Correcting the endpoint values gives \(F(-1)=-2\cdot1=-2\) and \(F(-2)=-2\cdot\frac{1}{4}=-\frac{1}{2}\). 4. Therefore, \(F(-1)-F(-2)=-2-\left(-\frac{1}{2}\right)=-\frac{3}{2}\).

The first error occurs in step 4 because \((-2)^{-2}=\frac{1}{4}\), not \(4\). The corrected value is \(-\frac{3}{2}\), or \(-1.5\).

- Simplify the integrand before finding an antiderivative. - Use the binomial formula and the distributive property to expand the product. - When substituting the negative lower limit, check the signs of its powers carefully.

1. Expand the integrand: \((x-2)^2(x+1)=(x^2-4x+4)(x+1)=x^3-3x^2+4\). 2. An antiderivative is \(F(x)=\frac{1}{4}x^4-x^3+4x\). 3. Evaluate at the limits: \(F(1)=\frac{1}{4}-1+4=\frac{13}{4}\) and \(F(-2)=4+8-8=4\). 4. Therefore, \(F(1)-F(-2)=\frac{13}{4}-4=-\frac{3}{4}\).

\(-\frac{3}{4}\), or \(-0.75\)

- Differentiate \(F\) to check whether it is an antiderivative. - Use the chain rule for the logarithm of a function. - Apply the Fundamental Theorem of Calculus at the two limits. - Use a logarithm property to simplify the difference.

1. Differentiate using the chain rule: \(F'(x)=\frac{1}{x^2+4}\cdot2x=\frac{2x}{x^2+4}\). Therefore, \(F\) is an antiderivative of the integrand. 2. By the Fundamental Theorem of Calculus, \(\int_{0}^{2}\frac{2x}{x^2+4}\,dx=F(2)-F(0)\). 3. Evaluate: \(F(2)=\ln(8)\) and \(F(0)=\ln(4)\). 4. Thus, \(\ln(8)-\ln(4)=\ln\left(\frac{8}{4}\right)=\ln(2)\).

\(\ln(2)\)

- Use the product rule to differentiate \(F\). - To verify an antiderivative, show that its derivative equals the integrand. - Evaluate the antiderivative at the upper and lower limits. - Use parentheses when subtracting the lower-limit value.

1. Differentiate \(F\) using the product rule: \(F'(x)=2e^x+(2x-1)e^x\). 2. Combine like terms: \(F'(x)=e^x(2+2x-1)=(2x+1)e^x=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 3. Apply the Fundamental Theorem of Calculus: \(\int_{0}^{1}f(x)\,dx=F(1)-F(0)\). 4. Since \(F(1)=e\) and \(F(0)=-1\), the integral is \(e-(-1)=e+1\).

\(e+1\)

- Which function is an antiderivative of \(f+g\)? - Recall the derivative rule for a sum. - Apply the Fundamental Theorem of Calculus to each integral. - Rearrange the resulting terms so the two sides have the same form.

1. Since \((F+G)'=F'+G'=f+g\), the function \(F+G\) is an antiderivative of \(f+g\). 2. Apply the Fundamental Theorem of Calculus to the left side: \(\int_a^b [f(x)+g(x)]\,dx=[F(x)+G(x)]_a^b=F(b)+G(b)-F(a)-G(a)\). 3. Apply the theorem to the right side: \(\int_a^b f(x)\,dx+\int_a^b g(x)\,dx=[F(b)-F(a)]+[G(b)-G(a)]\). 4. Rearranging the right side gives \(F(b)+G(b)-F(a)-G(a)\), which is the same expression as the left side.

Because \(F+G\) is an antiderivative of \(f+g\), the left side equals \(F(b)+G(b)-F(a)-G(a)\). Evaluating the two integrals on the right gives \([F(b)-F(a)]+[G(b)-G(a)]\), which simplifies to the same expression.

- Combine the square roots into one radical. - Rewrite the fraction using a negative exponent. - Apply the power rule to each term. - Use the Fundamental Theorem of Calculus to evaluate the definite integral.

1. On \([1,2]\), \(x>0\), so \(\sqrt{3x}\sqrt{12x}=\sqrt{36x^2}=6x\). Also, \(\frac{4}{x^3}=4x^{-3}\). Thus, the integrand is \(6x-4x^{-3}\). 2. An antiderivative is \(F(x)=3x^2+2x^{-2}=3x^2+\frac{2}{x^2}\). 3. Evaluate: \(F(2)=12+\frac{1}{2}=\frac{25}{2}\) and \(F(1)=3+2=5\). 4. Therefore, \(F(2)-F(1)=\frac{25}{2}-5=\frac{15}{2}=7.5\).

\(\frac{15}{2}\), or \(7.5\)

- Split the numerator so that each term is divided by the denominator. - Use the quotient rule for exponents with the same base. - Check the derivative of \(e^{-x}\) when finding an antiderivative. - Recall the value of a nonzero base raised to the zero power.

1. Split the fraction and subtract exponents: \(\frac{e^{3x}}{e^{2x}}+\frac{e^x}{e^{2x}}=e^x+e^{-x}\). 2. An antiderivative is \(F(x)=e^x-e^{-x}\). 3. Evaluate at the limits: \(F(1)=e-e^{-1}=e-\frac{1}{e}\) and \(F(0)=1-1=0\). 4. Therefore, the integral equals \(e-\frac{1}{e}\).

\(e-\frac{1}{e}\)

- Rewrite the radical and the fraction using rational and negative exponents. - Apply the power rule to find an antiderivative. - Evaluate at the upper limit and subtract the value at the lower limit. - Be careful when subtracting a negative endpoint value.

1. Rewrite the integrand as \(\frac{3}{2}x^{1/2}+2x^{-2}\). 2. An antiderivative is \(F(x)=x^{3/2}-2x^{-1}=x\sqrt{x}-\frac{2}{x}\). 3. Evaluate: \(F(4)=8-\frac{1}{2}=\frac{15}{2}\) and \(F(1)=1-2=-1\). 4. Thus, \(F(4)-F(1)=\frac{15}{2}-(-1)=\frac{17}{2}\).

\(\frac{17}{2}\), or \(8.5\)

- Recall antiderivatives of sine and the exponential function. - Use the exact sine and cosine values at \(0\) and \(\pi\). - Recall that \(e^0=1\). - Subtract the entire lower-limit value carefully.

1. An antiderivative is \(F(x)=-\cos(x)-\frac{1}{3}e^x\). 2. At the upper limit, \(F(\pi)=1-\frac{1}{3}e^{\pi}\). 3. At the lower limit, \(F(0)=-1-\frac{1}{3}=-\frac{4}{3}\). 4. Therefore, \(F(\pi)-F(0)=1-\frac{1}{3}e^{\pi}+\frac{4}{3}=\frac{7-e^{\pi}}{3}\).

\(\frac{7-e^{\pi}}{3}\)

- Rewrite the radical and fraction using exponents. - Use the power rule to find an antiderivative. - Evaluate the antiderivative at the upper and lower limits. - Subtract the lower-limit value from the upper-limit value.

1. Rewrite the integrand as \(x^{-1/2}+2x^{-2}\). 2. An antiderivative is \(F(x)=2x^{1/2}-2x^{-1}=2\sqrt{x}-\frac{2}{x}\). 3. Evaluate: \(F(4)=4-\frac{1}{2}=\frac{7}{2}\) and \(F(1)=2-2=0\). 4. Therefore, \(F(4)-F(1)=\frac{7}{2}\).

\(\frac{7}{2}\), or \(3.5\)

- Use the product rule to differentiate \(x\ln(x)\). - Recall the derivative of \(\ln(x)\). - Use the antiderivative to evaluate the definite integral. - Recall the values of \(\ln(e)\) and \(\ln(1)\).

1. For a), use the product rule: \(F'(x)=\ln(x)+x\cdot\frac{1}{x}+1=\ln(x)+2=f(x)\). Thus, \(F\) is an antiderivative of \(f\). 2. For b), apply the Fundamental Theorem of Calculus: \(\int_{1}^{e}f(x)\,dx=F(e)-F(1)\). 3. Since \(F(e)=e\ln(e)+e=2e\) and \(F(1)=\ln(1)+1=1\), the integral is \(2e-1\).

a) \(F'(x)=\ln(x)+2=f(x)\), so \(F\) is an antiderivative of \(f\). b) \(2e-1\)

- Apply the chain rule when differentiating the square of \(\ln(x)\). - Recall the derivative of \(\ln(x)\). - Simplify \(\ln(e^2)\) exactly. - Use the difference of the antiderivative values at the limits.

1. For a), differentiate using the chain rule: \(F'(x)=\frac{1}{2}\cdot2\ln(x)\cdot\frac{1}{x}=\frac{\ln(x)}{x}=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 2. For b), \(\int_{1}^{e^2}f(x)\,dx=F(e^2)-F(1)\). 3. Evaluate: \(F(e^2)=\frac{1}{2}(\ln(e^2))^2=\frac{1}{2}\cdot2^2=2\), and \(F(1)=0\). 4. The integral equals \(2\).

a) \(F'(x)=\frac{\ln(x)}{x}=f(x)\), so \(F\) is an antiderivative of \(f\). b) \(2\)

- Rewrite the integrand using a negative exponent. - Apply the bounds to the antiderivative. - Clear the fractions in the resulting equation. - Use the condition that \(a\) is a positive integer.

1. An antiderivative is \(F(x)=-\frac{8}{x+a}\). 2. Apply the bounds: \(\frac{8}{a}-\frac{8}{a+4}=1\). 3. Combine the fractions: \(\frac{32}{a(a+4)}=1\), so \(a^2+4a-32=0\). 4. Factor: \((a-4)(a+8)=0\). The positive integer solution is \(a=4\).

\(a=4\)

- Recall where the cosine function equals zero. - Adapt the general cosine-zero formula to the argument in this function. - Reverse the chain rule when finding an antiderivative. - Determine where the graph is above and below the \(x\)-axis. - Distinguish between net signed area and total geometric area.

1. For a), solve \(\cos\left(\frac{\pi}{3}x\right)=0\). Since cosine is zero when its argument is \(\frac{\pi}{2}+k\pi\), \(\frac{\pi}{3}x=\frac{\pi}{2}+k\pi\), so \(x=\frac{3}{2}+3k\), where \(k\in\mathbb{Z}\). Two consecutive zeros are \(x=1.5\) and \(x=4.5\). 2. For b), an antiderivative is \(F(x)=\frac{3}{\pi}\sin\left(\frac{\pi}{3}x\right)\). Therefore, \(\int_{0}^{3}f(x)\,dx=\frac{3}{\pi}\left(\sin(\pi)-\sin(0)\right)=0\). 3. The graph is above the \(x\)-axis on \([0,1.5]\) and below it on \([1.5,3]\). The definite integral adds signed areas, so the equal positive and negative contributions cancel. Total geometric area would add their magnitudes instead.

a) Two consecutive zeros are, for example, \(x=1.5\) and \(x=4.5\). b) \(\int_{0}^{3}f(x)\,dx=0\). The graph lies both above and below the \(x\)-axis, so the definite integral gives net signed area rather than total geometric area.

- Solve the exponential equation by comparing exponents or taking logarithms. - Use the antiderivative rule for a function of the form \(e^{ax}\). - Determine the sign of the function on each side of its zero. - Recall how regions below the \(x\)-axis contribute to a definite integral.

1. For a), solve \(e^{0.5x}-e=0\). Then \(e^{0.5x}=e^1\), so \(0.5x=1\) and \(x=2\). 2. For b), an antiderivative is \(G(x)=2e^{0.5x}-ex\). Thus, \(\int_{0}^{4}g(x)\,dx=G(4)-G(0)=\left(2e^2-4e\right)-2=2e^2-4e-2\approx 1.90\). 3. The function is negative on \([0,2)\) and positive on \((2,4]\). The definite integral subtracts the area below the \(x\)-axis from the area above it. Total geometric area would require adding the absolute values of the two contributions.

a) The zero is \(x=2\). b) \(\int_{0}^{4}g(x)\,dx=2e^2-4e-2\approx 1.90\). This is net signed area, not total geometric area, because the function changes sign at \(x=2\).

- In part a), consider the symmetry of the integrand. - For a linear expression in an exponent or angle, account for its derivative when finding an antiderivative. - Check signs carefully when integrating sine and cosine. - Use exponent and logarithm properties to simplify part b).

1. For a), an antiderivative is \(F(x)=\frac{1}{16}x^4-\frac{1}{2}x^2\). Because \(F(2)=F(-2)=-1\), the integral is \(0\). 2. For b), an antiderivative is \(F(x)=\frac{1}{3}e^{3x}\). Therefore, \(F(\ln(2))-F(0)=\frac{1}{3}e^{3\ln(2)}-\frac{1}{3}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}\). 3. For c), an antiderivative is \(F(x)=-2\cos(x)-\frac{1}{2}\sin(2x)\). Thus, \(F\left(\frac{\pi}{2}\right)-F(0)=0-(-2)=2\).

a) \(0\) b) \(\frac{7}{3}\) c) \(2\)

- Recall antiderivative rules for powers, exponential functions, and trigonometric functions. - Evaluate the antiderivative at the upper limit and subtract its value at the lower limit. - In part b), rewrite the fraction using a negative exponent. - In part c), account for the coefficient in the exponent.

1. For a), an antiderivative is \(F(x)=2\sin(x)+\cos(x)\). Thus, \(F\left(\frac{\pi}{2}\right)-F(0)=2-1=1\). 2. For b), an antiderivative is \(F(x)=-2x^{-2}+2x=-\frac{2}{x^2}+2x\). Thus, \(F(2)-F(1)=\frac{7}{2}-0=\frac{7}{2}\). 3. For c), an antiderivative is \(F(x)=\frac{1}{2}e^{2x}+\frac{1}{2}x^2\). Thus, \(F(1)-F(0)=\left(\frac{1}{2}e^2+\frac{1}{2}\right)-\frac{1}{2}=\frac{1}{2}e^2\).

a) \(1\) b) \(\frac{7}{2}\) c) \(\frac{1}{2}e^2\)

- Rewrite the radical as a power before using the power rule. - Recall the antiderivative of \(\frac{1}{x}\). - Use parentheses when substituting negative values into powers. - In part c), symmetry can simplify part of the integral.

1. For a), rewrite the integrand as \(3x^{1/2}-1\). An antiderivative is \(F(x)=2x^{3/2}-x\). Therefore, \(F(4)-F(1)=12-1=11\). 2. For b), an antiderivative is \(F(x)=2\ln(x)\) on the interval. Thus, \(F(e^2)-F(e)=4-2=2\). 3. For c), an antiderivative is \(F(x)=\frac{1}{6}x^6-x^3\). Therefore, \(F(1)-F(-1)=-\frac{5}{6}-\frac{7}{6}=-2\).

a) \(11\) b) \(2\) c) \(-2\)

- Express a general normal cumulative distribution function in terms of the standard normal function. - Compare the density values at each distribution's mean. - What cumulative probability is reached when the upper endpoint moves without bound? - A larger \(\sigma\) produces a wider distribution.

1. First stretch the graph horizontally by a factor of \(2\), producing \(\Phi_{0,1}\left(\frac{x}{2}\right)\). Then shift it \(5\) units to the right, giving \(F_2(x)=\Phi_{0,1}\left(\frac{x-5}{2}\right)\). 2. At the mean, the slope of a normal cumulative distribution function is \(\frac{1}{\sigma\sqrt{2\pi}}\). Since \(1<2\), \(F_1\) has the greater slope at its mean. 3. Both functions approach \(1\) as \(x\to\infty\). This represents the total probability of all possible values.

1. Stretch horizontally by a factor of \(2\), then shift \(5\) units right. 2. The slope at the mean is inversely proportional to \(\sigma\), so \(F_1\) is steeper. 3. The limit is \(1\), representing total probability.

- Rewrite the second fraction using a negative exponent. - The power rule does not apply to \(x^{-1}\); use a logarithm for that term. - Apply the Fundamental Theorem of Calculus using the two limits. - Recall the values of \(\ln(e)\) and \(\ln(1)\).

1. Rewrite \(g(x)=\frac{2}{x}+6x^{-2}\). 2. An antiderivative is \(G(x)=2\ln(x)-6x^{-1}=2\ln(x)-\frac{6}{x}\). 3. Apply the Fundamental Theorem of Calculus: \(\int_{1}^{e}g(x)\,dx=G(e)-G(1)\). 4. Since \(G(e)=2-\frac{6}{e}\) and \(G(1)=-6\), the value is \(2-\frac{6}{e}-(-6)=8-\frac{6}{e}\).

a) \(G(x)=2\ln(x)-\frac{6}{x}\) b) \(8-\frac{6}{e}\)

- Simplify the integrand before integrating. - Apply the power rule after expanding. - Check signs carefully when substituting the negative lower limit. - Use upper-limit value minus lower-limit value.

1. Expand the integrand: \((x+1)(x-2)^2=(x+1)(x^2-4x+4)=x^3-3x^2+4\). 2. An antiderivative is \(F(x)=\frac{1}{4}x^4-x^3+4x\). 3. Evaluate: \(F(2)=4-8+8=4\) and \(F(-1)=\frac{1}{4}+1-4=-\frac{11}{4}\). 4. Therefore, \(F(2)-F(-1)=4+\frac{11}{4}=\frac{27}{4}\).

\(\frac{27}{4}\), or \(6.75\)

- Use the power rule to find an antiderivative. - Simplify powers of \(\sqrt{6}\) carefully. - Subtract the lower-limit value from the upper-limit value.

1. An antiderivative is \(F(x)=\frac{1}{12}x^4-x^2\). 2. At the upper limit, \(F(\sqrt{6})=\frac{1}{12}(\sqrt{6})^4-(\sqrt{6})^2=3-6=-3\). 3. Since \(F(0)=0\), the integral is \(F(\sqrt{6})-F(0)=-3\).

\(-3\)

- Apply the Fundamental Theorem of Calculus. - Use the power rule to find each antiderivative. - Check signs carefully at a negative lower limit. - In part b), consider the symmetry of the integrand and interval.

1. For a), an antiderivative is \(F(x)=\frac{1}{3}x^3-2x^2+6x\). Since \(F(4)=\frac{40}{3}\) and \(F(1)=\frac{13}{3}\), the integral is \(9\). 2. For b), an antiderivative is \(G(x)=\frac{1}{4}x^4-\frac{3}{2}x^2\). Since \(G(2)=G(-2)=-2\), the integral is \(0\).

a) \(9\) b) \(0\)

- Find an antiderivative before substituting the limits. - When integrating \(x^n\), increase the exponent by one and divide by the new exponent. - Put the lower-limit value in parentheses when subtracting.

1. For a), an antiderivative is \(F(x)=\frac{1}{12}x^4-\frac{1}{3}x^3+2x\). Then \(F(3)-F(0)=\frac{15}{4}-0=\frac{15}{4}\). 2. For b), an antiderivative is \(G(x)=2x^3-x^2-4x\). Since \(G(2)=4\) and \(G(-1)=1\), the integral is \(3\).

a) \(\frac{15}{4}\), or \(3.75\) b) \(3\)

- Apply the power rule for derivatives. - Recognize an antiderivative of the second derivative. - Use the Fundamental Theorem of Calculus at the limits \(1\) and \(x\). - Differentiate the result and compare it with the integrand.

1. The derivatives are \(f'(x)=4x^3-4x\) and \(f''(x)=12x^2-4\). 2. Since \(f'\) is an antiderivative of \(f''\), \(G(x)=f'(x)-f'(1)\). Because \(f'(1)=4-4=0\), \(G(x)=4x^3-4x\). 3. Differentiating gives \(G'(x)=12x^2-4=f''(x)\). Integrating a derivative recovers the original function up to a constant, and differentiating an accumulation function recovers its integrand.

1. \(f'(x)=4x^3-4x\); \(f''(x)=12x^2-4\) 2. \(G(x)=4x^3-4x\) 3. \(G'(x)=12x^2-4=f''(x)\)

- Use the derivative rule for a constant multiple. - Identify an antiderivative of \(kf\) from an antiderivative of \(f\). - Apply the Fundamental Theorem at the endpoints. - Factor \(k\) from the resulting difference.

1. Let \(F\) be an antiderivative of \(f\), so \(F'(x)=f(x)\). 2. By the constant multiple rule for derivatives, \([kF(x)]'=kF'(x)=kf(x)\). Thus \(kF\) is an antiderivative of \(kf\). 3. By the Fundamental Theorem of Calculus, \(\int_a^b kf(x)\,dx=[kF(x)]_a^b=kF(b)-kF(a)\). 4. Factor out \(k\): \(k[F(b)-F(a)]=k\int_a^b f(x)\,dx\).

If \(F'=f\), then \((kF)'=kf\). Therefore, the Fundamental Theorem gives \(\int_a^b kf(x)\,dx=[kF(x)]_a^b=k[F(b)-F(a)]=k\int_a^b f(x)\,dx\).

- Differentiate the difference of two antiderivatives. - Use the derivative difference rule. - Apply the Fundamental Theorem to the difference. - Regroup the endpoint terms into two separate differences.

1. Let \(F'=f\) and \(G'=g\). By the difference rule for derivatives, \((F-G)'=F'-G'=f-g\). 2. Thus \(F-G\) is an antiderivative of \(f-g\). By the Fundamental Theorem, \(\int_a^b[f(x)-g(x)]\,dx=[F(x)-G(x)]_a^b\). 3. Expand the endpoint expression: \([F(b)-G(b)]-[F(a)-G(a)]\) \(=[F(b)-F(a)]-[G(b)-G(a)]\). 4. Applying the Fundamental Theorem to each difference gives the desired result.

Because \((F-G)'=f-g\), the Fundamental Theorem gives \(\int_a^b(f-g)\,dx=[F-G]_a^b=[F(b)-F(a)]-[G(b)-G(a)]=\int_a^b f\,dx-\int_a^b g\,dx\).

- Rewrite the fraction and radical using negative and rational exponents. - Apply the power rule term by term. - Evaluate each endpoint carefully. - Use the Fundamental Theorem of Calculus.

1. Rewrite the integrand as \(3x^2-4x^{-2}+x^{1/2}\). 2. An antiderivative is \(F(x)=x^3+4x^{-1}+\frac{2}{3}x^{3/2}\). 3. Evaluate: \(F(4)=64+1+\frac{16}{3}=\frac{211}{3}\), and \(F(1)=1+4+\frac{2}{3}=\frac{17}{3}\). 4. Therefore, \(F(4)-F(1)=\frac{194}{3}\).

\(\frac{194}{3}\), or \(64\frac{2}{3}\)

- Simplify the rational expression before integrating. - Divide each term in the numerator by the denominator. - Rewrite the remaining reciprocal term with a negative exponent. - Use parentheses when subtracting the lower-limit value.

1. Divide each numerator term by \(2x^2\): \(\frac{4x^4-3x^2+2}{2x^2}=2x^2-\frac{3}{2}+x^{-2}\). 2. An antiderivative is \(F(x)=\frac{2}{3}x^3-\frac{3}{2}x-x^{-1}\). 3. Evaluate: \(F(2)=\frac{16}{3}-3-\frac{1}{2}=\frac{11}{6}\), and \(F(1)=\frac{2}{3}-\frac{3}{2}-1=-\frac{11}{6}\). 4. Therefore, \(F(2)-F(1)=\frac{11}{6}+\frac{11}{6}=\frac{11}{3}\).

\(\frac{11}{3}\), or \(3\frac{2}{3}\)

- Rewrite reciprocal powers using negative exponents. - Treat \(\sqrt{x}\) as \(x^{1/2}\). - Apply the power rule term by term. - Check signs carefully when subtracting a negative antiderivative value.

1. For a), rewrite the integrand as \(3x^{-4}\). An antiderivative is \(F(x)=-x^{-3}\). Thus, \(F(2)-F(1)=-\frac{1}{8}-(-1)=\frac{7}{8}\). 2. For b), an antiderivative is \(F(x)=\frac{1}{2}x^2+\frac{2}{3}x^{3/2}\). Thus, \(F(4)-F(1)=\frac{40}{3}-\frac{7}{6}=\frac{73}{6}\). 3. For c), an antiderivative is \(F(x)=\frac{1}{4}x^4-\frac{1}{2}x^2\). Thus, \(F(2)-F(0)=2\).

a) \(\frac{7}{8}\) b) \(\frac{73}{6}\) c) \(2\)

- Expand the product into a polynomial before integrating. - Apply the power rule term by term. - Check signs carefully when substituting a negative lower limit. - Use the Fundamental Theorem of Calculus.

1. Expand the integrand: \((x^2-1)(x+2)=x^3+2x^2-x-2\). 2. An antiderivative is \(F(x)=\frac{1}{4}x^4+\frac{2}{3}x^3-\frac{1}{2}x^2-2x\). 3. Evaluate: \(F(2)=\frac{10}{3}\), and \(F(-1)=\frac{13}{12}\). 4. Therefore, \(F(2)-F(-1)=\frac{40}{12}-\frac{13}{12}=\frac{9}{4}\).

\(\frac{9}{4}\), or \(2.25\)

- Rewrite radicals and reciprocals using rational and negative exponents. - Integrate each term separately. - Check the signs produced by the power rule for negative exponents. - Differentiate your antiderivative to verify it before using the limits.

1. Rewrite the integrand as \(\frac{3}{2}t^{-1/2}-2t^{-2}\). 2. An antiderivative is \(F(t)=3t^{1/2}+2t^{-1}=3\sqrt{t}+\frac{2}{t}\). 3. Evaluate: \(F(4)=6+\frac{1}{2}=\frac{13}{2}\), and \(F(1)=3+2=5\). 4. Therefore, \(F(4)-F(1)=\frac{3}{2}\).

\(\frac{3}{2}\), or \(1.5\)

- Rewrite the radical as a rational exponent. - Apply the power rule symbolically. - Remember that \(1\) raised to any real power is \(1\). - Use upper bound minus lower bound.

1. Rewrite the integrand: \(\sqrt[n]{x}=x^{1/n}\). 2. By the power rule, an antiderivative is \(F(x)=\frac{x^{1+1/n}}{1+1/n}=\frac{n}{n+1}x^{(n+1)/n}\). 3. Apply the bounds: \(\int_1^k\sqrt[n]{x}\,dx=\frac{n}{n+1}\left(k^{(n+1)/n}-1\right)\).

\(\int_1^k\sqrt[n]{x}\,dx=\frac{n}{n+1}\left(k^{(n+1)/n}-1\right)\)

- Because the function is positive on the interval, its definite integral equals the area. - Find an antiderivative of each term separately. - Evaluate the antiderivative at \(1\) and \(0\).

1. Since \(f(x)>0\) on \([0,1]\), the area is \(A=\int_0^1\left(e^x+3x^2\right)\,dx\). 2. By the sum rule, an antiderivative is \(F(x)=e^x+x^3\). 3. Apply the fundamental theorem of calculus: \(A=F(1)-F(0)=(e+1)-1=e\).

\(e\) square units, or approximately \(2.718\) square units

- Split the interval where the function definition changes. - Evaluate each piece using its own antiderivative. - Add the two definite integrals.

1. Split the integral where the formula changes: \(\int_{-1}^{0}e^{2x}\,dx+\int_0^2(1-x)\,dx\). 2. The first part is \([\frac{1}{2}e^{2x}]_{-1}^{0}=\frac{1}{2}-\frac{1}{2e^2}\). 3. The second part is \([x-\frac{1}{2}x^2]_0^2=0\). 4. Therefore, the total is \(\frac{1}{2}-\frac{1}{2e^2}\approx 0.432\).

\(\frac{1}{2}-\frac{1}{2e^2}\approx 0.432\)

- Recall the antiderivative of the reciprocal function. - Use absolute value inside the logarithm for negative limits. - Simplify logarithms of powers of \(e\). - Relate the sign of a definite integral to whether the graph lies above or below the x-axis.

1. An antiderivative on either interval is \(F(x)=2\ln|x|\). 2. For a), \(I_1=2\ln(e^2)-2\ln(1)=4\). 3. For b), \(I_2=2\ln|-1|-2\ln|-e|=0-2=-2\). 4. The function is odd because \(f(-x)=-f(x)\). Its graph lies above the x-axis for \(x>0\) and below the x-axis for \(x<0\). Therefore, the signed area in part a) is positive, while the signed area in part b) is negative.

a) \(I_1=4\) b) \(I_2=-2\) c) The graph is symmetric about the origin. It is above the x-axis on the interval for \(I_1\) and below the x-axis on the interval for \(I_2\), so the integrals have opposite signs.

- Recall the general antiderivative of \(\frac{1}{x}\). - Keep the absolute value when logarithms involve negative limits. - Use logarithm properties to simplify the results. - Recall the values of \(\ln(e)\) and \(\ln(e^k)\).

1. For a), an antiderivative is \(F(x)=5\ln|x|\). Thus, \(F(-e)-F(-e^2)=5\ln(e)-5\ln(e^2)=5-10=-5\). 2. For b), an antiderivative is \(G(x)=\frac{1}{2}\ln|x|\). Thus, \(G(2)-G(0.5)=\frac{1}{2}\ln(2)-\frac{1}{2}\ln\left(\frac{1}{2}\right)=\ln(2)\).

a) \(-5\) b) \(\ln(2)\)

- Split the fraction into simpler terms. - Recall which function has derivative \(\frac{1}{x}\). - Rewrite the second term with a negative exponent. - Use logarithm properties to combine the final expression.

1. Split and simplify the integrand: \(\frac{x-2}{x^2}=\frac{1}{x}-2x^{-2}\). 2. An antiderivative is \(F(x)=\ln(x)+\frac{2}{x}\) on \([2,4]\). 3. Evaluate: \(F(4)=\ln(4)+\frac{1}{2}\), and \(F(2)=\ln(2)+1\). 4. Therefore, \(F(4)-F(2)=\ln(4)-\ln(2)-\frac{1}{2}=\ln(2)-\frac{1}{2}\).

\(\ln(2)-\frac{1}{2}\)

- Rewrite the numerator so it contains a multiple of the denominator. - Split the expression into a constant and a simple reciprocal term. - Recall the antiderivative of \(\frac{1}{x+k}\). - Use parentheses when subtracting the lower-limit value.

1. Rewrite the numerator: \(2x+3=2(x+2)-1\). Therefore, \(\frac{2x+3}{x+2}=2-\frac{1}{x+2}\). 2. An antiderivative is \(F(x)=2x-\ln(x+2)\) on \([0,1]\). 3. Evaluate: \(F(1)=2-\ln(3)\), and \(F(0)=-\ln(2)\). 4. Thus, \(F(1)-F(0)=2-\ln(3)+\ln(2)=2+\ln\left(\frac{2}{3}\right)\).

\(2+\ln\left(\frac{2}{3}\right)\)

- Expand the squared binomial before integrating. - Use exponent properties to simplify \(e^xe^{-x}\). - Find an antiderivative of each resulting term. - Evaluate carefully at \(x=0\), where exponential terms equal \(1\).

1. Expand the square: \((e^x+e^{-x})^2=e^{2x}+2e^xe^{-x}+e^{-2x}=e^{2x}+2+e^{-2x}\). 2. An antiderivative is \(F(x)=\frac{1}{2}e^{2x}+2x-\frac{1}{2}e^{-2x}\). 3. Evaluate: \(F(1)=\frac{1}{2}e^2+2-\frac{1}{2}e^{-2}\), and \(F(0)=\frac{1}{2}-\frac{1}{2}=0\). 4. Therefore, the integral is \(\frac{1}{2}e^2+2-\frac{1}{2}e^{-2}\).

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

- What sign must the derivative have for a function to be strictly increasing? - Multiply the two binomials and use a difference-of-squares identity. - Notice that \(c\) is an antiderivative of \(s\). - Use \(e^{\ln(a)}=a\) and \(e^{-\ln(a)}=\frac{1}{a}\).

1. \(s'(x)=\frac{1}{2}(e^x+e^{-x})\). Both exponential terms are positive for every real \(x\), so \(s'(x)>0\). Therefore, \(s\) is strictly increasing. 2. \(2s(x)c(x)=2\left[\frac{1}{2}(e^x-e^{-x})\right]\left[\frac{1}{2}(e^x+e^{-x})\right]\). Using the difference of squares, this becomes \(\frac{1}{2}(e^{2x}-e^{-2x})=s(2x)\). 3. Since \(s(x)\geq0\) on \([0,\ln(2)]\), the area is \(A=\int_0^{\ln(2)}s(x)\,dx\). An antiderivative of \(s\) is \(c\), so \(A=c(\ln(2))-c(0)\). 4. \(c(\ln(2))=\frac{1}{2}\left(2+\frac{1}{2}\right)=\frac{5}{4}\), and \(c(0)=1\). Therefore, \(A=\frac{5}{4}-1=\frac{1}{4}\).

1. \(s'(x)=\frac{1}{2}(e^x+e^{-x})>0\), so \(s\) is strictly increasing. 2. \(2s(x)c(x)=\frac{1}{2}(e^{2x}-e^{-2x})=s(2x)\) 3. \(A=\frac{1}{4}\) square units

- Use \(F'=f\) to connect zeros of \(f\) with critical points of \(F\). - The sign of \(f\) tells whether \(F\) increases or decreases. - Compare those features with each candidate curve. - Use the Fundamental Theorem of Calculus for the definite integral.

1. Because \(F'=f\), the zeros of \(f\) at \(x=0\) and \(x=2\) must be critical points of \(F\). 2. The graph of \(f\) is positive for \(x<0\), negative for \(0<x<2\), and positive for \(x>2\). Therefore, \(F\) increases, then decreases, then increases. 3. Thus, \(F\) has a local maximum at \(x=0\) and a local minimum at \(x=2\). Only curve g has this behavior. 4. By the Fundamental Theorem of Calculus, \(\int_0^2 f(x)\,dx=F(2)-F(0)\). 5. \(F(2)=\frac{1}{4}(8)-\frac{3}{4}(4)+1=0\), and \(F(0)=1\). 6. Therefore, \(\int_0^2 f(x)\,dx=0-1=-1\).

a) Curve g b) \(-1\)

- A local maximum occurs when the derivative changes from positive to negative. - Use the graph of \(f\) to locate that sign change. - Relate the given signed area to \(F_1(0)-F_1(-2)\). - Use the known value \(F_1(0)=3\).

1. A local maximum of \(F\) occurs where \(f=F'\) changes from positive to negative. The graph shows this sign change at \(x=1\). 2. By the Fundamental Theorem of Calculus, \(\int_{-2}^{0}f(x)\,dx=F_1(0)-F_1(-2)\). 3. The signed area on \([-2,0]\) is \(4\). Therefore, \(4=3-F_1(-2)\). 4. Solving gives \(F_1(-2)=-1\).

a) \(x=1\) b) \(F_1(-2)=-1\)

- Relate the definite integral of \(f\) to values of its antiderivative \(F\). - Identify the lower and upper limits in the integral. - Read \(F(0)\) and \(F(2)\) from the red graph. - Compute \(F(2)-F(0)\), paying attention to the sign.

1. By the Fundamental Theorem of Calculus, \(\int_{0}^{2}f(x)\,dx=F(2)-F(0)\). 2. From the red graph, \(F(2)=0\) and \(F(0)=2\). 3. Therefore, \(F(2)-F(0)=0-2=-2\).

\(-2\)

- What cumulative probability occurs at the mean of a normal distribution? - What cumulative probability occurs at \(\mu+\sigma\)? - How are a density function and its cumulative distribution function related? - What does a maximum derivative value mean for the original graph?

1. a) A normal cumulative distribution function satisfies \(F(\mu)=0.5\). Also, \(F(\mu+\sigma)\approx 0.8413\). From the graph, \(F_1(1)=0.5\) and \(F_1(2)\approx 0.8413\), so \(\mu_1=1\) and \(\sigma_1=1\). Also, \(F_2(4)=0.5\) and \(F_2(6)\approx 0.8413\), so \(\mu_2=4\) and \(\sigma_2=2\). 2. b) The maximum of a normal density is \(\frac{1}{\sigma\sqrt{2\pi}}\). Thus, the first maximum is \(\frac{1}{\sqrt{2\pi}}\approx 0.40\), and the second is \(\frac{1}{2\sqrt{2\pi}}\approx 0.20\). 3. c) Since \(F'(x)=f(x)\), the maximum density value is the maximum slope of the cumulative distribution graph. It is the slope of the tangent at the inflection point \((\mu, 0.5)\).

a) \(F_1\): \(\mu_1=1\), \(\sigma_1=1\); \(F_2\): \(\mu_2=4\), \(\sigma_2=2\) b) First density: approximately \(0.40\); second density: approximately \(0.20\) c) Each value is the maximum slope of the corresponding cumulative distribution function at \((\mu, 0.5)\).

- Use the Fundamental Theorem of Calculus with the graphed antiderivative. - Read \(F(0)\) and \(F(3)\) from the graph. - Find the change in \(F\) between the two endpoints.

1. By the Fundamental Theorem of Calculus, \(\int_{0}^{3}f(x)\,dx=F(3)-F(0)\). 2. From the graph, \(F(3)=4\) and \(F(0)=4\). 3. Therefore, \(F(3)-F(0)=4-4=0\).

\(0\)

- Apply the Fundamental Theorem of Calculus in general form. - Identify an antiderivative of \(h'\). - Treat \(a\) and \(h(a)\) as constants. - Compare the roles of antiderivatives and derivatives.

1. Since \(h\) is an antiderivative of \(h'\), \(I(x)=h(x)-h(a)\). 2. Because \(h(a)\) is constant with respect to \(x\), \(\frac{dI}{dx}=h'(x)\). 3. For a), integrating \(h'\) gives \(h(x)\) plus the constant \(-h(a)\). For b), differentiating the accumulation function gives its integrand \(h'(x)\).

1. \(I(x)=h(x)-h(a)\) 2. \(\frac{dI}{dx}=h'(x)\) 3. Integration recovers \(h\) up to a constant, and differentiation of \(I\) recovers \(h'\).

- Compare the numerator and denominator. - Rewrite the denominator as a perfect square. - Split the rational expression into simpler terms. - Express \(\frac{k}{(x+a)^2}\) using a negative exponent before integrating.

1. Recognize that \(x^2+2x+1=(x+1)^2\), and rewrite the numerator as \(x^2+2x+5=(x+1)^2+4\). 2. Then the integrand becomes \(1+\frac{4}{(x+1)^2}=1+4(x+1)^{-2}\). 3. An antiderivative is \(F(x)=x-\frac{4}{x+1}\). 4. Evaluate: \(F(2)=2-\frac{4}{3}=\frac{2}{3}\), and \(F(0)=-4\). 5. Therefore, \(F(2)-F(0)=\frac{2}{3}+4=\frac{14}{3}\).

\(\frac{14}{3}\)