- The given function \(F\) is connected to the integrand through differentiation. - For a definite integral, compare the antiderivative values at the two endpoints in the correct order.

1. By the Fundamental Theorem of Calculus, \(\int_1^5 f(x)\,\text{d}x=F(5)-F(1)\). 2. Therefore \(\int_1^5 f(x)\,\text{d}x=11-3=8\).

\(8\)

- Connect the integral to the change in an antiderivative. - Which endpoint value is known, and which is missing? - Use the given change to find the missing value.

1. By the Fundamental Theorem of Calculus, \(F(7)-F(2)=\int_2^7f(x)\,\text{d}x=-3\). 2. Thus \(F(7)=5-3=2\).

\(F(7)=2\)

- Check how the Fundamental Theorem orders the endpoint values. - Is a negative result reasonable for this positive integrand?

1. The student subtracts the antiderivative endpoint values in the wrong order. The Fundamental Theorem gives \(\int_1^2g(x)\,\text{d}x=G(2)-G(1)\). 2. Evaluate \(2^3-1^3=8-1=7\), which is also consistent with \(g(x)>0\) on this interval.

The subtraction is reversed. The correct value is \(G(2)-G(1)=7\).

- The Fundamental Theorem does not require you to write an elementary antiderivative first. - Which version of the theorem connects a changing integral with its derivative?

1. The integrand \(e^{-t^2}\) is continuous. 2. By the Fundamental Theorem of Calculus, differentiating this variable-upper-limit integral gives \(H'(x)=e^{-x^2}\).

\(H'(x)=e^{-x^2}\)

- 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) \(-2.25\)

- 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 \(\int_{0}^{4}f(x)\,\text{d}x=2\). On \([4, 8]\), it is below the axis and contributes signed area \(\int_{4}^{8}f(x)\,\text{d}x=-2\). 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.

- 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, \(\int_a^b f(x)\,\text{d}x=F(b)-F(a)\). 2. Reversing the limits gives \(\int_b^a f(x)\,\text{d}x=F(a)-F(b)\). 3. Therefore, \(-\int_b^a f(x)\,\text{d}x=-[F(a)-F(b)]=F(b)-F(a)\), which equals \(\int_a^b f(x)\,\text{d}x\).

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

- 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 \(3.5\)

- 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, \(F\left(\frac{\pi}{2}\right)=-3\cdot 0-2\cdot 1=-2\). 3. At the lower limit, \(F(0)=-3\cdot 1-2\cdot 0=-3\). 4. Therefore, \(F\left(\frac{\pi}{2}\right)-F(0)=-2-(-3)=1\).

\(1\)

- 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) \(-2\)

- 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: \(\int_{1}^{e}g(x)\,\text{d}x=G(e)-G(1)\). 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) \(8-\frac{6}{e}\)

- 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 \(0.375\)

- 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) \(3\)

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

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

- 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 \(64\frac{2}{3}\)

- 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) \(0\)

- 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) \(2\)

- 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 \(1.5\)

- 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, \(F(2)-F(-1)=4-1=3\).

\(3\)

- Split the integral into the sum of two simpler integrals. - Find one antiderivative by integrating each term separately. - Evaluate at \(1\) and \(0\). - Check the sign of the integrand before interpreting the result as geometric area.

1. By the sum rule, \(\int_0^1(e^x+3x^2)\,\text{d}x=\int_0^1e^x\,\text{d}x+\int_0^1 3x^2\,\text{d}x\). 2. An antiderivative of the full integrand is \(F(x)=e^x+x^3\). Therefore, the integral is \(F(1)-F(0)=(e+1)-1=e\). 3. Since \(e^x+3x^2>0\) throughout \([0, 1]\), the definite integral has no negative signed-area contribution. Thus, its value is also the geometric area between the graph and the x-axis.

The integral equals \(e\). Because the integrand is positive on \([0, 1]\), the corresponding geometric area is also \(e\) square units.

- 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, \(F(3)-F(1)=12\).

\(12\)

- 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, \(\int_{0}^{2}f(x)\,\text{d}x=F(2)-F(0)\). 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, \(F(e^2)-F(1)=10-e^2-(-1)=11-e^2\).

\(11-e^2\)

- 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) \(\ln(2)\)

- 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, \(F(4)-F(0)=2e+6\).

\(2e+6\)

- Relate the definite integral of \(f\) to values of its antiderivative \(F\). - Read \(F(0)\) and \(F(2)\) from the curve labeled \(F\). - Compute the upper-endpoint value minus the lower-endpoint value.

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

\(-2\)

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

1. By the Fundamental Theorem of Calculus, \(F(4)-F(1)=\int_1^4 f(x)\,\text{d}x\). 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 \(\frac{1}{2}\cdot 2\cdot 2=2\). 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 \(-\frac{1}{2}\). 4. Therefore, \(F(4)-F(1)=2-\frac{1}{2}=\frac{3}{2}\).

\(F(4)-F(1)=\frac{3}{2}\)

- Use \(F(3)-F(0)=\int_0^3 f(x)\,\text{d}x\). - 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, \(F(3)=F(0)+\int_0^3 f(x)\,\text{d}x\). 2. The graph of \(f\) is above the x-axis on \([0, 3)\) and equals \(0\) at \(x=3\). Therefore, \(\int_0^3 f(x)\,\text{d}x>0\). 3. Thus, \(F(3)=1+\int_0^3 f(x)\,\text{d}x>1\).

\(F(3)>1\)

- Find an antiderivative for each integrand. - Apply the Fundamental Theorem of Calculus as endpoint subtraction. - Keep track of the sign when substituting a negative lower bound.

1. For a), an antiderivative is \(F(x)=0.5x\). Thus \(F(6)-F(-4)=3-(-2)=5\). 2. For b), an antiderivative is \(F(x)=3x-\frac{1}{2}x^2\). Thus \(F(4)-F(0)=12-8=4\). 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, \(\int_0^3 f(x)\,\text{d}x=g(3)-g(0)\). 2. From the graph, \(g(3)=4\) and \(g(0)=1\). 3. Therefore, \(\int_0^3 f(x)\,\text{d}x=4-1=3\).

\(3\)

- The signed area under a derivative graph gives the difference between function values. - Consider which side of the x-axis the region lies on. - Use the signed change to work backward to the earlier value.

1. The region between \(f'\) and the x-axis from \(0\) to \(2\) is a trapezoid with heights \(1\) and \(3\), and width \(2\). It lies below the x-axis. 2. Its signed area is \(\int_0^2 f'(x)\,\text{d}x=-\frac{1+3}{2}\cdot2=-4\). 3. Since \(f(2)-f(0)=-4\) and \(f(2)=6\), the earlier value is \(f(0)=10\).

\(f(0)=10\)

- 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: \(I(x)=-\int_{10}^{x}\cos(z^2)\,\text{d}z\). 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. - Rewrite the negative exponent using a reciprocal before evaluating. - Pay close attention to both the negative base and the negative exponent. - After evaluating the antiderivative, compare its values at the two bounds in the correct order.

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. Ximena used \((-2)^{-2}=4\), but \((-2)^{-2}=\frac{1}{(-2)^2}=\frac14\). 3. Correcting the endpoint values gives \(F(-1)=-2\cdot1=-2\) and \(F(-2)=-2\cdot\frac14=-\frac12\). 4. Therefore, \(F(-1)-F(-2)=-2-\left(-\frac12\right)=-\frac32\).

The first error occurs in step 4 because \((-2)^{-2}=\frac14\), not \(4\). The corrected value is \(-\frac32\), 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: \(F'(x)=\frac{1}{x^2+4}\cdot 2x=\frac{2x}{x^2+4}\). Therefore, \(F\) is an antiderivative of the integrand. 2. By the Fundamental Theorem of Calculus, \(\int_{0}^{2}\frac{2x}{x^2+4}\,\text{d}x=F(2)-F(0)\). 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: \(\int_{0}^{1}f(x)\,\text{d}x=F(1)-F(0)\). 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: \(\int_a^b [f(x)+g(x)]\,\text{d}x=[F(x)+G(x)]_a^b=F(b)+G(b)-F(a)-G(a)\). 3. Apply the theorem to the right side: \(\int_a^b f(x)\,\text{d}x+\int_a^b g(x)\,\text{d}x=[F(b)-F(a)]+[G(b)-G(a)]\). 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 \(e-\frac{1}{e}\).

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

- 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: \(\int_{1}^{e}f(x)\,\text{d}x=F(e)-F(1)\). 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: \(F'(x)=\frac{1}{2}\cdot 2\ln(x)\cdot \frac{1}{x}=\frac{\ln(x)}{x}=f(x)\). Therefore, \(F\) is an antiderivative of \(f\). 2. For b), \(\int_{1}^{e^2}f(x)\,\text{d}x=F(e^2)-F(1)\). 3. Evaluate: \(F(e^2)=\frac{1}{2}(\ln(e^2))^2=\frac{1}{2}\cdot 2^2=2\), 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) \(2\)

- 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: \(\int_0^2\frac{k}{x+2}\,\text{d}x=k(\ln(4)-\ln(2))=k\ln(2)\). 3. Since \(\ln(16)=\ln(2^4)=4\ln(2)\), \(k\ln(2)=4\ln(2)\), so \(k=4\).

\(k=4\)

- 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, \(\int_{0}^{3}f(x)\,\text{d}x=\frac{3}{\pi}\left(\sin(\pi)-\sin(0)\right)=0\). 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) \(\int_{0}^{3}f(x)\,\text{d}x=0\). 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, \(\int_{0}^{4}g(x)\,\text{d}x=G(4)-G(0)=\left(2e^2-4e\right)-2=2e^2-4e-2\approx 1.90\). 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) \(\int_{0}^{4}g(x)\,\text{d}x=2e^2-4e-2\approx 1.90\). This is net signed area, not total geometric area, because the function changes sign at \(x=2\).

- Equal integration limits give an integral of \(0\). - First apply the Fundamental Theorem at the upper limit \(x^2\). - Then multiply by the derivative of the upper-limit function. - The exponential factor is always positive, so use the remaining factor to determine the sign of the derivative.

1. When \(x=0\), the upper and lower limits are equal, so \(H(0)=\int_0^0 e^{-t^2}\,\text{d}t=0\). 2. Let \(u=x^2\). The Fundamental Theorem of Calculus gives the derivative with respect to the upper limit as \(e^{-u^2}\). Multiplying by \(u'=2x\) gives \(H'(x)=2xe^{-x^4}\). 3. Since \(e^{-x^4}>0\) for every real \(x\), the sign of \(H'(x)\) is the sign of \(x\). Thus, \(H\) decreases on \((-\infty, 0)\) and increases on \((0, \infty)\). Therefore, \(H\) has an absolute minimum at \(x=0\), with minimum value \(0\).

1. \(H(0)=0\) 2. \(H'(x)=2xe^{-x^4}\) 3. \(H\) decreases on \((-\infty, 0)\), increases on \((0, \infty)\), and has an absolute minimum of \(0\) at \(x=0\).

- Differentiate the upper-limit contribution and subtract the lower-limit contribution. - Multiply each integrand value by the derivative of its limit. - For the explicit formula, evaluate one antiderivative at \(2x\) and at \(x\). - Factor the formula before solving for its zeros.

1. Differentiate the contribution from each variable limit. The upper limit contributes \(((2x)^2+1)\cdot 2\), and the lower limit contributes \(-(x^2+1)\cdot 1\). Thus, \(Q'(x)=2(4x^2+1)-(x^2+1)=7x^2+1\). 2. An antiderivative of \(t^2+1\) is \(\frac{1}{3}t^3+t\). Therefore, \(Q(x)=\left[\frac{1}{3}t^3+t\right]_x^{2x}=\frac{8}{3}x^3+2x-\frac{1}{3}x^3-x=\frac{7}{3}x^3+x\). Differentiating gives \(Q'(x)=7x^2+1\), which matches part 1. 3. Factor: \(Q(x)=x\left(\frac{7}{3}x^2+1\right)\). The second factor is positive for every real \(x\), so the only zero is \(x=0\).

1. \(Q'(x)=7x^2+1\) 2. \(Q(x)=\frac{7}{3}x^3+x\) 3. The only zero is \(x=0\).

- Distinguish values of the integrand from values of an antiderivative. - Simplify the integrand before integrating. - Apply the power rule term by term. - After finding an antiderivative, compare its endpoint values as required by the Fundamental Theorem.

1. The values of the integrand at the endpoints do not determine a definite integral. The integral depends on the accumulated values of the function throughout the interval. 2. Expand the integrand: \((x+1)(x-2)^2=(x+1)(x^2-4x+4)=x^3-3x^2+4\). 3. An antiderivative is \(F(x)=\frac14x^4-x^3+4x\). 4. Evaluate: \(F(2)=4-8+8=4\) and \(F(-1)=\frac14+1-4=-\frac{11}{4}\). Therefore, \(F(2)-F(-1)=4+\frac{11}{4}=\frac{27}{4}\).

The reasoning is incorrect because endpoint values of the integrand do not determine the accumulated integral. The integral equals \(\frac{27}{4}\).

- 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 \(k=2\).

\(k=2\)

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

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

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

- 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 \(3\frac{2}{3}\)

- 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, \(F(2)-F(0)=\frac{2}{3}+4=\frac{14}{3}\).

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

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

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

- 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: \(\int_{-1}^{0}e^{2x}\,\text{d}x+\int_0^2(1-x)\,\text{d}x\). 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 \(\frac{1}{2}-\frac{1}{2e^2}\approx 0.432\).

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

- 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 \(a=4\).

\(a=4\)

- 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.

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

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

- Apply the Fundamental Theorem of Calculus and then the chain rule to the upper limit. - An integral with identical lower and upper limits equals zero. - Check the sign of \((3x-2)^2+1\) for all real \(x\). - A strictly increasing function can cross the x-axis at most once.

1. The upper limit is the composite function \(3x-2\), so the chain rule contributes its derivative, \(3\). Therefore, \(H'(x)=3[(3x-2)^2+1]\). 2. When \(x=1\), the upper limit is \(1\), so \(H(1)=\int_1^1(t^2+1)\,\text{d}t=0\). Also, \(H'(1)=3(1^2+1)=6\). 3. Because \((3x-2)^2+1>0\) for every real \(x\), \(H'(x)>0\) for every real \(x\). Thus, \(H\) is strictly increasing. Since \(H(1)=0\), a strictly increasing function cannot have another zero.

a) The missing factor is the derivative of the upper limit: \(H'(x)=3[(3x-2)^2+1]\). b) \(H(1)=0\) and \(H'(1)=6\) c) \(H'(x)>0\) for every \(x\), so \(H\) is strictly increasing and \(x=1\) is its only zero.

- Zeros of \(f\) correspond to horizontal tangents of an antiderivative. - Use the sign of \(f\) to decide where an antiderivative must increase or decrease. - After choosing the curve, read its endpoint values from the coordinate grid. - The definite integral is the change in the antiderivative between the two endpoints.

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 \(x>2\). Thus \(F\) must increase, then decrease, then increase. 3. Only curve g has a local maximum at \(x=0\) and a local minimum at \(x=2\), so g is the possible antiderivative. 4. From curve g, \(F(0)=1\) and \(F(2)=0\). 5. By the Fundamental Theorem of Calculus, \(\int_0^2 f(x)\,\text{d}x=F(2)-F(0)=0-1=-1\).

a) Curve g b) \(F(0)=1\), \(F(2)=0\), and \(\int_0^2 f(x)\,\text{d}x=-1\).

- 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, \(\int_{-2}^{0}f(x)\,\text{d}x=F_1(0)-F_1(-2)\). 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) \(F_1(-2)=-1\)

- Apply the Fundamental Theorem of Calculus to the variable upper limit. - Include the derivative of \(2x\) from the chain rule. - Read \(f(0)\) and \(f(2)\) from the graph. - Use the stated positivity of \(f\), not just the visible part of its graph, to explain the result for all real numbers.

1. The Fundamental Theorem of Calculus gives the integrand evaluated at the upper limit, and the chain rule contributes the derivative of \(2x\). Thus, \(H'(x)=2f(2x)\). 2. From the graph, \(f(0)=1\), so \(H'(0)=2f(0)=2\). Also, \(f(2)=5\), so \(H'(1)=2f(2)=10\). 3. The problem states that \(f\) is positive for every real input. Therefore, \(H'(x)=2f(2x)>0\) for every real \(x\), so \(H\) is strictly increasing on \(\mathbb{R}\).

a) \(H'(x)=2f(2x)\) b) \(H'(0)=2\) and \(H'(1)=10\) c) Since \(f(2x)>0\) for every real \(x\) by the given assumption, \(H'(x)>0\); hence \(H\) is strictly increasing.

- What does the slope of the derivative graph tell you about the concavity of the original function? - Areas under the derivative graph give changes in the original function. - Use the derivative's sign to locate a maximum; use its direction of change to locate an inflection point.

1. The derivative increases until \(x=1\) and decreases afterward. Therefore \(f\) changes from concave up to concave down at \(x=1\). 2. The region under \(f'\) from \(0\) to \(3\) is a triangle with base \(3\) and height \(2\). Thus \(f(3)=f(0)+\frac{1}{2}\cdot3\cdot2=3\). 3. The area from \(0\) to \(1\) is \(1\), so the inflection point is \((1,1)\). Since \(f'>0\) for \(0<x<3\) and \(f'<0\) for \(3<x<6\), \(f\) increases, then decreases, with an absolute maximum at \((3,3)\). It is concave up on \((0,1)\) and concave down on \((1,6)\).

a) \(x=1\), where \(f'\) changes from increasing to decreasing. b) \(f(3)=3\) c) Increasing on \((0,3)\), decreasing on \((3,6)\); concave up on \((0,1)\), concave down on \((1,6)\). Inflection point: \((1,1)\). Absolute maximum: \((3,3)\).

- Use \(\int_a^b f(x)\,\text{d}x=F(b)-F(a)\). - The sign of \(F'\) is determined by whether \(F\) is increasing or decreasing. - A zero definite integral represents zero net change, not necessarily a zero integrand.

1. By the Fundamental Theorem of Calculus, \(\int_{0}^{3}f(x)\,\text{d}x=F(3)-F(0)\). The graph shows \(F(3)=4\) and \(F(0)=4\), so the integral is \(0\). 2. Since \(f=F'\), \(f\) is negative where \(F\) decreases and positive where \(F\) increases. The graph decreases on \((0, 2)\) and increases on \((2, 3)\). Therefore, \(f<0\) on \((0, 2)\) and \(f>0\) on \((2, 3)\). 3. The decrease in \(F\) from \(x=0\) to \(x=2\) is exactly balanced by its increase from \(x=2\) to \(x=3\). Equivalently, the negative and positive signed contributions of \(f\) cancel.

a) \(0\) b) \(f<0\) on \((0, 2)\), and \(f>0\) on \((2, 3)\). c) The negative and positive signed contributions cancel, so the net integral is \(0\).

- For the first method, read \(g(-2)\) and \(g(2)\) from the graph. - For the second method, compare the portions of the graph of \(f\) on opposite sides of the origin. - At matching x-values on opposite sides of the origin, how do the heights of \(f\) compare?

1. By the Fundamental Theorem of Calculus, \(\int_{-2}^{2}f(x)\,\text{d}x=g(2)-g(-2)\). The graph shows \(g(2)=0\) and \(g(-2)=0\), so the integral is \(0\). 2. The graph of \(f\) has origin symmetry, so \(f\) is odd. The signed areas on \([-2, 0]\) and \([0, 2]\) have equal magnitudes and opposite signs. Therefore, the integral is also \(0\) by symmetry.

a) \(g(2)-g(-2)=0-0=0\) b) Since \(f\) is odd and the interval is symmetric about \(0\), the integral is \(0\).

- View the expression as an accumulation with a moving endpoint. - The endpoint is itself a function of \(x\). - Account for both the value of the length density and the rate at which the endpoint moves.

1. The integrand evaluated at the upper limit is \(\sqrt{1+[f'(x^2)]^2}\). 2. The upper limit \(x^2\) has derivative \(2x\). 3. By the Fundamental Theorem of Calculus and the chain rule, \(S'(x)=2x\sqrt{1+[f'(x^2)]^2}\). The factor \(2x\) accounts for how quickly the horizontal endpoint \(x^2\) moves.

\(S'(x)=2x\sqrt{1+[f'(x^2)]^2}\). The factor \(2x\) is the rate at which the upper endpoint moves.

- 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. a) Integrating \(h'\) recovers \(h\) up to a constant. b) Differentiating \(I\) recovers its integrand \(h'\).