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AP Calculus AB : Techniques of antidifferentiation

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Example Questions

Example Question #2 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{27}^{31}(\frac{(8x)}{51})dx\end{align*}$

Possible Answers:

7.00

100.08

18.20

2.19

Correct answer:

18.20

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{27}^{31}(\frac{(8x)}{51})dx=\frac{(4x^{2})}{51}|_{27}^{31}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{27}^{31}(\frac{(8x)}{51})dx=\frac{(4(31)^{2})}{51}-(\frac{(4(27)^{2})}{51})\\&\int_{27}^{31}(\frac{(8x)}{51})dx=18.20\end{align*}$

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Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx\end{align*}$

Possible Answers:

0.64

0.34

0.24

1.99

Correct answer:

1.99

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=\frac{(40\cdot 2^{(\frac{x}{4})})}{(7ln(2))}|_{-6}^{-3}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=\frac{(40\cdot 2^{(\frac{(-3)}{4})})}{(7ln(2))}-(\frac{(40\cdot 2^{(\frac{(-6)}{4})})}{(7ln(2))})\\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=1.99\end{align*}$

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Example Question #3 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx\end{align*}$

Possible Answers:

-10.41

-0.16

-5.21

-1.53

Correct answer:

-1.53

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=\frac{(59ln(x))}{9}|_{-24}^{-19}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=\frac{(59ln((-19)))}{9}-(\frac{(59ln((-24)))}{9})\\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=-1.53\end{align*}$

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Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-6}^{-2}(\frac{11}{59})dx\end{align*}$

Possible Answers:

0.75

6.71

0.11

0.20

Correct answer:

0.75

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{-6}^{-2}(\frac{11}{59})dx=\frac{(11x)}{59}|_{-6}^{-2}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-6}^{-2}(\frac{11}{59})dx=\frac{(11(-2))}{59}-(\frac{(11(-6))}{59})\\&\int_{-6}^{-2}(\frac{11}{59})dx=0.75\end{align*}$

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Example Question #2 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx\end{align*}$

Possible Answers:

1.25

0.15

3.50

8.25

Correct answer:

1.25

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=\frac{(9sin(x + 2))}{11}|_{-16}^{-14}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=\frac{(9sin((-14) + 2))}{11}-(\frac{(9sin((-16) + 2))}{11})\\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=1.25\end{align*}$

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Example Question #11 : Antiderivatives Following Directly From Derivatives Of Basic Functions

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-6}^{-2.5}(10e^{(x)})dx\end{align*}$

Possible Answers:

0.23

0.80

6.61

0.14

Correct answer:

0.80

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{-6}^{-2.5}(10e^{(x)})dx=10e^{(x)}|_{-6}^{-2.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-6}^{-2.5}(10e^{(x)})dx=10e^{((-2.5))}-(10e^{((-6))})\\&\int_{-6}^{-2.5}(10e^{(x)})dx=0.80\end{align*}$

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Example Question #81 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx\end{align*}$

Possible Answers:

55.77

340.17

223.07

540.93

Correct answer:

55.77

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx=\frac{(3\cdot 3^{(\frac{x}{3})})}{(4ln(3))}|_{7.5}^{12.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx=\frac{(3\cdot 3^{(\frac{(12.5)}{3})})}{(4ln(3))}-(\frac{(3\cdot 3^{(\frac{(7.5)}{3})})}{(4ln(3))})\\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx=55.77\end{align*}$

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Example Question #83 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx\end{align*}$

Possible Answers:

0.24

3.00

1.43

0.16

Correct answer:

1.43

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx=\frac{(3\cdot 2^{(\frac{x}{3})})}{(20ln(2))}|_{14}^{15}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx=\frac{(3\cdot 2^{(\frac{(15)}{3})})}{(20ln(2))}-(\frac{(3\cdot 2^{(\frac{(14)}{3})})}{(20ln(2))})\\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx=1.43\end{align*}$

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Example Question #84 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx\end{align*}$

Possible Answers:

-20.52

-5.64

-2.56

-0.42

Correct answer:

-2.56

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx=\frac{(20sin(3x))}{9}|_{9}^{12.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx=\frac{(20sin(3(12.5)))}{9}-(\frac{(20sin(3(9)))}{9})\\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx=-2.56\end{align*}$

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Example Question #85 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{10}^{15}(\frac{51}{(2x)})dx\end{align*}$

Possible Answers:

55.83

36.19

10.34

1.26

Correct answer:

10.34

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{10}^{15}(\frac{51}{(2x)})dx=\frac{(51ln(x))}{2}|_{10}^{15}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{10}^{15}(\frac{51}{(2x)})dx=\frac{(51ln((15)))}{2}-(\frac{(51ln((10)))}{2})\\&\int_{10}^{15}(\frac{51}{(2x)})dx=10.34\end{align*}$

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AP Calculus AB : Techniques of antidifferentiation

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #2 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #3 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #1 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #2 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #11 : Antiderivatives Following Directly From Derivatives Of Basic Functions

Example Question #81 : Integrals

Example Question #83 : Integrals

Example Question #84 : Integrals

Example Question #85 : Integrals

All AP Calculus AB Resources

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