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Calculus 2 : Definite Integrals

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #562 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-25}^{-20.5}(\frac{4}{(3x)})dx\end{align*}$

Possible Answers:

-0.03

-0.05

-0.26

-0.53

Correct answer:

-0.26

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int_{-25}^{-20.5}(\frac{4}{(3x)})dx=\frac{(4ln(x))}{3}|_{-25}^{\frac{-41}{2}}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-25}^{-20.5}(\frac{4}{(3x)})dx=\frac{(4ln(-\frac{41}{2}))}{3}-(\frac{(4ln(-25))}{3})\\&\int_{-25}^{-20.5}(\frac{4}{(3x)})dx=-0.26\end{align*}$

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Example Question #212 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-8}^{-6}(\frac{(43e^{(x)})}{5})dx\end{align*}$

Possible Answers:

0.02

0.15

0.12

0.05

Correct answer:

0.02

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{-8}^{-6}(\frac{(43e^{(x)})}{5})dx=\frac{(43e^{(x)})}{5}|_{-8}^{-6}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-8}^{-6}(\frac{(43e^{(x)})}{5})dx=\frac{(43e^{(-6)})}{5}-(\frac{(43e^{(-8)})}{5})\\&\int_{-8}^{-6}(\frac{(43e^{(x)})}{5})dx=0.02\end{align*}$

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Example Question #2311 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{18}^{22}(\frac{19}{(6x)})dx\end{align*}$

Possible Answers:

4.45

1.78

0.06

0.64

Correct answer:

0.64

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int_{18}^{22}(\frac{19}{(6x)})dx=\frac{(19ln(x))}{6}|_{18}^{22}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{18}^{22}(\frac{19}{(6x)})dx=\frac{(19ln(22))}{6}-(\frac{(19ln(18))}{6})\\&\int_{18}^{22}(\frac{19}{(6x)})dx=0.64\end{align*}$

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Example Question #2312 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-3.5}^{0.5}(\frac{(5e^{(-x)})}{9})dx\end{align*}$

Possible Answers:

45.15

2.05

18.06

2.96

Correct answer:

18.06

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{-3.5}^{0.5}(\frac{(5e^{(-x)})}{9})dx=-\frac{(5e^{(-x)})}{9}|_{-3.5}^{0.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-3.5}^{0.5}(\frac{(5e^{(-x)})}{9})dx=-\frac{(5e^{(-0.5)})}{9}-(-\frac{(5e^{(--3.5)})}{9})\\&\int_{-3.5}^{0.5}(\frac{(5e^{(-x)})}{9})dx=18.06\end{align*}$

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Example Question #201 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-26}^{-21}(\frac{25}{(9x)})dx\end{align*}$

Possible Answers:

-1.36

-0.59

-5.58

-0.10

Correct answer:

-0.59

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int_{-26}^{-21}(\frac{25}{(9x)})dx=\frac{(25ln(x))}{9}|_{-26}^{-21}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-26}^{-21}(\frac{25}{(9x)})dx=\frac{(25ln(-21))}{9}-(\frac{(25ln(-26))}{9})\\&\int_{-26}^{-21}(\frac{25}{(9x)})dx=-0.59\end{align*}$

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Example Question #565 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-26}^{-\frac{45}{2}}(\frac{(7cos(4x))}{37})dx\end{align*}$

Possible Answers:

-0.06

-0.01

-0.52

-0.21

Correct answer:

-0.06

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-26}^{-\frac{45}{2}}(\frac{(7cos(4x))}{37})dx=\frac{(7sin(4x))}{148}|_{-26}^{-\frac{45}{2}}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-26}^{-\frac{45}{2}}(\frac{(7cos(4x))}{37})dx=\frac{(7sin(4(-\frac{45}{2})))}{148}-(\frac{(7sin(4(-26)))}{148})\\&\int_{-26}^{-\frac{45}{2}}(\frac{(7cos(4x))}{37})dx=-0.06\end{align*}$

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Example Question #2313 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-12}^{-8.5}(\frac{(21sin(x + 1))}{4})dx\end{align*}$

Possible Answers:

-0.82

-0.26

-1.80

-0.21

Correct answer:

-1.80

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{-12}^{-8.5}(\frac{(21sin(x + 1))}{4})dx=-\frac{(21cos(x + 1))}{4}|_{-12}^{-8.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-12}^{-8.5}(\frac{(21sin(x + 1))}{4})dx=-\frac{(21cos((-8.5) + 1))}{4}-(-\frac{(21cos((-12) + 1))}{4})\\&\int_{-12}^{-8.5}(\frac{(21sin(x + 1))}{4})dx=-1.80\end{align*}$

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Example Question #201 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{28}^{31}(3x)dx\end{align*}$

Possible Answers:

2150.55

1672.65

265.50

78.09

Correct answer:

265.50

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{28}^{31}(3x)dx=\frac{(3x^{2})}{2}|_{28}^{31}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{28}^{31}(3x)dx=\frac{(3(31)^{2})}{2}-(\frac{(3(28)^{2})}{2})\\&\int_{28}^{31}(3x)dx=265.50\end{align*}$

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Example Question #2321 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-14}^{-10}(\frac{(71sin(3x))}{7})dx\end{align*}$

Possible Answers:

-1.87

-6.37

-0.20

-10.31

Correct answer:

-1.87

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{-14}^{-10}(\frac{(71sin(3x))}{7})dx=-\frac{(71cos(3x))}{21}|_{-14}^{-10}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-14}^{-10}(\frac{(71sin(3x))}{7})dx=-\frac{(71cos(3(-10)))}{21}-(-\frac{(71cos(3(-14)))}{21})\\&\int_{-14}^{-10}(\frac{(71sin(3x))}{7})dx=-1.87\end{align*}$

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Example Question #221 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-30}^{-26}(\frac{(17cos(x + 1))}{3})dx\end{align*}$

Possible Answers:

-0.53

-3.01

-0.36

-8.73

Correct answer:

-3.01

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-30}^{-26}(\frac{(17cos(x + 1))}{3})dx=\frac{(17sin(x + 1))}{3}|_{-30}^{-26}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-30}^{-26}(\frac{(17cos(x + 1))}{3})dx=\frac{(17sin((-26) + 1))}{3}-(\frac{(17sin((-30) + 1))}{3})\\&\int_{-30}^{-26}(\frac{(17cos(x + 1))}{3})dx=-3.01\end{align*}$

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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #562 : Integrals

Example Question #212 : Finding Integrals

Example Question #2311 : Calculus Ii

Example Question #2312 : Calculus Ii

Example Question #201 : Definite Integrals

Example Question #565 : Integrals

Example Question #2313 : Calculus Ii

Example Question #201 : Definite Integrals

Example Question #2321 : Calculus Ii

Example Question #221 : Finding Integrals

All Calculus 2 Resources

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