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Calculus 2 : Calculus II

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

Example Questions

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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Example Question #2322 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx\end{align*}$

Possible Answers:

-14.57

-278.82

-392.88

-42.25

Correct answer:

-42.25

Explanation:

$\begin{align*}\\&\text{the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx=-\frac{ (13cos(3x))}{6}-\frac{ (76cos(x + 1))}{3}|_{28}^{31}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx=-\frac{ (13cos(3(31)))}{6}-\frac{ (76cos((31) + 1))}{3}-(-\frac{ (13cos(3(28)))}{6}-\frac{ (76cos((28) + 1))}{3})\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx=-42.25\end{align*}$

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Example Question #2323 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx\end{align*}$

Possible Answers:

-0.77

-16.05

-0.52

-4.86

Correct answer:

-4.86

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx=-\frac{ (47cos(x + 1))}{7}-\frac{ 5}{(34x)}|_{-23}^{-18}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx=-\frac{ (47cos((-18) + 1))}{7}-\frac{ 5}{(34(-18))}-(-\frac{ (47cos((-23) + 1))}{7}-\frac{ 5}{(34(-23))})\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx=-4.86\end{align*}$

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Example Question #2324 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx\end{align*}$

Possible Answers:

286.75

2494.72

102.41

44.12

Correct answer:

286.75

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx=\frac{(31x^{2})}{12}-\frac{ (3e^{(-2x)})}{8}|_{17}^{20}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx=\frac{(31(20)^{2})}{12}-\frac{ (3e^{(-2(20))})}{8}-(\frac{(31(17)^{2})}{12}-\frac{ (3e^{(-2(17))})}{8})\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx=286.75\end{align*}$

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Example Question #2321 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx\end{align*}$

Possible Answers:

-8.01

-0.91

-4.55

-3.09

Correct answer:

-0.91

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx=\frac{sin(4x)}{47}+\frac{ (10sin(x + 2))}{3}|_{-12}^{-8}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx=\frac{sin(4(-8))}{47}+\frac{ (10sin((-8) + 2))}{3}-(\frac{sin(4(-12))}{47}+\frac{ (10sin((-12) + 2))}{3})\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx=-0.91\end{align*}$

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Example Question #2325 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx\end{align*}$

Possible Answers:

0.58

38.35

1.24

3.95

Correct answer:

3.95

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx=\frac{(2sin(4x))}{47}+\frac{ (79ln(x))}{5}|_{9}^{11.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx=\frac{(2sin(4(11.5)))}{47}+\frac{ (79ln((11.5)))}{5}-(\frac{(2sin(4(9)))}{47}+\frac{ (79ln((9)))}{5})\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx=3.95\end{align*}$

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Example Question #2322 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-14}^{-10.5}(\frac{(19cos(x + 2))}{10}+\frac{ (8sin(x + 1))}{57})dx\end{align*}$

Possible Answers:

-2.27

-0.41

-0.23

-7.03

Correct answer:

-2.27

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-14}^{-10.5}(\frac{(19cos(x + 2))}{10}+\frac{ (8sin(x + 1))}{57})dx=\frac{(19sin(x + 2))}{10}-\frac{ (8cos(x + 1))}{57}|_{-14}^{-10.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\frac{(19sin((-10.5) + 2))}{10}-\frac{ (8cos((-10.5) + 1))}{57}-(\frac{(19sin((-14) + 2))}{10}-\frac{ (8cos((-14) + 1))}{57})\\&-2.27\end{align*}$

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Example Question #580 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{8}^{9.5}(\frac{(31cos(4x))}{2}+\frac{ (4e^{(-2x)})}{27})dx\end{align*}$

Possible Answers:

-0.18

-0.11

-0.99

-2.67

Correct answer:

-0.99

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{8}^{9.5}(\frac{(31cos(4x))}{2}+\frac{ (4e^{(-2x)})}{27})dx=\frac{(31sin(4x))}{8}-\frac{ (2e^{(-2x)})}{27}|_{8}^{9.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\frac{(31sin(4(9.5)))}{8}-\frac{ (2e^{(-2(9.5))})}{27}-(\frac{(31sin(4(8)))}{8}-\frac{ (2e^{(-2(8))})}{27})\\&-0.99\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #201 : Definite Integrals

Example Question #2321 : Calculus Ii

Example Question #221 : Finding Integrals

Example Question #2322 : Calculus Ii

Example Question #2323 : Calculus Ii

Example Question #2324 : Calculus Ii

Example Question #2321 : Calculus Ii

Example Question #2325 : Calculus Ii

Example Question #2322 : Calculus Ii

Example Question #580 : Integrals

All Calculus 2 Resources

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