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Calculus 2 : Finding Integrals

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Example Questions

Example Question #195 : Definite Integrals

Evaluate the following definite integral:

$\int^5_4 \frac{10x-24}{x^2-5x+6}\,dx$

Possible Answers:

6.98

5.78

7.13

2.60

6.24

Correct answer:

5.78

Explanation:

This integral can be solved by using partial fractions. First, we have to factor the denominator write the fraction as a sum of two fractions:

$\int^5_4 \frac{10x-24}{x^2-5x+6}\,dx=\int^5_4 \frac{10x-24}{(x-2)(x-3)}\,dx=\int^5_4(\frac{A}{x-2}+\frac{B}{x-3})dx$

Next, we can solve for A and B:

10x-24=A(x-3)+B(x-2)

When we let x=3:

$30-24=0+B\rightarrow B=6$

When x=2:

$20-24=-A+0\rightarrow A=4$

Replacing A and B in the integral, we can now solve it:

$\int^5_4(\frac{4}{x-2}+\frac{6}{x-3})dx=[4\ln(x-2)+6\ln(x-3) ] |^5_4$

$[4\ln(3)+6\ln(2)]-[4\ln(2)+6\ln(1)]=5.78$

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Example Question #561 : Integrals

Evaluate the following integral:

$\int_{\pi}^{0}Sin(x) dx$

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

Recall that given two real numbers b and a with b>a the integral $\int_{b}^{a}f(x)dx$ is equivalent to $-\int_{a}^{b}f(x)dx$ .

Therefore,

$\int_{\pi}^{0}Sin(x) dx$

is equivalent

$-\int_{0}^{\pi}Sin(x) dx$ .

Evaluating the integral one finds

$-\int_{0}^{\pi}Sin(x) dx= -(-Cos(\pi)-(-Cos(0))))=-(-(-1)-1))=-2$

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Example Question #193 : Definite Integrals

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

Possible Answers:

15.22

166.25

79.17

9.90

Correct answer:

79.17

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{10}^{15}(\frac{x^{2}}{10})dx=\frac{x^{3}}{30}|_{10}^{15}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{10}^{15}(\frac{x^{2}}{10})dx=\frac{15^{3}}{30}-(\frac{10^{3}}{30})\\&\int_{10}^{15}(\frac{x^{2}}{10})dx=79.17\end{align*}$

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Example Question #194 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-4}^{-2}(\frac{1}{(6\cdot 2^x)})dx\end{align*}$

Possible Answers:

6.64

0.42

26.26

2.89

Correct answer:

2.89

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{-4}^{-2}(\frac{1}{(6\cdot 2^x)})dx=-\frac{1}{(6\cdot 2^xln(2))}|_{-4}^{-2}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{-4}^{-2}(\frac{1}{(6\cdot 2^x)})dx=-\frac{1}{(6\cdot 2^{-2}ln(2))}-(-\frac{1}{(6\cdot 2^{-4}ln(2))})\\&\int_{-4}^{-2}(\frac{1}{(6\cdot 2^x)})dx=2.89\end{align*}$

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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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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #195 : Definite Integrals

Example Question #561 : Integrals

Example Question #193 : Definite Integrals

Example Question #194 : Definite Integrals

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

All Calculus 2 Resources

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