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Calculus 2 : Calculus II

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Example Questions

Example Question #2301 : Calculus Ii

Solve the following definite integral.

$\int_0^5 \,2xy+3\,dx$

Possible Answers:

xy^2+3y

x^2y+3x

25y+15

Correct answer:

25y+15

Explanation:

$\int_0^5 \,2xy+3\,dx=x^2y+3x|_{x=0}^{x=5}$

=5^2y+3(5)-[0^2y+3(0)]

=25y+15

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Example Question #2302 : Calculus Ii

If $\int_{1}^{7}f(x)dx=5, \int_{3}^{7}f(x)dx=2, \int_{1}^{3}f(x)dx=?$

Possible Answers:

Not enough information.

Correct answer:

Explanation:

If we think of the anti-derivative as computing the area under the curve, then between x=1 and x=7 , the area under our function equals . We also know that between x=3 and x=7 , the area under our function equals . Then, to find the area under the function between x=1 and x=3 , we must simply subtract the other two areas. Therefore:

$\int_{1}^{7}f(x)dx- \int_{3}^{7}f(x)dx=\int_{1}^{3}f(x)dx=5-2=3.$

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Example Question #189 : Definite Integrals

If f(x)=x^2-3x+5 , what is $\int_{5}^{5}f(x)dx$ ?

Possible Answers:

$\frac{1}{2}$

$\frac{2}{3}$

Correct answer:

Explanation:

Remember that the anti-derivative computes the area under the curve of the function between the values specified by the upper and lower integral limits. In this question, the upper and lower integral limits match! This, by definition, means that the integral equals There is no area if you start and end at the same point!

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Example Question #552 : Integrals

If $\int_{1}^{3} f(x)dx=5, \int_{7}^{3} f(x)dx=-4$ , and $\int_{7}^{10}f(x)dx=1,$ what is $\int_{1}^{10} f(x)dx?$

Possible Answers:

Correct answer:

Explanation:

To find the total integral on our interval from x=1 to x=10 , we need to find the area under each subinterval and add them all up. The first and last part are given in the problem statement, but the middle interval has backwards limits.

Remember:

$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx.$

Therefore, we need to switch the sign on the middle area. Therefore,

$\int_{1}^{10} f(x)dx=\int_{1}^{3} f(x)dx+-\int_{7}^{3} f(x)dx+\int_{7}^{10} f(x)dx$

=5+--4+1=10.

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Example Question #204 : Finding Integrals

In exponentially decaying systems, often times the solutions to differential equations take on the form of an integral called Duhamel's Integral. This is given by:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Where is a constant and f(b) is a function that represents an external force. g(t) represents the overall solution to the differential equation.

Determine g(t) if the external force is given by f(b)=1

Possible Answers:

$g(t)=1-e^{-st}$

$g(t)=\frac{1}{s}[1+e^{-st}]$

$g(t)=\frac{1}{s}[1-e^{-st}]$

$g(t)=-\frac{1}{s}[1-e^{-st}]$

Correct answer:

$g(t)=\frac{1}{s}[1-e^{-st}]$

Explanation:

We can rewrite the integral as

$g(t)=\int_{0}^{t}e^{-s(t-b)}*1 db$

Using substitution, we let

u=-st+sb

$\frac{du}{db}=s$

$db=\frac{du}{s}$

We can therefore write the integral as:

$g(t)=\frac{1}{s}\int_{0}^{t}e^udu$

Since $\int e^u=e^u$ .

$g(t)=\frac{1}{s}[e^u]_{0}^{t}= \frac{1}{s}[e^{(-st+sb)}]_{0}^{t}= \frac{1}{s}[1-e^{-st}]$

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Example Question #193 : Definite Integrals

In exponentially decaying systems, often times the solutions to differential equations take on the form of an integral called Duhamel's Integral. This is given by:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Where is a constant and f(b) is a function that represents an external force.

Determine the value at t=0 if f(b)=b .

Possible Answers:

$\frac{1}{s^2}$

s^2-1

$\frac{s+1-e^{-s}}{s^2}$

Correct answer:

Explanation:

Plugging f(b) into the equation:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*b db$

Integrating by parts, we can assign values for , , , and :

U=b $dV=e^{-s(t-b)}db$

dU=db $V=\int e^{-s(t-b)}db=\frac{1}{s}e^{-s(t-b)}$

Plugging into our equation:

$\int_{0}^{t}UdV= [UV]_{0}^{t}-\int_{0}^{t}VdU$

$\int_{0}^{t}e^{-s(t-b)}*b db=\frac{1}{s}[e^{-s(t-b)}b]_{0}^{t}-\frac{1}{s}\int_{0}^{t}e^{-s(t-b)}db$

The first term becomes:

$\frac{1}{s}[e^{-s(t-b)}b]_{0}^{t}= \frac{1}{s}[t-0]=\frac{t}{s}$

The integral term becomes $\int_{0}^{t}e^{-s(t-b)}db=\frac{1}{s}[e^{-s(t-b)}]_{0}^{t}=\frac{1}{s}[e^{-s(t-t)}-e^{-st}]=\frac{1-e^{-st}}{s}$

Plugging everything else in we get:

$g(t)=\frac{t}{s}+\frac{e^{-st}-1}{s^2}$

$g(0)=\frac{0}{s}+\frac{e^{0}-1}{s^2}=\frac{0+1-1}{s}=0$

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Example Question #2302 : Calculus Ii

Evaluate the following definite integral:

$\int^1_{-1}(4x+1)dx$

Possible Answers:

Correct answer:

Explanation:

This integral requires use of the power rule for antiderivatives, which simplifies as follows:

$\int^1_{-1}(4x+1)dx=(2x^2+x)|^1_{-1}=(2(1)^2+1)-(2(-1)^2-1)=5-3=2$

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Example Question #192 : Definite Integrals

Evaluate the following definite integral:

$\int^1_0 \frac{x^4}{8-x^5}dx$

Possible Answers:

0.793

0.0561

0.0324

0.0267

0.610

Correct answer:

0.0267

Explanation:

Make the substitution:

u=8-x^5

Where

$du=-5x^4dx \rightarrow x^4dx=-\frac{du}{5}$

The limits of the integral, 0 and 1, are also changed in the substitution:

$0\rightarrow 8-0^5=8$

$1\rightarrow 8-1^5=7$

Using this in the original expression:

$\int^1_0 \frac{x^4}{8-x^5}dx=-\frac{1}{5}\int^7_8\frac{du}{u}= -\frac{1}{5}\cdot \ln(u)|^7_8=-\frac{1}{5}[\ln(7)-\ln(8)]$

$=-\frac{1}{5}\ln(\frac{7}{8})=0.0267$

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Example Question #193 : Definite Integrals

Evaluate the following definite integral:

$\int^2_0 xe^{2x}dx$

Possible Answers:

$\frac{e^2+2}{4}$

$e^{4}$

$\frac{e^4}{2}$

6e^4+4

$\frac{3e^4+1}{4}$

Correct answer:

$\frac{3e^4+1}{4}$

Explanation:

To solve this with integration by parts, we rewrite the expression in the form

$\int u \: dv$

where

$u=x, \: \: \:dv=e^{2x}\, dx$

and

$du=dx,\: \: \:v=\frac{e^{2x}}{2}$

To integrate, apply the formula for integration by parts:

$\int u \: dv=uv-\int vdu=\frac{xe^{2x}}{2}|^2_0-\int^2_0\frac{e^{2x}}{2}dx =(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4})|^2_0$

$=\frac{e^{2x}}{4}(2x-1)|^2_0=\frac{e^{4}}{4}(3)-(\frac{1}{4})(-1)=\frac{3e^{4}+1}{4}$

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Example Question #194 : Definite Integrals

Evaluate the following definite integral:

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

Possible Answers:

5.78

6.98

6.24

7.13

2.60

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2301 : Calculus Ii

Example Question #2302 : Calculus Ii

Example Question #189 : Definite Integrals

Example Question #552 : Integrals

Example Question #204 : Finding Integrals

Example Question #193 : Definite Integrals

Example Question #2302 : Calculus Ii

Example Question #192 : Definite Integrals

Example Question #193 : Definite Integrals

Example Question #194 : Definite Integrals

All Calculus 2 Resources

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