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Calculus 2 : Integrals

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Example Questions

Example Question #31 : Improper Integrals

Evaluate $\int_{1}^{\infty} \frac{1}{x^{2}}dx$

Possible Answers:

$\infty$

$\frac{1}{2}$

$\frac{1}{3}$

Correct answer:

Explanation:

Since $\infty$ is not an actual number, this is an improper integral. Therefore we have to rewrite it as a limit.

$\lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^{2}}dx$

We replaced the improper bound, $\infty$ , with a new variable, t. Then the limit will allow us to evalute the integral as this new bound, t, approaches $\infty$ .

Now we can evaluate the integral normally.

$\lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^{2}}dx \Rightarrow \lim_{t \to \infty} [\frac{-1}{x}]_{1}^{t}\Rightarrow \lim_{t \to \infty}[(\frac{-1}{t})-(\frac{-1}{1})]\Rightarrow$

$\Rightarrow \lim_{t \to \infty} [\frac{-1}{t}+1]$

Then we evaluate the limit.

$\lim_{t \to \infty} [\frac{-1}{t}+1] \Rightarrow [\frac{-1}{\infty} + 1] \Rightarrow 0+1\Rightarrow 1$

Thus the answer is .

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Example Question #32 : Improper Integrals

Classify each improper integral as convergent or divergent.

A) $\int_{0}^{\infty}\frac{2}{x+1}dx$

B) $\int_{-2}^{\infty} \frac{2x}{\sqrt{4-x^2}}dx$

C) $\int_1^{\infty}\frac{1}{x^7}dx$

Possible Answers:

A) Converges

B) Diverges

C) Converges

A) Diverges

B) Converges

C) Converges

A) Diverges

B) Diverges

C) Converges

A) Converges

B) Converges

C) Diverges

A) Converges

B) Converges

C) Converges

Correct answer:

A) Diverges

B) Diverges

C) Converges

Explanation:

A) $\int_{0}^{\infty}\frac{2}{x+1}dx$

Write the integral as a limit,

$\lim_{t-\infty}\int_0^t \frac{2}{x+1}dx =\lim_{t->\infty}\left [ 2\ln(x+1)\right ]_{x=0}^{x=t}=\lim_{t->\infty}2\ln(t+1)$

Because we know that the natural logarithm is an increasing function, it blows up as $t->\infty$ , therefore (A) is divergent.

B) $\int_{-2}^{\infty} \frac{2x}{\sqrt{4-x^2}}dx$

Write as a limit,

$\lim_{t->\infty }\int_{-2}^{t} \frac{2x}{\sqrt{4-x^2}}dx$

We can use a substitution to integrate,

u = 4-x^2

du = -2xdx

$\int\frac{2x}{\sqrt{4-x^2}}dx = -\int\frac{du}{\sqrt{u}}=-2\sqrt{u}$

Now write the indefinite integral in terms of .

$\int\frac{2x}{\sqrt{4-x^2}} dx= -2\sqrt{4-x^2}$

$\lim_{t->\infty }\int_{-2}^{t} \frac{2x}{\sqrt{4-x^2}}dx=\lim_{t->\infty} \left[-2\sqrt{4-x^2}\right]_{x=-2}^{x=t}=\lim_{t->\infty} -2\sqrt{4-t^2}$

This limit does not exist, and therefore the integral diverges.

C) $\int_1^{\infty}\frac{1}{x^7}dx$

Recall that improper integrals of the form $\int_a^{\infty} \frac{1}{x^p}dx$ will converge if p>1 and will diverge if $p\leq 1$ provided that a>1 .

In this case we have p=7 . Therefore (C) Converges.

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Example Question #32 : Improper Integrals

Integrate:

$\int_ {2}^{4} 4x^5$

Possible Answers:

2688

2268

2668

2688

Correct answer:

2688

Explanation:

Step 1: Integrate...

$\large \large \int 4x^5=\frac {4x^{5+1}}{5+1}$

Step 2: Add up the exponent and the denominator:

$\large \frac {4x^{5+1}}{5+1}\implies \frac {4x^6}{6}$

Step 3: Simplify the coefficients...

$\large \frac {4x^6}{6}=\frac {{\color{Red} 2}(2x^6)}{{\color{Red} 2}(3)}=\frac {2x^6}{3}$

Step 4: Plug in the upper limit, $\large x=4$ ...

$\large \large \frac {{2(4)^6}}{3}=\frac {2(4096)}{3}=\frac {8192}{3}$

Step 5: Plug in the lower limit, $\large x=2$ ...

$\large \frac {2(2)^6}{3}=\frac {2^7}{3}=\frac {128}{3}$

Step 6: To find the value of the integral, subtract the value of step 5 from step 4...

$\large \frac {8192}{3}-\frac {128}{3}=\frac {8064}{3}=2688$

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Example Question #33 : Improper Integrals

he Laplace Transform is an integral transform that converts functions from the time domain to the complex frequency domain . The transformation of a function f(t) into its Laplace Transform F(s) is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where s=a+bi , where and are constants and is the imaginary number.

Determine the Laplace Transform of f(t)=1

Possible Answers:

$\frac{1}{s}$

$\infty$

Correct answer:

$\frac{1}{s}$

Explanation:

$F(s)= \int_{0}^{\infty}e^{-st}dt$

Let U=st

$\frac{dU}{dt}=s$

$dt=\frac{dU}{s}$

$F(s)=\frac{1}{s}\int_{0}^{\infty}e^{-u} du$

$F(s)=\frac{1}{s}(-1(e^{-\infty}-e^0))$

$F(s)=-\frac{1}{s}[0-1]=\frac{1}{s}$

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Example Question #34 : Improper Integrals

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where s=a+bi , where and are constants and is the imaginary number.

Give the Laplace Transform of $\frac{\mathrm{d} }{\mathrm{d} t}[f(t)]$ .

Possible Answers:

sF(k)-F(k)

sF(k)-f(0)

sF(k)+f(0)

sF(k)

Correct answer:

sF(k)-f(0)

Explanation:

The Laplace Transform of the derivative is given by:

$\int_{0}^{\infty}e^{-st}*f'(t) dt$

Using integration by parts,

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

Let dV=f'(t)dt and $U=e^{-st}$

$[e^{-st}*f(t) ]_{0}^{\infty}-\int_{0}^{\infty}(-s)e^{-st}*f(t) dt$

The first term becomes

0-f(0)

and the second term becomes

s*F(k)

The Laplace Transform therefore becomes:

sF(k)-f(0)

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Example Question #32 : Improper Integrals

Evaluate the following improper integral:

$\int^1_{-\infty }e^{2x}dx$

Possible Answers:

$\infty$

$-\infty$

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

2e^x

e^2

Correct answer:

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

Explanation:

To evaluate an improper integral, we first evaluate it with a finite quantity a in place of negative infinity:

$\int^1_{-\infty }e^{2x}dx\rightarrow\int^1_{a }e^{2x}dx=\frac{e^{2x}}{2}|^1_a= \frac{e^2}{2}-\frac{e^{2a}}{2}$

Now, we take the limit of this quantity as a approaches negative infinity:

$\lim_{a\rightarrow -\infty}(\frac{e^2}{2}-\frac{e^{2a}}{2})=\frac{e^2}{2}$

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Example Question #35 : Improper Integrals

The Laplace Transform is an integral transform that converts functions from the time domain to the complex frequency domain . The transformation of a function f(t) into its Laplace Transform F(s) is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where s=a+bi , where and are constants and is the imaginary number.

Determine the Laplace Transform of $e^{it}$ .

Possible Answers:

$\frac{1}{i+s}$

$\infty$

$\frac{1}{i-s}$

$\frac{1}{s-i}$

Correct answer:

$\frac{1}{s-i}$

Explanation:

$F(s)= \int_{0}^{\infty}e^{-st}*e^{it} dt= \int_{0}^{\infty}e^{(-s+i)t}dt$

Let U=(s-i)t

$\frac{dU}{s-i}=dt$

$F(s)= \frac{1}{s-i}\int_{0}^{\infty}e^{-U}dU$

$F(s)= \frac{1}{s-i}(-1*(e^{-\infty}-e^0))=\frac{1}{s-i}$

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Example Question #921 : Integrals

Evaluate

$\int_{0}^{\infty }e^{-x}dx$

Possible Answers:

e^2

$\infty$

Correct answer:

Explanation:

We must solve this as an improper integral.

$\lim_{t\rightarrow \infty }\int_{0}^{t}e^{-x}dx=\lim_{t\rightarrow \infty }-e^{-t}-(-e^0)$

=0+1=1

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Example Question #572 : Finding Integrals

Give the indefinite integral:

$\int \frac{\ln 3x \; \mathrm{d}x}{x}$

Possible Answers:

$\frac{3\left ( \ln 3 x \right )^2}{2} + C$

$\frac{x}{3}+C$

$\frac{ \ln 3x }{2} + C$

$\frac{\left ( \ln 3x \right )^2}{2} + C$

$\frac{3}{x}+C$

Correct answer:

$\frac{\left ( \ln 3x \right )^2}{2} + C$

Explanation:

Let $u = \ln 3x$ . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln 3x = \frac{3}{3x} = \frac{1}{x}$

and

$\mathrm{d}u = \frac{\mathrm{d}x }{x }$ .

The integral can be rewritten as

$\int u \; \mathrm{d}u$

$=\frac{ u^2}{2} + C$

$=\frac{\left ( \ln 3x \right )^2}{2} + C$

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Example Question #581 : Finding Integrals

Evaluate:

$\int_{-2}^{2} 6^{x}\; \mathrm{d}x$

Possible Answers:

$72 \ln 6$

$\frac{1,295 \ln 6 }{36}$

$\frac{72}{\ln 6}$

$\frac{1,295}{36 \ln 6}$

Correct answer:

$\frac{1,295}{36 \ln 6}$

Explanation:

$6^{x} = (e^{\ln 6} )^{x} = e^{x \ln 6}$ , so the integral can be rewritten as $\int_{-2}^{2} e^{x \ln 6} \; \mathrm{d}x$ .

Substitute $u = x \ln 6$ .

Therefore, $\frac{\mathrm{d} u}{\mathrm{d} x} = \ln 6$ , $\mathrm{d} u = \ln 6 \cdot \mathrm{d} x$ , and $\frac{\mathrm{d} u }{\ln 6}= {d} x$ .

The bounds of integration become $u = 2 \ln 6 = \ln 36$ and $u = - 2 \ln 6 = \ln \frac{1}{36}$ .

$\int_{-2}^{2} e^{x \ln 6} \; \mathrm{d}x$

$= \int_{\ln \frac{1}{36}}^{\ln 36} e^{u} \;\cdot \frac{1}{\ln 6} \mathrm{d}u$

$= \frac{1}{\ln 6} \int_{\ln \frac{1}{36}}^{\ln 36} e^{u} \; \mathrm{d}u$

$= \left.\begin{matrix} \frac{1}{\ln 6} e^{u} \\ \textrm{ } \end{matrix}\right|$ $\begin{matrix} \ln 36 \\ \\\ln \frac{1}{36} \end{matrix}$

$= \frac{1}{\ln 6} e ^{\ln 36}- \frac{1}{\ln 6} e ^{\ln \frac{1}{36}}$

$= \frac{1}{\ln 6} \cdot 36 - \frac{1}{\ln 6} \cdot \frac{1}{36}$

$= \frac{1}{\ln 6} \cdot \frac{1, 296}{36} - \frac{1}{\ln 6} \cdot \frac{1}{36} = \frac{1}{\ln 6} \cdot \frac{1, 295}{36} = \frac{1,295}{36 \ln 6}$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #31 : Improper Integrals

Example Question #32 : Improper Integrals

Example Question #32 : Improper Integrals

Example Question #33 : Improper Integrals

Example Question #34 : Improper Integrals

Example Question #32 : Improper Integrals

Example Question #35 : Improper Integrals

Example Question #921 : Integrals

Example Question #572 : Finding Integrals

Example Question #581 : Finding Integrals

All Calculus 2 Resources

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