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Calculus 2 : Improper Integrals

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Example Question #1 : Improper Integrals

Evaluate:

$\int_{1}^{\infty} \frac{2}{x^3}\: dx$

Possible Answers:

$\infty$

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

Explanation:

Write the formula rule for the case:

$\int_{a}^{\infty} F(x)\:dx = \lim_{b \to \infty} \int_{a}^{b}f(x)\:dx$

Apply this rule for the following question.

$\int_{1}^{\infty} \frac{2}{x^3}\: dx= 2\int_{1}^{\infty} \frac{1}{x^3}\: dx$

$2\int_{1}^{\infty} \frac{1}{x^3}\: dx = 2\lim_{b \to \infty} \int_{1}^{b}{x}^-^3}\:dx= 2\left[-\frac{x^-^2}{2}\right]_{1}^{b}$

$\lim_{b \to \infty} 2\left[-\frac{x^-^2}{2}\right]_{1}^{b}= \lim_{b \to \infty}2\left[-\frac{1}{2b^2}+\frac{1}{2}\right]$

$=\lim_{b \to \infty}{-\frac{1}{b^2}}+1 = 0+1 = 1$

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Example Question #2 : Improper Integrals

Evaluate $\int_{1}^{\infty}\frac{3}{x^{2}}dx$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

By the Formula Rule, we know that $\int_{a}^{\infty}F(x)dx=\lim_{b\rightarrow \infty}\int_{a}^{b}f(x)dx$ . We therefore know that $\int_{1}^{\infty}\frac{3}{x^{2}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{3}{x^{2}}dx$ .

Continuing the calculation:

$\int_{1}^{b}\frac{3}{x^{2}}dx$

$=3\int_{1}^{b}\frac{1}{x^{2}}dx$

$=3\int_{1}^{b}{x^{-2}}dx$

By the Power Rule for Integrals, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ with an arbitrary constant of integration . Therefore:

$3\int_{1}^{b}{x^{-2}}dx$

$=3({\frac{x^{-2+1}}{-2+1}})_{1}^{b}$

$=3(\frac{x^{-1}}{-1}})_{1}^{b}$

$=3(-{\frac{1}{x}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }3[-{\frac{1}{x}}]_{1}^b$

$=\lim_{b\rightarrow \infty }3[-{\frac{1}{b}-(-\frac{1}{1}})]$

$=\lim_{b\rightarrow \infty }3[-{\frac{1}{b}+1]$

As $b\rightarrow \infty$ , $\lim_{b\rightarrow \infty }3[-{\frac{1}{b}+1]=3(0+1)=3$

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Example Question #3 : Improper Integrals

Evaluate $\int_{1}^{\infty}\frac{7}{x^{8}}dx$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

By the Formula Rule, we know that $\int_{a}^{\infty}F(x)dx=\lim_{b\rightarrow \infty}\int_{a}^{b}f(x)dx$ . We therefore know that $\int_{1}^{\infty}\frac{7}{x^{8}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{7}{x^{8}}dx$ .

Continuing the calculation:

$\int_{1}^{b}\frac{7}{x^{8}}dx$

$=7\int_{1}^{b}\frac{1}{x^{8}}dx$

$=7\int_{1}^{b}{x^{-8}}dx$

By the Power Rule for Integrals, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ with an arbitrary constant of integration . Therefore:

$7\int_{1}^{b}{x^{-8}}dx$

$=7({\frac{x^{-8+1}}{-8+1}})_{1}^{b}$

$=7(\frac{x^{-7}}{-7}})_{1}^{b}$

$=7(-{\frac{1}{7x^{7}}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }7(-{\frac{1}{7x^{7}}})_{1}^b$

$=\lim_{b\rightarrow \infty }7[-{\frac{1}{7b^{7}}-(-\frac{1}{7}})]$

$=\lim_{b\rightarrow \infty }7[-{\frac{1}{7b^{7}}+\frac{1}{7}})]$

$=\lim_{b\rightarrow \infty }[-{\frac{1}{b^{7}}+1}]$ .

=0+1

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Example Question #4 : Improper Integrals

Evaluate $\int_{1}^{\infty}\frac{4}{x^{5}}dx$ .

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

By the Formula Rule, we know that $\int_{a}^{\infty}F(x)dx=\lim_{b\rightarrow \infty}\int_{a}^{b}f(x)dx$ . We therefore know that $\int_{1}^{\infty}\frac{4}{x^{5}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{4}{x^{5}}dx$ .

Continuing the calculation:

$\int_{1}^{b}\frac{4}{x^{5}}dx$

$=4\int_{1}^{b}\frac{1}{x^{5}}dx$

$=4\int_{1}^{b}{x^{-5}}dx$

By the Power Rule for Integrals, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ with an arbitrary constant of integration . Therefore:

$4\int_{1}^{b}{x^{-5}}dx$

$=4({\frac{x^{-5+1}}{-5+1}})_{1}^{b}$

$=4(\frac{x^{-4}}{-4}})_{1}^{b}$

$=4(-{\frac{1}{4x^{4}}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }4(-{\frac{1}{4x^{4}}})_{1}^b$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }[-{\frac{1}{b^{4}}+1}]$ .

=0+1

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Example Question #5 : Improper Integrals

Evaluate:

$\int_{2}^{5}\frac{1}{2x-8} \:dx$

Possible Answers:

$\frac{1}{2}ln(-2)$

$\frac{1}{2}ln\left(-\frac{1}{2}\right)$

$\frac{1}{2}ln(8)$

Diverge

ln(5)-ln(2)

Correct answer:

Diverge

Explanation:

To evaluate $\int_{2}^{5}\frac{1}{2x-8} \:dx$ , notice that the denominator does not exist at x=4 . The integral must be rewritten so that it is in terms of a limit. This is an improper integral.

$\int_{2}^{5}\frac{1}{2x-8} \:dx= \lim_{a \to 4^-}\int_{2}^{a}\frac{1}{2x-8} \:dx+\lim_{b \to 4^+}\int_{b}^{5}\frac{1}{2x-8} \:dx$

Evaluate the integrals.

u=2x-8

$du=2\:dx$

$\frac{1}{2}du=dx$

$\frac{1}{2}\lim_{a \to 4^-}ln\left | \frac{1}{2x-8} \right |_{2}^{a} +\frac{1}{2}\lim_{b \to 4^+}ln\left | \frac{1}{2x-8} \right |_{b}^{5}$

By evaluating the terms and substituting the limits, we will notice that the integral diverges as a result since the terms $\frac{1}{2}ln\left | \frac{1}{0} \right |-\frac{1}{2}ln\left | \frac{1}{0} \right |$ cannot be cancelled as a result.

The correct answer is: Diverge

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Example Question #6 : Improper Integrals

Evaluate the following integral:

$\int_{0}^{1}\frac{3}{x-1}dx$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

$-\infty$

Explanation:

The integral we were given in the problem statement is improper because the upper limit of integration creates 0 in the denominator of the fraction.

To fix this, we use the following technique:
$3\lim_{t\rightarrow 1^{-}}\int_{0}^{t}\frac{dx}{x-1}$

Note that we moved the three outside of the integral, and then outside of the limit altogether. This can be done with coefficients and makes things far easier to look at!

Next, keeping the coefficient and limit, integrate:

$3\lim_{t\rightarrow 1-}\ln \left | x-1 \right | |_{0}^{t}$

The integration was done using the following rule:

$\int \frac{dx}{x}=\ln\left | x \right |+C$

Next, solve the result:

$3\lim_{t\rightarrow 1-} \ln(t-1)-0=3\cdot -\infty =-\infty$

We chose 1 from the left in our limit because 1 is the upper limit of integration.

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Example Question #1 : Improper Integrals

Evaluate the improper integral

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

Possible Answers:

The integral does not exist.

$\frac{1}{2}$

Correct answer:

Explanation:

First, rewrite the function using a negative exponent.

$\int_{0}^{\infty}(1+x)^{-2}dx$ .

Next make the substitution u=1+x. Then du=dx. We can also rewrite the limits of integration in terms of u. When x approaches infinity, u approaches infinity. When x=0, u =1+0 = 1 .

Rewriting and evaluating the integral using the power rule, $\int u^ndu=\frac{u^{n+1}}{n+1}+c$ , gives us:

$\int_{1}^{\infty}u^{-2}du=\lim_{u\to\infty}-u^{-1}-(-(1)^{-1})=0-(-1)=1$

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Example Question #8 : Improper Integrals

Evaluate the following integral:

$\int_{1}^{\infty }\frac{dx}{x^3}$

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

The integral is improper because of the upper limit of integration (infinity). To evaluate the integral, we must do the following:

$\lim_{t\rightarrow \infty }\int_{1}^{t}\frac{dx}{x^3}$

Now integrate and evaluate the integral from t to 1:

$\lim_{t\rightarrow \infty }\frac{-1}{2}t^{-2}+\frac{1}{2}(1^{-2})=\lim_{t\rightarrow \infty }\frac{-1}{2}t^{-2}+\frac{1}{2}=\frac{1}{2}$

The integration was done using the following rule:

$\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

but because this was definite integration, we plug in the upper limit and subtract from what we get from plugging in the lower limit.

When evaluating the limit, the term containing t is a fraction with infinity in the denominator, which equals zero.

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Example Question #9 : Improper Integrals

Evaluate the following integral:

$\int_{1}^{3}\frac{2}{x-1}dx$

Possible Answers:

$\infty$

$-\infty$

$\ln(2)$

Correct answer:

$\infty$

Explanation:

The integral is improper because of the lower limit of integration (creates $\frac{1}{0}$ which is equal to infinity).

Therefore, we must do the following:

$\lim_{t\rightarrow 1+}\int_{t}^{3}\frac{2}{x-1}dx$

We evaluate the limit from the right because 1 is the lower limit of integration.

Next, we move the constant 2 in front of the limit, and keeping the limit, integrate:

$2\lim_{t\rightarrow 1+}(\ln(2)-\ln\left | t-1\right |)$

The following rule was used in the integration:

$\int \frac{dx}{x}=\ln(x)+C$

Because this was a definite integral, we plug in the upper and lower limits into the function we get from integrating and subract the two (as shown above).

Finally, evaluate the limit:

$2\lim_{t\rightarrow 1+}(\ln(2)-\ln\left | t-1\right |)\equiv -(-\infty )=\infty$

The dominant term in the limit is when natural log approaches zero, which is negative infinity. There is a negative sign in front of this term, so it becomes positive infinity.

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Example Question #10 : Improper Integrals

Evaluate the improper integral

$\int_{-\infty}^{0}\frac{1}{\sqrt{2-x}}dx$

Possible Answers:

The integral diverges

$\frac{9}{2}$

$2\sqrt{2}$

Correct answer:

The integral diverges

Explanation:

A substitution makes evaluating this antiderivative easier. Let u=2-x , and du=-dx . We also need to rewrite the limits of integration in terms of . When $x=0,\ u=2$ and when $x\to -\infty,\ u\to \infty$ .

After making these substitutions, the integral becomes

$\int_{2}^{\infty}\frac{-1}{\sqrt{u}}du=-\int_{2}^{\infty}u^{-\frac{1}{2}}du$

Next the power rule is needed to find the antiderivative. The general case of the power rule is

$\int{x^n}dx=\frac{x^{n+1}}{n+1}+c$ .

Therefore, the integral evaluates to

$-\int_{2}^{\infty}u^{-\frac{1}{2}}du=-\frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right{|_{2}^{\infty}}$

$=\lim_{u\to \infty}-2u^{\frac{1}{2}}-(-2(2)^{\frac{1}{2}})=-\infty$

Since the integral does not approach a finite value, it diverges.

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Calculus 2 : Improper Integrals

Study concepts, example questions & explanations for Calculus 2

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Example Question #1 : Improper Integrals

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