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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #2781 : Calculus Ii

Evaluate the integral:

$\int xcos(\frac{1}{2}x^2)dx$

Possible Answers:

$\frac{1}{2}sin(\frac{1}{2}x^2)+C$

$sin(\frac{1}{2}x^2)+C$

$cos(\frac{1}{2}x)+C$

2sin(x^2)+C

2sin(x)+C

Correct answer:

$sin(\frac{1}{2}x^2)+C$

Explanation:

You must use u substitution to solve this problem.

In this case

$u=\frac{1}{2}x^2$

and

du=xdx .

So the integral simplifies to

$\int cos(u)du=sin(u)+C=sin(\frac{1}{2}x^2)+C$

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Example Question #1032 : Integrals

Calculate the following integral:

$\int_{1}^{9}x\sqrt[3]{x-1}dx$

Possible Answers:

384/7

125/4

572

468/7

Correct answer:

468/7

Explanation:

Use $t=\sqrt[3]{x-1}$ as a substitute, then:

$t^3=x-1\:\Rightarrow\:x=t^3+1\:\Rightarrow\:dx=3t^2dt$ ;

Now, rewrite the boundaries of integration in terms of t:

$\newline t_1=\sqrt[3]{x_1-1}=\sqrt[3]{1-1}=0;\newline t_2=\sqrt[3]{x_2-1}=\sqrt[3]{9-1}=2;$

Rewrite the integral in terms of t:

$\\\int_{1}^{9}x\sqrt[3]{x-1}dx\\\\=\int_{0}^{2}(t^3+1)t\:3t^2\:dt\\\\=3\int_{0}^{2}t^6\:dt +3\int_{0}^{2}t^3\:dt\\\\=\frac{3*128}{7}+\frac{3*16}{4}\\\\=\frac{468}{7}$

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Example Question #681 : Finding Integrals

Evaluate the following integral:

$\int \frac{e^{\frac{1}{x}}}{x^2}dx$

Possible Answers:

$e^{\frac{1}{x^2}}+c$

$e^{\frac{1}{x}}+c$

$\frac{1}{e^x}+c$

$-e^{\frac{1}{x}}+c$

$2x+e^{\frac{1}{x}}+c$

Correct answer:

$-e^{\frac{1}{x}}+c$

Explanation:

We must use substitution to solve this integrand.

Let

$u=\frac{1}{x}$

and

$du=-\frac{1}{x^2}dx$

don't forget to divide buy -1 to isolate the right side

$-du=\frac{1}{x^2}dx$

putting all the values back in and pulling the negative sign out of the integral we get:

$\int \frac{e^{\frac{1}{x}}}{x^2}dx=-\int e^udu=-e^u+c=-e^{\frac{1}{x}}+c$

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Example Question #1 : Geometric Series

Consider: $\sum_{n=0}^{\infty} \left (\frac{\sqrt 2}{2}\right )^n$ . Will the series converge or diverge? If converges, where does this coverge to?

Possible Answers:

$2-\sqrt{2}$

Diverge

$\sqrt{2}$

$2+\sqrt2$

$\sqrt{2}-2$

Correct answer:

$2+\sqrt2$

Explanation:

This is a geometric series. Use the following formula, where is the first term of the series, and is the ratio that must be less than 1. If is greater than 1, the series diverges.

$\frac{a}{1-r} = \frac{1}{1-\frac{\sqrt2}{2}}= \frac{1}{\frac{2-\sqrt2}{2}} = \frac{2}{2-\sqrt2}$

Rationalize the denominator.

$\frac{2}{2-\sqrt2} \cdot \frac{2+\sqrt2}{2+\sqrt2} = \frac{2(2+\sqrt2)}{4-2}= 2+\sqrt2$

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Example Question #1 : Series In Calculus

Consider the following summation: $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...$ . Does this converge or diverge? If it converges, where does it approach?

Possible Answers:

$\frac{7}{4}$

Diverges

1.33

1.35

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

The problem can be reconverted using a summation symbol, and it can be seen that this is geometric.

$1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...=\sum_{n=0}^{\infty} \left ( \frac{1}{4} \right )^n$

Since the ratio is less than 1, this series will converge. The formula for geometric series is:

$\frac{a}{1-r}$

where is the first term, and is the common ratio. Substitute these values and solve.

$\frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

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Example Question #3 : Geometric Series

A worm crawls up a wall during the day and slides down slowly during the night. The first day the worm crawls one meter up the wall. The first night the worm slides down a third of a meter. The second day the worm regains one third of the lost progress and slides down one third of that distance regained on the second night. This pattern of motion continues...

Which of the following is a geometric sum representing the distance the worm has travelled after 12-hour periods of motion? (Assuming day and night are both 12 hour periods).

Possible Answers:

$\sum_{k=0}^{n-1} (\frac{-1}{3})^{k}$

$\sum_{k=0}^{\infty} (\frac{-1}{3})^{k}$

$\sum_{k=0}^{n} (\frac{1}{3})^{k}$

$1+ \sum_{k=0}^{n} (\frac{-1}{3})^{k}$

$1+\sum_{k=1}^{n} (\frac{-1}{3})^{k}$

Correct answer:

$\sum_{k=0}^{n-1} (\frac{-1}{3})^{k}$

Explanation:

The sum must be alternating, and after one period you should have the worm at 1m. After two periods, the worm should be at 2/3m. There is only one sum for which that is true.

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Example Question #1 : Series In Calculus

Which of the following is a correct representation of the sum $\sum_{k=4}^{\infty} \frac{10}{4^{k}}$ in +... notation?

Possible Answers:

$10\left(\frac{1}{4^{4}}+\frac{1}{4^{5}}+\frac{1}{4^{6}}+...\right)$

$\frac{10}{4^{1}}+\frac{10}{4^{2}}+\frac{10}{4^{3}}+\frac{10}{4^{4}}$

$10(1+\frac{10}{4^{4}}+\frac{10}{4^{16}}+\frac{10}{4^{64}}+...)$

$\frac{10}{4^{1}}+\frac{10}{4^{2}}+\frac{10}{4^{3}}+...$

Correct answer:

$10\left(\frac{1}{4^{4}}+\frac{1}{4^{5}}+\frac{1}{4^{6}}+...\right)$

Explanation:

By writing out the first few terms and factoring out a 10, we can arrive at a correct representation of the sum in +... notation.

The sum of the first term where k=4 is $\frac{10}{4^4}$ .

The term when k=5 is $\frac{10}{4^5}$ .

The term when k=6 is $\frac{10}{4^6}$ .

Creating the summation we get,

$\frac{10}{4^4}+\frac{10}{4^5}+\frac{10}{4^6}+...=10\left(\frac{1}{4^4}+\frac{1}{4^5}+\frac{1}{4^6}+... \right )$

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Example Question #2 : Series In Calculus

One technique for determining whether a series converges is comparing it to a related improper integral.

Which of the following is the most correct proof of convergence for the series $\sum^{\infty}_{k=1} \frac{1}{k^{3/2}}$ ?

Possible Answers:

$$Since$ \int_{1}^{\infty} \frac{1}{x^{3/2}}dx$ $$converges and $ \int_{1}^{\infty} \frac{1}{x^{3/2}}dx \leq \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} $ then the sum $ \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} $ must converge.$

$$Since$ \int_{1}^{\infty} \frac{1}{x^{3/2}}dx$ $$converges and $ 0\leq \int_{0}^{\infty} \frac{1}{x^{3/2}}dx \leq \sum_{k=0}^{\infty} \frac{1}{k^{3/2}} $ then the sum $ \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} $ must converge.$

$$Since$ \int_{1}^{\infty} \frac{1}{x^{3/2}}dx$ $$converges and $0\leq \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \leq \int_{1}^{\infty} \frac{1}{x^{3/2}}dx $ then the sum $ \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} $ must converge.$

$$Since$ \int_{1}^{\infty} \frac{1}{x^{3/2}}dx$ $$converges and $0\leq \int_{1}^{\infty} \frac{1}{x^{3/2}}dx \leq \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} $ then the sum $ \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} $ must converge.$

Correct answer:

Explanation:

In the spirit of the direct comparison test, we can compare the convergent improper integral to the infinite sum, such that the sum is less than the convergent integral by interpreting the sum as a right hand Riemann sum. Note that you can depict the sum as either, with the property that in one case you get information about its convergence or divergence and in the other case you cannot say anything. We want to make sure for correctness that we assert the sum and integral are both positive, as a condition for making the comparison.

By definition of the Comparison Test

Given that $a_k\leq b_k$ where a_k, b_k>0 .

If $\sum b_k$ converges, then $\sum a_k$ also converges.

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Example Question #1 : Series In Calculus

To what value does the sum $\sum_{k=4}^{\infty}\frac{10}{4^{k}}$ converge?

Possible Answers:

$\frac{4}{3}$

$\frac{1}{.75}$

$\frac{10}{3\cdot4^{3}}$

$\frac{3}{4}$

Correct answer:

$\frac{10}{3\cdot4^{3}}$

Explanation:

The first term of the series is $\frac{10}{4^{4}}$ , and the common ratio is $\frac{1}{4}$ , so we can rewrite the series in the form $\sum_{k=0}^{\infty} ar^{k}$ and use the formula $\frac{a}{1-r}$ to arrive at the correct answer.

$=\frac{\frac{10}{4^4}}{1-\frac{1}{4}}=\frac{\frac{10}{4^4}}{\frac{3}{4}}=\frac{10}{4^4}\cdot \frac{4}{3}=\frac{10}{4^3\cdot 3}$

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Example Question #1 : Series In Calculus

Given that $\sum_{k=1}^{\infty} a_{k}$ converges, what can you say about $\sum_{k=1}^{\infty} b_{k}$ where $b_{k} = \frac{a_{k}}{2}$ and $a_{k},b_{k}\geq0$ .

Possible Answers:

Not enough information.

The sum is zero.

The sum must converge.

The sum must diverge.

Correct answer:

The sum must converge.

Explanation:

The sum must converge by the direct comparison test. All terms are positive and $b_{k}$ is smaller than $a_{k}$ for all k. Therefore by definition the sum must converge.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2781 : Calculus Ii

Example Question #1032 : Integrals

Example Question #681 : Finding Integrals

Example Question #1 : Geometric Series

Example Question #1 : Series In Calculus

Example Question #3 : Geometric Series

Example Question #1 : Series In Calculus

Example Question #2 : Series In Calculus

Example Question #1 : Series In Calculus

Example Question #1 : Series In Calculus

All Calculus 2 Resources

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