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Calculus 2 : Definite Integrals

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Example Questions

Example Question #2261 : Calculus Ii

$\int_{0}^{5}\frac{x^2+x-1}{x}dx$

Possible Answers:

15.8986

15.1906

15.8906

18.5906

15.8726

Correct answer:

15.8906

Explanation:

First, chop up the fraction into three separate terms:

$\int x+1-\frac{1}{x}dx$

Next, integrate:

$\frac{x^2}{2}+x-ln\left | x \right |$

Evaluate at 5 and then 0. Subtract the results:

$(\frac{25}{2}+5-ln5)-(0)=15.8906$

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Example Question #2262 : Calculus Ii

What is the integral of $\int_{4}^{8} \frac{2}{x^2}dx$ ?

Possible Answers:

$\frac{3}{4}$

$\frac{1}{2}$

Correct answer:

$\frac{3}{4}$

Explanation:

Rewriting the integral as an easy-to-integrate indefinite integral gives us $\int 2(x^{-2})$ , for which we can just use the power rule to get: $-2x^{-1}=-\frac{2}{x}$ .

Plugging in the values gives us $-\frac{2}{8}+\frac{2}{4}=-\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$ .

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Example Question #2263 : Calculus Ii

Which of the following is $\int_{0}^{1} x^{\tfrac{1}{3}}dx$ ?

Possible Answers:

$\frac{3}{4}$

$\frac{1}{3}$

$\frac{4}{3}$

Correct answer:

$\frac{3}{4}$

Explanation:

If we solve for the indefinite integral $\int x^\frac{1}{3}$ using the power rule, we get $\frac{3}{4}x^\frac{4}{3}$ . If we then plug in the values, we get $\frac{3}{4}(1)^\frac{4}{3}-\frac{3}{4}(0)^\frac{4}{3}=\frac{3}{4}(1)-\frac{3}{4}(0)=\frac{3}{4}-0=\frac{3}{4}$ .

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Example Question #2264 : Calculus Ii

Evaluate the definite integral

$\int_0^{\pi/4}sec^2x\,dx$

Possible Answers:

$\frac{1}{2}$

$\sqrt{2}$

Correct answer:

Explanation:

The antiderivative of sec^2x is $tan\,x$ .

Using the Fundamental Theorem of Calculus,

$\int_0^{\pi/4}sec^2x\,dx=tan\,x|_0^{\pi/4}$

$=tan\,(\frac{\pi}{4})-tan(0)$

=1-0

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Example Question #2265 : Calculus Ii

Evaluate the definite integral

$\int_0^{1}\frac{1}{x^2+1}\,dx$

Possible Answers:

$\pi$

$\frac{3}{2}$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

Correct answer:

$\frac{\pi}{4}$

Explanation:

For this problem we use the fact that

$\int_a^b\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}(x)|_a^b$

As such,

$\int_0^1\frac{1}{x^2+1}dx=\frac{1}{1}tan^{-1}(x)|_0^1$

$=tan^{-1}(x)|_0^1$

$=tan^{-1}(1)-tan^{-1}(0)$

$=\frac{\pi}{4}-0$

$=\frac{\pi}{4}$

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Example Question #2266 : Calculus Ii

Evaluate

$\small \int_0^1 e^{5x+1}dx$

Possible Answers:

$\small \frac{1}{5}(e^6-e^2)$

$\small \frac{1}{5}(e^6-e)$

$\small \frac{1}{5}(e^6-e^5)$

$\small \frac{1}{5}(e^5-e)$

Correct answer:

$\small \frac{1}{5}(e^6-e)$

Explanation:

To evaluate

$\small \int_0^1 e^{5x+1}dx$

we take its antiderivative $\small F(x)$ and calculate $\small F(1)-F(0)$ . With $\small F(x)=\frac{1}{5}e^{5x+1}$ , we get

$\small \small \int_0^1 e^{5x+1}dx=\frac{1}{5}(e^{5+1}-e^{0+1})=\frac{1}{5}(e^6-e)$

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Example Question #2267 : Calculus Ii

Evaluate

$\small \small \int_0^\pi 6x^2dx$

Possible Answers:

$\small 6\pi^2$

$6\pi^3$

$\small 3\pi^2$

$2\pi^3$

Correct answer:

$2\pi^3$

Explanation:

To evaluate

$\small \small \int_0^\pi 6x^2dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small F(\pi)-F(0)$ . With $\small \small F(x)=2x^3$ , we get

$\small \small \small \int_0^\pi 6x^2dx=2(\pi^3-0^3)=2\pi^3$

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Example Question #151 : Definite Integrals

Evaluate

$\small \small \small \small \small \int_2^\pi e^{\pi x-1}dx$

Possible Answers:

$\frac{1}{\pi}(e^{\pi^2-1}-e^{2\pi-1})$

$\small \frac{1}{\pi}e^{2\pi-1}$

$\small e^{\pi^2-1}-e^{2\pi-1}$

$\frac{1}{\pi}e^{\pi^2-1}$

Correct answer:

$\frac{1}{\pi}(e^{\pi^2-1}-e^{2\pi-1})$

Explanation:

To evaluate

$\small \small \small \small \small \int_2^\pi e^{\pi x-1}dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small F(\pi)-F(2)$ . With $\small \small \small \small F(x)=\frac{e^{\pi x-1}}{\pi}$ , we get

$\small \small \small \int_2^\pi e^{\pi x-1}dx=\frac{1}{\pi}(e^{\pi^2-1}-e^{2\pi-1})$

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Example Question #152 : Definite Integrals

Evaluate

$\small \small \small \int_1^{e^2} \frac{3}{x}dx$

Possible Answers:

2e^2

$\small 4$

$\small 2(e^2-1)$

Correct answer:

Explanation:

To evaluate

$\small \small \small \int_1^{e^2} \frac{3}{x}dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small F(e^2)-F(1)$ . With $\small \small F(x)=3\ln x$ , we get

$\small \small \small \small \int_0^{e^2} \frac{3}{x}dx=3(\ln e^2-\ln 1)=3(2-0)=6$

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Example Question #153 : Definite Integrals

Evaluate

$\small \small \small \small \int_5^6 x^2dx$

Possible Answers:

$\frac{92}{3}$

$\frac{91}{3}$

Correct answer:

$\frac{91}{3}$

Explanation:

To evaluate

$\small \small \small \small \int_5^6 x^2dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small F(6)-F(5)$ . With $\small \small \small F(x)=\frac{x^3}{3}$ , we get

$\small \small \int_5^{6} x^2dx=\frac{1}{3}(6^3-5^3)=\frac{1}{3}(216-125)=\frac{91}{3}$

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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2261 : Calculus Ii

Example Question #2262 : Calculus Ii

Example Question #2263 : Calculus Ii

Example Question #2264 : Calculus Ii

Example Question #2265 : Calculus Ii

Example Question #2266 : Calculus Ii

Example Question #2267 : Calculus Ii

Example Question #151 : Definite Integrals

Example Question #152 : Definite Integrals

Example Question #153 : Definite Integrals

All Calculus 2 Resources

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