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Calculus 2 : Definite Integrals

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Example Questions

Example Question #161 : Definite Integrals

Evaluate

$\small \small \small \small \small \small \int_2^5 \frac{1}{x}dx$

Possible Answers:

$\ln\frac{5}{2}$

$\small \ln 3$

$\small \frac{\ln 5}{\ln 2}$

$\small \ln 5$

Correct answer:

$\ln\frac{5}{2}$

Explanation:

To evaluate

$\small \small \small \small \small \small \int_2^5 \frac{1}{x}dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small F(5)-F(2)$ . With $\small \small \small \small \small F(x)=\ln x$ , we get

$\small \small \small \small \small \int_2^5 \frac{1}{x}dx=\ln 5-\ln 2=\ln \frac{5}{2}$

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Example Question #162 : Definite Integrals

Evaluate

$\small \small \small \int_{-1}^1 5x^5 dx$

Possible Answers:

$\small \frac{25}{2}$

$\small 2$

$\small 0$

$\small 10$

Correct answer:

$\small 0$

Explanation:

To evaluate

$\small \small \small \int_{-1}^1 5x^5 dx$

we can see that $\small f(x)=5x^5$ is odd (what this means is that $\small f(-x)=-f(x)$ ), which means that the integral evaluates to $\small 0$ , so we get

$\small \small \small \small \int_{-1}^1 5x^5 dx=0$

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Example Question #163 : Definite Integrals

Evaluate

$\small \small \int_1^3 4x dx$

Possible Answers:

$\small 8$

$\small 18$

$\small 16$

$\small 14$

Correct answer:

$\small 16$

Explanation:

To evaluate

$\small \small \int_1^3 4x dx$

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

$\small \small \int_1^3 4x dx=2(3^2-1^2)=2(9-1)=16$

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Example Question #164 : Definite Integrals

Evaluate

$\small \small \small \int_0^1 \pi x dx$

Possible Answers:

$\small 4\pi$

$\small \small \pi$

$\small \frac{\pi}{2}$

$\small 2\pi$

Correct answer:

$\small \frac{\pi}{2}$

Explanation:

To evaluate

$\small \small \small \int_0^1 \pi x dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small \small \small F(1)-F(0)$ . With $\small \small \small \small \small \small \small F(x)=\frac{\pi}{2}x^2$ , we get

$\small \small \small \int_0^1 \pi x dx=\frac{\pi}{2}(1^2-0^2)=\frac{\pi}{2}$

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Example Question #165 : Definite Integrals

Evaluate

$\small \small \small \small \int_0^1 \frac{3}{x+1} dx$

Possible Answers:

$\small \small 2\ln 3$

$\small \small \ln \sqrt{8}$

$\small 3\ln 2$

$\small \small \ln 2^{\frac{1}{3}}$

Correct answer:

$\small 3\ln 2$

Explanation:

To evaluate

$\small \small \small \small \int_0^1 \frac{3}{x+1} dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small \small \small F(1)-F(0)$ . With $\small \small \small \small \small \small \small \small F(x)=3\ln(x+1)$ , we get

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

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Example Question #166 : Definite Integrals

Evaluate

$\small \small \int_0^1 ex^2 dx$

Possible Answers:

$\frac{e^3}{3}$

$\frac{e}{2}$

$\frac{e}{3}$

$\frac{e^2}{3}$

Correct answer:

$\frac{e}{3}$

Explanation:

To evaluate

$\small \small \int_0^1 ex^2 dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small \small \small F(1)-F(0)$ . With $\small \small F(x)=\frac{e}{3}x^3$ , we get

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

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Example Question #167 : Definite Integrals

Evaluate

$\small \small \small \int_0^1 e^2x^3 dx$

Possible Answers:

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

$\small \small \frac{e^2}{5}$

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

$\small \small \frac{e^2}{3}$

Correct answer:

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

Explanation:

To evaluate

$\small \small \small \int_0^1 e^2x^3 dx$

we take its antiderivative $\small F(x)$ and calculate $\small \small \small \small \small F(1)-F(0)$ . With $\small \small \small F(x)=\frac{e^2}{4}x^4$ , we get

$\small \small \small \int_0^1 e^2x^3dx=\frac{e^2}{4}(1^4-0^4)=\frac{e^2}{4}$

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Example Question #2281 : Calculus Ii

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

Possible Answers:

Correct answer:

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{3x^3}{3}-\frac{x^2}{2}+x=x^3-\frac{x^2}{2}+x$

Now, evaluate at 3 and then 1. Subtract the results:

$(27-\frac{9}{2}+3)-(1-\frac{1}{2}+1)=28-4=24$

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Example Question #2282 : Calculus Ii

$\int_{0}^{2}x^{\frac{3}{2}}dx$

Possible Answers:

2.2627

2.2692

2.4727

3.2627

2.2197

Correct answer:

2.2627

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^{\frac{5}{2}}}{\frac{5}{2}}$

Simplify:

$\frac{2}{5}x^{\frac{5}{2}}$

Now evaluate at 2 and then 0. Subtract the results:

$\frac{2}{5}(2^{\frac{5}{2}})-\frac{2}{5}(o^{\frac{5}{2}})=2.2627-0=2.2627$

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Example Question #168 : Definite Integrals

$\int_{1}^{3}3x^2+2x+2dx$

Possible Answers:

Correct answer:

Explanation:

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{3x^3}{3}+\frac{2x^2}{2}+2x$

Simplify:

x^3+x^2+2x

Evaluate at 3 and then at 1. Subtract the results:

(27+9+6)-(1+1+2)=42-4=38

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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #161 : Definite Integrals

Example Question #162 : Definite Integrals

Example Question #163 : Definite Integrals

Example Question #164 : Definite Integrals

Example Question #165 : Definite Integrals

Example Question #166 : Definite Integrals

Example Question #167 : Definite Integrals

Example Question #2281 : Calculus Ii

Example Question #2282 : Calculus Ii

Example Question #168 : Definite Integrals

All Calculus 2 Resources

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