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Calculus 2 : Integrals

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Example Questions

Example Question #92 : Riemann Sums

Compute the following integral by using four equal left hand Riemann sums:

$\int_{0}^{4}x^2dx.$

Possible Answers:

$\frac{64}{3}$

Correct answer:

Explanation:

Riemann sums are simply estimating the area under the curve of our function using an approximation of a rectangle. The area of a rectangle is given by A=length*width . Therefore, we need to figure both of these values for our four boxes and sum them all up. Since the boxes are equal, we can simply find the width by the following equation:

$width=\frac{b-a}{n}=\frac{4-0}{4}=1.$

and are your integration limits, and refers to the number of boxes.

The length of all of our boxes is given by the function. Since, this is a left Riemann sums, we need to consider the left side of each integration interval.

$\int_{0}^{4}x^2=0^2*1+1^2*1+2^2*1+3^2*1=0+1+4+9=14.$

For the lengths, we used -values of 0,1,2,3 , since those are the starting points of our four intervals.

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Example Question #91 : Riemann Sums

Approximate the following integral with four equal right Riemann sums:

$\int_{2}^{10}x^3dx.\textup{}$

Possible Answers:

3584

1592

796

1792

Correct answer:

3584

Explanation:

$width=\frac{b-a}{n}=\frac{10-2}{4}=2.$

and are your integration limits, and refers to the number of boxes.

The length of all of our boxes is given by the function. Since, this is a right Riemann sums, we need to consider the right side of each integration interval.

$\int_{2}^{10}x^3=4^3*2+6^3*2+8^3*2+10^3*2$

=2(4^3+6^3+8^3+10^3)=2(1792)=3584.

For the lengths, we used -values of 4,6,8,10 , since those are the ending points of our four intervals.

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Example Question #1 : Fundamental Theorem Of Calculus With Definite Integrals

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} t \ln t \; \mathrm{d}t$

Possible Answers:

$xe^{2x}$

$xe^{x }+ e$

2xe^x + 1

2xe^x

$xe^{x+1}$

Correct answer:

$xe^{2x}$

Explanation:

Set u = e^x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = e^x$ , and by the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} t \ln t \; \mathrm{d}t$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} t \ln t \; \mathrm{d}t$

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

By the fundamental theorem of Calculus, the above can be rewritten as

$e^x \cdot u \ln u$

$= e^x \cdot e^x \ln e^x$

$= e^{2x} \cdot x = xe^{2x}$

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Example Question #141 : Integrals

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{\ln x} t \log e^t \; \mathrm{d}t$

Possible Answers:

$\frac{ 2 \ln x }{ \ln 10}$

$\frac{\left (\ln x \right )^2}{ \ln 10}$

$\frac{\left (\ln x \right )^2}{x \ln 10}$

$\left (\ln x \right )^2$

$\frac{ 2 \ln x }{x \ln 10}$

Correct answer:

$\frac{\left (\ln x \right )^2}{x \ln 10}$

Explanation:

Set $u = \ln x$ . Then by the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{\ln x} t \log e^t \; \mathrm{d}t$

$= \frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} t \log e^t \; \mathrm{d}t$

$= \frac{1}{x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} t \log e^t \; \mathrm{d}t$

By the Fundamental Theorem of Calculus, the above is equal to

$= \frac{1}{x} \cdot u \log e^u$

$= \frac{1}{x} \cdot \ln x \cdot \log e^ {\ln x}$

$= \frac{1}{x} \cdot \ln x \cdot \log x$

$= \frac{1}{x} \cdot \ln x \cdot \frac{\ln x}{\ln 10}$

$= \frac{\left (\ln x \right )^2}{x \ln 10}$

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Example Question #142 : Integrals

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{10x} \cos \frac{\pi t}{4} \log t \; \mathrm{d}t$

Possible Answers:

$10 \cos \frac{ 5\pi x}{2}$

$10 \cos \frac{ 5\pi x}{2} \log x$

$10 \cos \frac{ 5\pi x}{2} \left (1 - \log x \right )$

$10 \cos \frac{ 5\pi x}{2} \left (-1 + \log x \right )$

$10 \cos \frac{ 5\pi x}{2} \left (1 + \log x \right )$

Correct answer:

$10 \cos \frac{ 5\pi x}{2} \left (1 + \log x \right )$

Explanation:

Set u = 10x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 10$ and by the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{10x} \cos \frac{\pi t}{4} \log t \; \mathrm{d}t$

$=\frac{\mathrm{d} u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{0}^{u} \cos \frac{\pi t}{4}\log t \; \mathrm{d}t$

$=10 \cos \frac{\pi u}{4} \log u$

$=10 \cos \frac{\pi \cdot 10x}{4} \log 10x$

$=10 \cos \frac{ 5\pi x}{2} \log 10x$

$=10 \cos \frac{ 5\pi x}{2} \left (\log 10 + \log x \right )$

$=10 \cos \frac{ 5\pi x}{2} \left (1 + \log x \right )$

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Example Question #1 : Fundamental Theorem Of Calculus

Evaluate f'(0) :

$f(x)=\int_{0}^{x} \sin\left(\sqrt{t^2+\frac{\pi^2}{4}} \right )dt$

Possible Answers:

$-\frac{\pi}{2}$

$\frac{\pi}{2}$

Correct answer:

Explanation:

By the Fundamental Theorem of Calculus, we have that ${f'(x) = \frac{d}{dx} \int_0^x \sin \sqrt{t^2 + \frac{\pi ^2}{4}} dt = \sin \sqrt{x^2 + \frac{\pi ^2}{4}}$ . Thus, $f'(0) = \sin \sqrt{0^2 + \frac{\pi ^2}{4}} =1$ .

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Example Question #143 : Integrals

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} (\ln t)^3 \; \mathrm{d}t$

Possible Answers:

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

3x^2 e^x

x^3 e^x

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

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

Correct answer:

x^3 e^x

Explanation:

Let u = e^x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = e^x$ , and by the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x} \int_{1}^{e^x} (\ln t)^3 \; \mathrm{d}t$

$=\frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \int_{1}^{u} (\ln t)^3 \; \mathrm{d}t$

$=\frac{\mathrm{d}u }{\mathrm{d} x} \cdot (\ln u)^3$

$=e^x \cdot (\ln e^x)^3$

$=e^x \cdot x^3 = x^3 e^x$

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Example Question #144 : Integrals

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{x^2} t^{3} \; \mathrm {d}t$

Possible Answers:

2x^8

x^7

x^6

2 x^6

2 x^7

Correct answer:

2 x^7

Explanation:

Set u = x^2 . Then $\frac{\mathrm{d}u }{\mathrm{d} x} = 2x$ .

By the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{x^2} t^{3} \; \mathrm {d}t$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} } {\mathrm{d}u } \int_{0}^{u} t^{3} \; \mathrm {d}t$

$=2x \cdot \frac{\mathrm{d} } {\mathrm{d}u } \int_{0}^{u} t^{3} \; \mathrm {d}t$

By the Fundamental Theorem of Calculus, the above is equal to

$2x \cdot u^3 = 2x \cdot (x^2)^3 = 2x \cdot x^6 = 2 x^7$ .

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Example Question #145 : Integrals

Find the result:

$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{4x} e^t \; \mathrm {d}t$

Possible Answers:

$e^{4x}$

$4 \ln x$

$\ln 4x$

$4e^{4x}$

$\frac{e^{4x}}{4}$

Correct answer:

$4e^{4x}$

Explanation:

Set u = 4t . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 4$ .

By the chain rule,

$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{4x} e^t \; \mathrm {d}t$

$= \frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}\int_{0}^{u} e^t \; \mathrm {d}t$

$= 4 \cdot \frac{\mathrm{d} }{\mathrm{d} u}\int_{0}^{u} e^t \; \mathrm {d}t$

By the Fundamental Theorem of Calculus, the above is equal to

$4 e^u = 4e^{4x}$ .

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Example Question #1 : Fundamental Theorem Of Calculus And Techniques Of Antidifferentiation

Evaluate f'(4) when $f(x)=\int_{0}^{x}(2t^{2}+5t-4)dt$ .

Possible Answers:

Correct answer:

Explanation:

Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_{0}^{x}(2t^{2}+5t-4)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(2t^{2}+5t-4)dt=2x^{2}+5x-4$ .

Therefore $f'(4)=2(4)^{2}+5(4)-4=32+20-4=52-4=48$ .

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #92 : Riemann Sums

Example Question #91 : Riemann Sums

Example Question #1 : Fundamental Theorem Of Calculus With Definite Integrals

Example Question #141 : Integrals

Example Question #142 : Integrals

Example Question #1 : Fundamental Theorem Of Calculus

Example Question #143 : Integrals

Example Question #144 : Integrals

Example Question #145 : Integrals

Example Question #1 : Fundamental Theorem Of Calculus And Techniques Of Antidifferentiation

All Calculus 2 Resources

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