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Calculus 2 : Integrals

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

Example Questions

Example Question #161 : Introduction To Integrals

Evaluate the definite integral using the Fundamental Theorem of Calculus.

$\int_{0}^{3}(x^4 - 3x^2)dx$

Possible Answers:

75.6

21.6

162

-26.6

Correct answer:

21.6

Explanation:

The antiderivative of f(x) = x^4 - 3x^2 is $F(x) = \frac{1}{5}x^5 - x^3$ .

Evaluating F(3) - F(0) (by the fundamental theorem of calculus) gives us...

$\\F(3) - F(0)\\ = 21.6 - 0\\ = 21.6$

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Example Question #161 : Introduction To Integrals

Solve

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

Possible Answers:

Correct answer:

Explanation:

The antiderivative of f(x) = x^3 + 1 is $F(x) = \frac{1}{4}x^4 + x$ .

Evaluating F(2) - F(-2) (by the fundamental theorem of calculus) gives us...

$\\F(2) - F(-2)\\ = 6-2\\ = 4$

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Example Question #161 : Integrals

Evaluate the definite integral using the Fundamental Theorem of Calculus.

$\int_{1}^{e^2}\left(\frac{1}{x} + 1\right)dx$

Possible Answers:

1-e^2

2+e^2

$1+\frac{e^3}{3}$

2-e^2

1+e^2

Correct answer:

1+e^2

Explanation:

The antiderivative of $f(x) = \frac{1}{x} + 1$ is $F(x) = ln\left | x\right | + x$ .

By evaluating F(e^2) - F(1) (by the fundamental theorem of calculus) we get...

F(e^2) - F(1)

= (2 + e^2) - (0+1)

= 1+ e^2

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Example Question #162 : Introduction To Integrals

Evaluate the definite integral using the Fundamental Theorem of Calculus.

$\int_{1}^{2}\left(2 - \frac{1}{x^2}\right)dx$

Possible Answers:

-2.5

1.5

-1.5

2.5

Correct answer:

1.5

Explanation:

The antiderivative of $f(x) = 2 - \frac{1}{x^2}$ is $F(x) = 2x + \frac{1}{x}$ .

By evaluating F(2) - F(1) (by the fundamental theorem of calculus) we get...

F(2) - F(1)

$= (4 + \frac{1}{2}) - (2+1)$

= 4.5 - 3

= 1.5

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Example Question #163 : Introduction To Integrals

Given $f(x)=\int_{0}^{x}(t^{2}-2t+5)dt$ , what is f'(4) ?

Possible Answers:

Correct answer:

Explanation:

According to the Fundamental Theorem of Calculus, if is a continuous function on the interval [a,b] with as the function defined for all on [a,b] as

$F(x)=\int_{a}^{x}f(t)dt$ , then F'(x)=f(x) .

Therefore, if

$f(x)=\int_{0}^{x}(t^{2}-2t+5)dt$ , then

$f'(x)=\frac{d}{dx}\int_{0}^{x}(t^{2}-2t+5)dt=x^{2}-2x+5$ .

Thus,

$f'(4)=4^{2}-2(4)+5=16-8+5=13$ .

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Example Question #13 : Fundamental Theorem Of Calculus

Given $f(x)=\int_{0}^{x}(3t^{2}+8t-6)dt$ , what is f'(6) ?

Possible Answers:

130

110

150

140

120

Correct answer:

150

Explanation:

According to the Fundamental Theorem of Calculus, if is a continuous function on the interval [a,b] with as the function defined for all on [a,b] as

$F(x)=\int_{a}^{x}f(t)dt$ , then F'(x)=f(x) .

Therefore, if

$f(x)=\int_{0}^{x}(3t^{2}+8t-6)dt$ , then

$f'(x)=\frac{d}{dx}\int_{0}^{x}(3t^{2}+8t-6)dt=3x^{2}+8x-6$ .

Thus,

$f'(6)=3(6)^{2}+8(6)-6=108+48-6=156-6=150$ .

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Example Question #164 : Introduction To Integrals

Given $f(x)=\int_{0}^{x}(-2t^{2}+4t+7)dt$ , what is f'(1) ?

Possible Answers:

Correct answer:

Explanation:

According to the Fundamental Theorem of Calculus, if is a continuous function on the interval [a,b] with as the function defined for all on [a,b] as

$F(x)=\int_{a}^{x}f(t)dt$ , then F'(x)=f(x) .

Therefore, if

$f(x)=\int_{0}^{x}(-2t^{2}+4t+7)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(-2t^{2}+4t+7)dt=-2t^2+4t+7$ .

Thus,

f'(1)=-2(1)^2+4(1)+7=9 .

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Example Question #165 : Introduction To Integrals

Write $P_{final}-P_{initial}$ in integral form, if is position and P'(t)=v(t) where v(t) is velocity at time .

Possible Answers:

$P_{final}-P_{initial}=\int_{initial}^{final}v(t)dt$

$P_{final}-P_{initial}=\int_{initial}^{final}P(t)v(t)dt$

$P_{final}-P_{initial}=\frac{1}{2}\int_{initial}^{final}v(t)dt$

$P_{final}-P_{initial}=\int_{final}^{initial}v(t)dt$

Correct answer:

$P_{final}-P_{initial}=\int_{initial}^{final}v(t)dt$

Explanation:

To write position in integral form, we can take advantage of the fundamental theorem of calculus. Since the bounds are initial and final , and P'(t)=v(t) ,

$P_{final}-P_{initial}=\int_{initial}^{final}v(t)dt$

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Example Question #161 : Integrals

Given that F'(x)=f(x) , determine:

$\int_{0}^{\infty}f'(x)dx$

Possible Answers:

$F(\infty)-f(0)$

$F(\infty)-F(0)$

$F(0)-F(\infty)$

$f(\infty)-f(0)$

Correct answer:

$F(\infty)-F(0)$

Explanation:

Since F'(x)=f(x) , we know that

$\int f(x)dx=F(x)+C$

By the fundamental theorem of calculus:

$\int_{0}^{\infty} f(x)dx=[F(\infty)+C]-[F(0)+C]=F(\infty)-F(0)$

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Example Question #167 : Introduction To Integrals

Find $\frac{dy}{dx}$ of $\int_{0}^{8lnx}e^tdt.$

Possible Answers:

8x^8

x^8

$e^{x^{8}}$

$e^{x^{7}}$

8x^7

Correct answer:

8x^7

Explanation:

This is a Second Fundamental Theorem of Calculus. Since derivatives and anti-derivatives annihilate each other, we simply need to plug in the bounds into the function and multiply each by their derivative, respectively.

$\frac{dy}{dx} \int_{0}^{8lnx} e^tdt=e^{8lnx}*(8lnx)'-e^0*(0)'.$

The second term drops out since the derivative of zero is zero.

In the first term, we again have two functions that annihilate each other:

$e^{8lnx}*(8lnx)'=e^{lnx^8}*\frac{8}{x}={x^8}*\frac{8}{x}=\frac{8x^8}{x}=8x^7.$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #161 : Introduction To Integrals

Example Question #161 : Introduction To Integrals

Example Question #161 : Integrals

Example Question #162 : Introduction To Integrals

Example Question #163 : Introduction To Integrals

Example Question #13 : Fundamental Theorem Of Calculus

Example Question #164 : Introduction To Integrals

Example Question #165 : Introduction To Integrals

Example Question #161 : Integrals

Example Question #167 : Introduction To Integrals

All Calculus 2 Resources

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