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Calculus 2 : Fundamental Theorem of Calculus

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Example Questions

Example Question #2 : Fundamental Theorem Of Calculus With Definite Integrals

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 #2 : Fundamental Theorem Of Calculus

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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Example Question #42 : Integrals

Evaluate f'(1) when $f(x)=\int_{0}^{x}(6t^{2}-3t+10)dt$ .

Possible Answers:

Correct answer:

Explanation:

Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_{0}^{x}(6t^{2}-3t+10)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(6t^{2}-3t+10)dt=6x^{2}-3x+10$ . Therefore, $f'(1)=6(1)^{2}-3(1)+10=6-3+10=3+10=13$ .

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Example Question #2 : Fundamental Theorem Of Calculus

Evaluate f'(7) when $f(x)=\int_{0}^{x}(t^{2}-2t+5)dt$ .

Possible Answers:

Correct answer:

Explanation:

Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_{0}^{x}(t^{2}-2t+5)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(t^{2}-2t+5)dt=x^{2}-2x+5$ . Therefore, $f'(7)=(7)^{2}-2(7)+5=49-14+5=35+5=40$ .

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Example Question #2 : Fundamental Theorem Of Calculus

Suppose we have the function

$\small g(x)=\int_{0}^{\cos x}(1+\sin t)dt$

What is the derivative, $\small g'(x)$ ?

Possible Answers:

$\small \small \small \small \sin x\left[1+\sin\left(\cos x \right )\right]$

$\small \small \small 1+\sin\left(\cos(x) \right )$

$\small \small \small 1-\sin\left(\cos(x) \right )$

$\small \small -\sin x\left[1+\sin\left(\cos(x) \right )\right]$

Correct answer:

$\small \small -\sin x\left[1+\sin\left(\cos(x) \right )\right]$

Explanation:

We can view the function $\small g(x)$ as a function of $\small \cos x$ , as so

$\small g(x)=F(\cos x)$

where $\small F(x)=\int_0^x (1+\sin t)dt$ .

We can find the derivative of $\small g(x)$ using the chain rule:

$\small g'(x)=F'(\cos x)\cdot (\cos x)'=F'(\cos x)\cdot (-\sin x)$

where $\small F'(z)$ can be found using the fundamental theorem of calculus:

$\small \small F'(x)=\frac{d}{dz}\int_0^z (1+\sin t)dt=1+\sin z$

So we get

$\small \small \small \small g'(x)=-\sin x\cdot F'(\cos x)=-\sin x\cdot[1+\sin(\cos x)]$

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Example Question #1 : Fundamental Theorem Of Calculus

If $f(x)=\int_{0}^{x}(4t^{2}-6t+9)dt$ , what is f'(3) ?

Possible Answers:

Correct answer:

Explanation:

By the Fundamental Theorem of Calculus, we know that given a function $f(x)=\int_{0}^{x}(4t^{2}-6t+9)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(4t^{2}-6t+9)dt=4x^{2}-6x+9$ .

Thus, $f'(3)=4(3)^{2}-6(3)+9=36-18+9=18+9=27$

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Example Question #1 : Fundamental Theorem Of Calculus

If $f(x)=\int_{0}^{x}(t^{2}+7t-12)dt$ , what is f'(2) ?

Possible Answers:

Correct answer:

Explanation:

By the Fundamental Theorem of Calculus, we know that given a function $f(x)=\int_{0}^{x}(t^{2}+7t-12)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(t^{2}+7t-12)dt=t^{2}+7t-12$ .

Thus, $f'(2)=(2)^{2}+7(2)-12=4+14-12=4+2=6$

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Example Question #2 : Fundamental Theorem Of Calculus

If $f(x)=\int_{0}^{x}(8t^{2}-t+14)dt$ , what is f'(5) ?

Possible Answers:

189

209

219

199

229

Correct answer:

209

Explanation:

By the Fundamental Theorem of Calculus, we know that given a function $f(x)=\int_{0}^{x}(8t^{2}-t+14)dt$ , $f'(x)=\frac{d}{dx}\int_{0}^{x}(8t^{2}-t+14)dt=8x^{2}-x+14$ .

Thus, $f'(5)=8(5)^{2}-(5)+14=200-5+14=200+9=209$

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Example Question #2 : Fundamental Theorem Of Calculus

Given

$f(x)=\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt$ , what is f'(6) ?

Possible Answers:

None of the above.

$-\frac{36}{323}$

$\frac{323}{36}$

$\frac{36}{323}$

$-\frac{323}{36}$

Correct answer:

$-\frac{323}{36}$

Explanation:

By the Fundamental Theorem of Calculus, for all functions that are continuously defined on the interval [a,b] with in [a,b] and for all functions defined by by $F(x)=\int_{a}^{x}f(t)dt$ , we know that F(x)=f'(x) .

Thus, for

$f(x)=\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt$ ,

$f'(x)=\frac{d}{dx}\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt=\frac{1}{x^{2}}-2x+3$ .

Therefore,

$f'(6)=\frac{1}{6^{2}}-2(6)+3=\frac{1}{36}-12+3=\frac{1}{36}-9=\frac{1}{36}-\frac{324}{36}=-\frac{323}{36}$

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Example Question #161 : Ap Calculus Bc

Given

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

Possible Answers:

$-\frac{2}{7}$

$-\frac{7}{2}$

$\frac{7}{2}$

$\frac{2}{7}$

None of the above.

Correct answer:

$\frac{2}{7}$

Explanation:

Given

$f(x)=\int_{0}^{x}\left(\frac{2}{t^{2}+4t-5}\right)dt$ , then

$f'(x)=\frac{d}{dx}\int_{0}^{x}\left(\frac{2}{t^{2}+4t-5}\right)dt=\frac{2}{x^{2}+4x-5}$ .

Therefore,

$f'(2)=\frac{2}{2^{2}+4(2)-5}=\frac{2}{4+8-5}=\frac{2}{4+3}=\frac{2}{7}$ .

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Calculus 2 : Fundamental Theorem of Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Fundamental Theorem Of Calculus With Definite Integrals

Example Question #2 : Fundamental Theorem Of Calculus

Example Question #42 : Integrals

Example Question #2 : Fundamental Theorem Of Calculus

Example Question #2 : Fundamental Theorem Of Calculus

Example Question #1 : Fundamental Theorem Of Calculus

Example Question #1 : Fundamental Theorem Of Calculus

Example Question #2 : Fundamental Theorem Of Calculus

Example Question #2 : Fundamental Theorem Of Calculus

Example Question #161 : Ap Calculus Bc

All Calculus 2 Resources

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