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Calculus 2 : Fundamental Theorem of Calculus

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Example Questions

Example Question #161 : Integrals

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

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

Possible Answers:

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

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

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

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

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 #162 : Integrals

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

Possible Answers:

$e^{x^{8}}$

8x^7

$e^{x^{7}}$

8x^8

x^8

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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Example Question #21 : Fundamental Theorem Of Calculus

$\frac{dy}{dx} \int_{0}^{sinx}\sqrt{t}dt$

Possible Answers:

$cosx\sqrt{sinx}$

$\frac{-cosx}{2\sqrt{sinx}}$

$sinx\sqrt{cosx}$

$\frac{sinx}{2\sqrt{cosx}}$

$\frac{cosx}{2\sqrt{sinx}}$

Correct answer:

$cosx\sqrt{sinx}$

Explanation:

This is a Second Fundamental Theorem of Calculus problem. Since the derivative cancels out the integral, we just need to plug in the bounds into the function (top bound - bottom bottom) and multiply each by their derivative.

$\frac{dy}{dx} \int_{0}^{sinx}\sqrt{t}dt=\sqrt{sinx}*(sinx)'-\sqrt(0)*()0)'$

$=\sqrt{sinx}*(cosx)$ .

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Example Question #21 : Fundamental Theorem Of Calculus

Differentiate:

$F(x)=\int_{0}^{x^2+2x-1}\sin(t)dt$

Possible Answers:

F'(x)=(2x+2)cos(x^2+2x-1)

$F'(x)=(2x+2)\sin(x^2+2x-1)$

$F'(x)= - \cos(x^2+2x+1)-1$

$F'(x)=\sin(x^2+2x-1)$

$F'(x)= -\cos(x^2+2x-1)$

Correct answer:

$F'(x)=(2x+2)\sin(x^2+2x-1)$

Explanation:

Differentiate:

$F(x)=\int_{0}^{x^2+2x-1}\sin(t)dt$

Use the Second Fundamental Theorem of Calculus:

________________________________________________________

If the function is continuous on an open interval and if is in , then for a function defined by $F(x)=\int_a^{x}f(t)dt$ , we have .

________________________________________________________________

To understand how to apply the second FTOC, notice that F(x) in our case is a function of a function. The chainrule will therefore have to be applied as well. To see how this works, let u = x^2+2x-1 . The function can now be written in the form:

$\frac{du}{dx}\times \underset{\sin(u)}{ \underbrace{\frac{d}{du}\int_0^u\sin(t)dt}}=\frac{du}{dx}\sin(u)$

Now we can go back to writing in terms of , the derivative is,

$F'(x)=(2x+2)\sin(x^2+2x-1)$

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Example Question #25 : Fundamental Theorem Of Calculus

In exponentially decaying systems, often times the solutions to differential equations take on the form of an integral called Duhamel's Integral. This is given by:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Where is a constant and f(b) is a function that represents an external force.

Equations of the form $e^{it}$ are sinusoidal functions, where is the imaginary constant. Determine the external force f(b) needed to produce solutions g(t) of sinusoidal behavior $g(t)=e^{ik}$

Possible Answers:

$f(t)=e^{it}+f(0)e^{-st}$

$f(t)=ie^{it}+f(0)e^{st}$

$f(t)=e^{it}+f(0)e^{st}$

$f(t)=ie^{it}+f(0)e^{-st}$

Correct answer:

$f(t)=ie^{it}+f(0)e^{-st}$

Explanation:

We plug in $g(t)=e^{it}$ into our formula and determine the external force f(b) .

$e^{it}=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Taking the derivative of each side with respect to , and applying the fundamental theorem of calculus:

$ie^{it}=[e^{-s(t-t)}f(t)-e^{-st}f(0)]$

Solving for f(t) ,

$f(t)=ie^{it}+f(0)e^{-st}$

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Example Question #171 : Integrals

In harmonic systems, often times the solutions to differential equations take on the form of an integral called Duhamel's Integral. This is given by:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Where is a constant and f(b) is a function that represents an external force.

What external force is needed in order to obtain g(t)=1 if f(0)=2 ?

Possible Answers:

$e^{2st}=f(t)$

$e^{st}=f(t)$

$2e^{st}=f(t)$

$2e^{-st}=f(t)$

Correct answer:

$2e^{-st}=f(t)$

Explanation:

In order to solve this, we substitute g(t)=1 into our equation and solve for f(b) :

$1=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Taking the derivative of each side with respect to , we get that:

$0=\frac{d}{dt} \int_{0}^{t}e^{-s(t-b)}f(b) db$

Using the fundamental theorem of calculus:

$0= [e^{-s(t-t)}*f(t)-e^{(-st)}f(0)]$

Solving for f(t) :

$f(0)e^{-st}=f(t)$

Since f(0)=2 ,

$2e^{-st}=f(t)$

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Example Question #171 : Introduction To Integrals

Let

$f(x)=\int_{0}^{x}\frac{cos(x)+sin(x)}{x^{2}+cos(x)}dx$

Find f'(0) .

Possible Answers:

$\pi$

Correct answer:

Explanation:

The Fundamental Theorem of Calculus tells us that

$f(x)=\int f'(x)$

therefore,

$f'(0)=\frac{cos(x)+sin(x)}{x^{2}+cos(x)}$

From here plug in 0 into this equation.

$f'(0)=\frac{cos(0)+sin(0)}{0^{2}+cos(0)}$

$f'(0)=\frac{1+0}{0+1}=\frac{1}{1}=1$

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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 #161 : Integrals

Example Question #162 : Integrals

Example Question #21 : Fundamental Theorem Of Calculus

Example Question #21 : Fundamental Theorem Of Calculus

Example Question #25 : Fundamental Theorem Of Calculus

Example Question #171 : Integrals

Example Question #171 : Introduction To Integrals

All Calculus 2 Resources

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