Anti-derivative problems.

**Steelers72** · January 29th 2013, 04:11 PM

I'm stumped on these problems:

Find the anti-derivative

a) f ''(t) = 2e^t+2 sint, f(0)=0 f(pi)=0

b) f'(x) = 3/(sqrt(1-x^2)), f(1/2)=6

If someone could explain how to do it? I got stuck on the first one toward the end of the problem. I'm stumped on b). Thanks for any help!

**topsquark** · January 29th 2013, 04:36 PM

[QUOTE=Steelers72;766663]I'm stumped on these problems:

Find the anti-derivative

a) f ''(t) = 2e^t+2 sint, f(0)=0 f(pi)=0

$f'(t) = \int_0^{\pi} \left ( 2e^{t} + 2~sin(t) \right ) dt$

Now split the integrand into two integrals:

$= \int_0^{\pi} 2e^{t}dt + \int_0^{\pi} 2~sin(t) \right ) dt$

$= 2 \int_0^{\pi} e^{t}dt + 2 \int_0^{\pi} sin(t) \right ) dt$

Do you know the integrals of e^t and sin(t)?

-Dan

Originally Posted by Steelers72

I'm stumped on these problems:

Find the anti-derivative

b) f'(x) = 3/(sqrt(1-x^2)), f(1/2)=6

If someone could explain how to do it? I got stuck on the first one toward the end of the problem. I'm stumped on b). Thanks for any help!

Not sure about this one. It should have two constraints...f(1/2) = 6 is only one. I'll leave it to be an indefinite integral, though the answer with the second limit may have a large effect on the solution.

$f(x) = \frac{3~dx}{\sqrt{1-x^2}}$

Use the substitution $x = sin(t)$ (or you can choose cosine. It makes no difference.) Then we know that $dx = cos(t)~dt$ . So the integral becomes

$\frac{3~(cos(t))~dt}{\sqrt{1-(sin^2t)}}$

Using the fact that $sin^2(t) = 1 - cos^2(t)$ can you finish this?

-Dan

**Steelers72** · January 29th 2013, 05:06 PM

Thanks Dan,

My problem with part a is finding the D constant.
Here's what I have so far:
f ' (x)= 2e^t - 2cost +c =0
2e^0 -2cos0+c=0
2(1)-2(1)+c=0
2-2+c=0
c=0

f '(x)= 2e^t-2cost+0
f(x)=2e^t+2sint+D=0

antiderivative of -cosx is -sinx right? If so, it is a double negative making it +2sint like above.

so then you plug in pi to find the constant "D".

f(pi)=2e^pi+2sin(pi)+D=0
So does D=0?

Is my final answer 2^et+2sint+0 ?

for part b), I honestly am stuck and I guess I have to go to my TA tomorrow and ask him to enlighten me on that one haha.

**HallsofIvy** · January 29th 2013, 06:14 PM

Originally Posted by Steelers72

Thanks Dan,

My problem with part a is finding the D constant.
Here's what I have so far:
f ' (x)= 2e^t - 2cost +c =0

This is correct for a "first integral"

2e^0 -2cos0+c=0

This is NOT correct. The condition you gave before was "f(0)= 0" NOT "f'(0)= 0".

2(1)-2(1)+c=0
2-2+c=0
c=0

f '(x)= 2e^t-2cost+0

No, f'(x)= 2e^t- 2cos(t)+ C and integrating again, f(x)= 2e^t- 2 sin(t)+ Cx+ D

f(x)=2e^t+2sint+D=0

antiderivative of -cosx is -sinx right? If so, it is a double negative making it +2sint like above.

No, there is no "double" negative. The integral of - cos(x) is -sin(x) but that is only one "-" in that.

so then you plug in pi to find the constant "D".

f(pi)=2e^pi+2sin(pi)+D=0
So does D=0?

Is my final answer 2^et+2sint+0 ?

No. f'= 2e^t- 2cos(t)+ c so f= 2e^t- 2sin(t)+ ct+ d.
NOW f(0)= 2e^0 - 2sin(0)+ c(0)+ d= 2+ d= 0 and f(pi)= 2e^pi-2 sin(pi)+ c\pi+ d= 2e^\pi+ c\pi+ d= 0
That gives you two equations to solve for c and .

part b), I honestly am stuck and I guess I have to go to my TA tomorrow and ask him to enlighten me on that one haha.

He'll probably tell you that $\int \frac{dx}{\sqrt{1- x^2}}$ is a pretty standard integral- you could find it in a table of integrals.

For me, that " $\sqrt{1- x^2}$ " reminds me of " $\sqrt{1- sin^2(\theta)}= cos(\theta)$ ". So let $x=sin(\theta)$ and then $dx= cos(x)dx$ . The integral becomes $\int \frac{dx}{\sqrt{1- x^2}}=\int \frac{cos(\theta)d\theta}}{\sqrt{1- sin^2(\theta)}}= \int \frac{cos(\theta)d\theta}{cos(\theta)}= \int d\theta$

That should be easy!

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