Anti-derivative problems.

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!

Re: Anti-derivative problems.

Re: Anti-derivative problems.

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.

Re: Anti-derivative problems.

Quote:

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"

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

Quote:

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

Quote:

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.

Quote:

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 .

Quote:

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 is a pretty standard integral- you could find it in a table of integrals.

For me, that " " reminds me of " ". So let and then . The integral becomes

That should be easy!