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Math Help - Anti-derivative problems.

  1. #1
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    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!
    Last edited by Steelers72; January 29th 2013 at 04:21 PM.
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    Re: Anti-derivative problems.

    [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

    Quote Originally Posted by Steelers72 View Post
    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
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    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.
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    Re: Anti-derivative problems.

    Quote Originally Posted by Steelers72 View Post
    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!
    Last edited by HallsofIvy; January 29th 2013 at 06:17 PM.
    Thanks from topsquark
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