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Thread: Integral Problem involving FTC and IBP

  1. #1
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    Integral Problem involving FTC and IBP

    The FTC for $\displaystyle a=0$

    $\displaystyle \int_0^b f'(x)dx=f(b)-f(0) $

    Is used in the problem to obtain the obvious:

    $\displaystyle f(b)=f(0)+ \int_0^b f'(x)dx $

    The task is to now use IBP to show that:

    $\displaystyle f(b)=f(0)+f'(0)\cdot b +\int_0^b f''(x) (b-x) dx$

    My calculation is slightly different however. I define the change in variables:

    $\displaystyle u=f'(x)$
    $\displaystyle du=f''(x)dx$

    $\displaystyle dv=dx$
    $\displaystyle v=x$

    $\displaystyle \int_0^b f'(x)dx=[x f'(x)]_0^b -\int_0^b xf''(x) dx$

    Substituting the first equation on the left side of the last equation I obtain:

    $\displaystyle f(b)=f(0)+f'(0)\cdot b -\int_0^b xf''(x)dx$

    The last term isn't the same as shown. What happens to the sign, and why is there b-x instead of just x in the integrand?
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  2. #2
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    Re: Integral Problem involving FTC and IBP

    Quote Originally Posted by adkinsjr View Post
    The FTC for $\displaystyle a=0$

    $\displaystyle \int_0^b f'(x)dx=f(b)-f(0) $

    Is used in the problem to obtain the obvious:

    $\displaystyle f(b)=f(0)+ \int_0^b f'(x)dx $

    The task is to now use IBP to show that:

    $\displaystyle f(b)=f(0)+f'(0)\cdot b +\int_0^b f''(x) (b-x) dx$

    My calculation is slightly different however. I define the change in variables:

    $\displaystyle u=f'(x)$
    $\displaystyle du=f''(x)dx$

    $\displaystyle dv=dx$
    $\displaystyle v=x$

    $\displaystyle \int_0^b f'(x)dx=[x f'(x)]_0^b -\int_0^b xf''(x) dx$

    Substituting the first equation on the left side of the last equation I obtain:

    $\displaystyle f(b)=f(0)+f'(0)\cdot b -\int_0^b xf''(x)dx$

    The last term isn't the same as shown. What happens to the sign, and why is there b-x instead of just x in the integrand?
    Your answer isn't wrong but there is a little trick you can use in your integration by parts. As you have done take$$
    u = f'(x),~dv = dx$$
    Now remember when you take the antiderivative for dv you don't have to take the constant of integration $=0$. Since we
    always do, we forget that we don't have to. so try:$$
    du = f'(x)dx,~v = x-b$$
    Try it. You will like it.
    Thanks from adkinsjr and HallsofIvy
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    Re: Integral Problem involving FTC and IBP

    Quote Originally Posted by adkinsjr View Post
    The FTC for $\displaystyle a=0$

    $\displaystyle \int_0^b f'(x)dx=f(b)-f(0) $

    Is used in the problem to obtain the obvious:

    $\displaystyle f(b)=f(0)+ \int_0^b f'(x)dx $

    The task is to now use IBP to show that:

    $\displaystyle f(b)=f(0)+f'(0)\cdot b +\int_0^b f''(x) (b-x) dx$

    My calculation is slightly different however. I define the change in variables:

    $\displaystyle u=f'(x)$
    $\displaystyle du=f''(x)dx$

    $\displaystyle dv=dx$
    $\displaystyle v=x$

    $\displaystyle \int_0^b f'(x)dx=[x f'(x)]_0^b -\int_0^b xf''(x) dx$

    Substituting the first equation on the left side of the last equation I obtain:

    $\displaystyle f(b)=f(0)+f'(0)\cdot b -\int_0^b xf''(x)dx$

    The last term isn't the same as shown. What happens to the sign, and why is there b-x instead of just x in the integrand?
    There is a term missing in your calculation of

    $\displaystyle \int_0^b f'(x)dx $

    namely $\displaystyle b f ' (b) $
    Last edited by Idea; Feb 3rd 2019 at 12:58 AM.
    Thanks from adkinsjr
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    Re: Integral Problem involving FTC and IBP

    Quote Originally Posted by Walagaster View Post
    Your answer isn't wrong but there is a little trick you can use in your integration by parts. As you have done take$$
    u = f'(x),~dv = dx$$
    Now remember when you take the antiderivative for dv you don't have to take the constant of integration $=0$. Since we
    always do, we forget that we don't have to. so try:$$
    du = f'(x)dx,~v = x-b$$
    Try it. You will like it.
    Thanks, that definitely works.

    Quote Originally Posted by Idea View Post
    There is a term missing in your calculation of

    $\displaystyle \int_0^b f'(x)dx $

    namely $\displaystyle b f ' (b) $
    Yes, that was copying mistake. I have the term written f'(0) b that should be f'(b) b
    Last edited by adkinsjr; Feb 3rd 2019 at 07:33 PM.
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