# Thread: Integral Problem involving FTC and IBP

1. ## 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?

2. ## Re: 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?
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.

3. ## Re: 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?
There is a term missing in your calculation of

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

namely $\displaystyle b f ' (b)$

4. ## Re: Integral Problem involving FTC and IBP

Originally Posted by Walagaster
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.

Originally Posted by Idea
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