# Evaluating an Integral

• May 1st 2013, 01:50 PM
Walshy
Evaluating an Integral
Assuming ∫ = the integral from (0-3)

I am having trouble deciding what to sub in as u in this integration problem:

∫ x(sqrt(x+1)) dx

Thanks.
• May 1st 2013, 02:16 PM
Re: Evaluating an Integral
Let $u = x+1$, $du = dx$

The integral can then be rewritten as:

$\int_{0}^{3} (u-1)(\sqrt{u})du$

I'm sure you can take it from here. :)
• May 1st 2013, 02:17 PM
Plato
Re: Evaluating an Integral
Quote:

Originally Posted by Walshy
Assuming ∫ = the integral from (0-3)

I am having trouble deciding what to sub in as u in this integration problem:

∫ x(sqrt(x+1)) dx

The integral $\int_0^3 {x\sqrt {x + 1} dx}$ becomes $\int_1^4 {(u - 1)\sqrt u du}$.

Do you see how?
• May 1st 2013, 02:23 PM
Walshy
Re: Evaluating an Integral
Yes, I see. Thanks for the help guys.
• May 1st 2013, 02:31 PM
Plato
Re: Evaluating an Integral
Quote:

Let $u = x+1$, $du = dx$

The integral can then be rewritten as:

$\color{red}\int_{0}^{3} (u-1)(\sqrt{u})du$

The limits of integration are not correct.
• May 2nd 2013, 07:49 PM
Re: Evaluating an Integral
Given this, I should clarify that my bounds were regards to X, so when you integrate, you should back substitute U so you are in respect to X before evaluation of the bounds.

You can change the bounds so that they are with respect to U and integrate.

Or, you can leave the bounds with regard to X, integrate with respect to U and then back substitute so that you end up with the original function (f(x), not not f'(x)) with respect to X.

Same result.