# Definite Integral By Substitution

• Jun 4th 2011, 05:40 PM
RezMan
Definite Integral By Substitution
hi can you guys help me with this please:

Evaluate the definite integral, if it exists. $\int_{-1}^{0} x^2(x^3+1)^8\! \, \mathrm{d}x.$

My teacher says there's a way that you don't have to bring back the original function, that you can just plug in for u. I got this far:

u+x^3+1

$\frac{du}{3}=x^2 dx$

$\frac{1}{3} \int_{-1}^{0} u^8\! \, \mathrm{d}u.$

where do I go from here to solve using only u?
• Jun 4th 2011, 05:46 PM
chisigma
Quote:

Originally Posted by RezMan
hi can you guys help me with this please:

Evaluate the definite integral, id it exists. $\int_{-1}^{0} x^2(x^3+1)^8\! \, \mathrm{d}x.$

My teacher says there's a way that you don't have to bring back the original function, that you can just plug in for u. I got this far:

u+x^3+1

$\frac{du}{3}=x^2 dx$

$\frac{1}{3} \int_{-1}^{0} u^8\! \, \mathrm{d}u.$

where do I go from here to solve using only u?

... only a little detail: after the substitution $1+x^{3}=u$ the integration interval is not any more [-1,0] , it has to be changed...

Kind regards

$\chi$ $\sigma$
• Jun 4th 2011, 05:49 PM
RezMan
I have to plug -1 and 0 into x^3 +1?
• Jun 4th 2011, 05:50 PM
Ackbeet
Quote:

Originally Posted by RezMan
I have to plug -1 and 0 into x^3 +1?

Correct. What do you get for your new limits?
• Jun 4th 2011, 06:03 PM
RezMan
would it be something like this:

$\frac{1}{3} [\frac{u^9}{9}]^{1}_{0}$
• Jun 4th 2011, 07:17 PM
RezMan
can I now plug 0 and 1 into u?
• Jun 4th 2011, 07:48 PM
CaptainBlack
Quote:

Originally Posted by RezMan
can I now plug 0 and 1 into u?

Yes

CB
• Jun 5th 2011, 03:38 AM
RezMan
so would it be
$\frac{1}{27}[1-0]$

1/27?
• Jun 5th 2011, 04:02 AM
Prove It
Quote:

Originally Posted by RezMan
so would it be
$\frac{1}{27}[1-0]$

1/27?

Yes