Definite integral problem

• Oct 26th 2010, 03:10 AM
Glitch
Definite integral problem
The question:

Find $\displaystyle \int_{0}^{1} 2x(1 + x^2)^3 dx$

My attempt:
Let u = $\displaystyle 1 + x^2$

$\displaystyle \frac{du}{dx} = 2x$

$\displaystyle du = 2x dx$

$\displaystyle \int_{0}^{1} u^3 du = _{0}^{1}[3u^2]$

Find bounds in terms of u...
Sub 1 into u:
$\displaystyle 1 + (1)^2 = 2$
Sub 0 into u:
$\displaystyle 1 + 0 = 1$

$\displaystyle _{1}^{2}[3u^2]$
$\displaystyle [3(2)^2] - [3(1)^2] = 9$

However, the answer is 15/4. :/

What am I doing wrong? Thanks!
• Oct 26th 2010, 03:15 AM
Plato
What is the derivative of $\displaystyle \dfrac{\left(1+x^2\right)^4}{4}~?$
• Oct 26th 2010, 03:20 AM
Glitch
$\displaystyle 2x(1+x^2)^3$
• Oct 26th 2010, 03:41 AM
Plato
Well there you have it. Do you not?
• Oct 26th 2010, 03:45 AM
Glitch
Not really, because you've solved the integral and I just proved it. I can't always make that leap like you did, which is why I tried the substitution method to help me. I'm still not sure what I've actually done wrong.
• Oct 26th 2010, 03:50 AM
earboth
Quote:

Originally Posted by Glitch
The question:

Find $\displaystyle \int_{0}^{1} 2x(1 + x^2)^3 dx$

My attempt:
Let u = $\displaystyle 1 + x^2$

$\displaystyle \frac{du}{dx} = 2x$

$\displaystyle du = 2x dx$

$\displaystyle \int_{0}^{1} u^3 du = _{0}^{1}[3u^2]$ <=== This is the derivative not the integral which should be $\displaystyle \frac14 u^4$

Find bounds in terms of u...
Sub 1 into u:
$\displaystyle 1 + (1)^2 = 2$
Sub 0 into u:
$\displaystyle 1 + 0 = 1$

$\displaystyle _{1}^{2}[3u^2]$
$\displaystyle [3(2)^2] - [3(1)^2] = 9$

However, the answer is 15/4. :/

What am I doing wrong? Thanks!

...
• Oct 26th 2010, 03:53 AM
Glitch
Nevermind, I was finding the derivative of u^3 instead of the anti-derivative like an idiot. All solved.

EDIT: Yeah, just realised earboth. Thanks.