# Integration by substitution

• December 16th 2007, 03:59 AM
Hoopy
Integration by substitution
Evaluate this Integral

x(x-1)^4 with upper limit 2 and lower limit 1.

Use the substitution u=x-1

Answer should be 11/30, but I get no where near that, I think I may be integrating it wrong

• December 16th 2007, 05:00 AM
Soroban
Hello, Hoopy!

We could find your error if you would show your work . . .

Quote:

Evaluate: . $\int^2_1x(x-1)^4\,dx$

Let: $u = x-1\quad\Rightarrow\quad x = u+1\quad\Rightarrow\quad dx = du$

Substitute: . $\int^1_0(u+1)u^4\,du \;=\;\int^1_0\left(u^5 + u^4\right)\,du\;= \;\frac{1}{6}u^6 + \frac{1}{5}u^5\,\bigg]^1_0 \;=\;\frac{11}{30}$

• December 16th 2007, 05:01 AM
Simplicity
Quote:

Originally Posted by Hoopy
Evaluate this Integral

x(x-1)^4 with upper limit 2 and lower limit 1.

Use the substitution u=x-1

Answer should be 11/30, but I get no where near that, I think I may be integrating it wrong

$\int^2_1 x(x-1)^4 \, \mathrm{d}x$
$u=x-1 \therefore \frac{\mathrm{d}u}{\mathrm{d}x} = 1 \implies {\mathrm{d}u}={\mathrm{d}x}$

Replace $x(x-1)^4$ with $u$ and $x$ with $u + 1$.

$\therefore limits: x=2 \implies u= x-1 = 2-1 =1$
$\therefore limits: x=1 \implies u= x-1 = 1-1 =0$

$\therefore \int^1_0 (u+1)(u)^4 \, \mathrm{d}u$
$= \int^1_0 (u^5 + u^4) \, \mathrm{d}u$
$= [\frac{u^6}{6} + \frac{u^5}{5}]^1_0$
$=[\frac{11}{30}] - [0] = \frac{11}{30}$
• December 16th 2007, 05:02 AM
Simplicity
Oops, A minute late. :(
• December 16th 2007, 09:10 AM
Hoopy
Cheers guys, I just forgot to change everything to u.