# Thread: [SOLVED] Definite integral involving parts

1. ## [SOLVED] Definite integral involving parts

$\displaystyle \int_2^{6} \sqrt {t} ln t dt$

$\displaystyle u = ln t$
$\displaystyle dv = \sqrt{t}$
$\displaystyle du = 1/t$
$\displaystyle v = 2t^{\frac{3}{2}}$
$\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}$
$\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}$
$\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}$

Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.

EDIT: Nevermind guys, I have my answer. Sorry for the waste of topic. How do I delete my topic?

2. Originally Posted by Archduke01
$\displaystyle \int_2^{6} \sqrt {t} ln t dt$

$\displaystyle u = ln t$
$\displaystyle dv = \sqrt{t}$
$\displaystyle du = 1/t$
$\displaystyle v = 2t^{\frac{3}{2}}$
$\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}$
$\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}$
$\displaystyle \frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}$

Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.
The fourth line of yours should be:
$\displaystyle v = \frac{2}{3}t^{\frac{3}{2}}$

and you should write $\displaystyle du = \frac{1}{t}dt$

So,

$\displaystyle uv - \int v du$

$\displaystyle = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \int \frac{2}{3}t^{\frac{3}{2}} \times \frac{1}{t} dt$

$\displaystyle = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{\frac{3}{2}} \times \frac{1}{t} dt$

$\displaystyle = \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{{\frac{3}{2}}-1} dt$