
Pessimist Singularitarian
#1
February 7th, 2013,
01:34
Here is the question:
Quote:
Calculus integral help?
Evaluate: the integral from 2(top) to 1(bottom) and the function is: x^2(lnx) dx
Here is a link to the question:
I have posted a link there to this topic so the OP can find my response.

February 7th, 2013 01:34
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Pessimist Singularitarian
#2
February 7th, 2013,
01:35
Thread Author
Hello Henry,
We are given to evaluate:
$\displaystyle \int_1^2x^2\ln(x)\,dx$
Using integration by parts, we may let:
$\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$
$\displaystyle dv=x^2\,dx\,\therefore\,v=\frac{1}{3}x^3$
and so we have:
$\displaystyle \int_1^2x^2\ln(x)\,dx=\left[\frac{1}{3}x^3\ln(x) \right]_1^2\frac{1}{3}\int_1^2x^2\,dx$
$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)\frac{1}{3}\left[\frac{1}{3}x^3 \right]_1^2$
$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)\frac{1}{9}\left(81 \right)$
$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)\frac{7}{9}$