MHB Integration by Parts: Calculus Integral Help?

[MarkFL]

Here is the question:

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:

Calculus integral help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

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:

$\displaystyle \int_1^2x^2\ln(x)\,dx=\left[\frac{1}{3}x^3\ln(x) \right]_1^2-\frac{1}{3}\int_1^2x^2\,dx$

$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)-\frac{1}{3}\left[\frac{1}{3}x^3 \right]_1^2$

$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)-\frac{1}{9}\left(8-1 \right)$

$\displaystyle \int_1^2x^2\ln(x)\,dx=\frac{8}{3}\ln(2)-\frac{7}{9}$