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    #1
    Here is the question:

    Quote 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.

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    #2 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:

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

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

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

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

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