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Thread: Integration by parts.. Find values of a and b.

  1. #1
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    Integration by parts.. Find values of a and b.

    Find the values of a and b if:

    ∫2,1 xlnx dx = lna -(b/a)

    btw, "∫2,1" means upper limit is 2, lower limit is 1

    Help is very appreciated!!! Thanks!
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  2. #2
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    Quote Originally Posted by Yehia View Post
    Find the values of a and b if:

    ∫2,1 xlnx dx = lna -(b/a)

    btw, "∫2,1" means upper limit is 2, lower limit is 1

    Help is very appreciated!!! Thanks!
    Let $\displaystyle u'(x)=x~\implies~u(x)=\frac12 x^2$

    and $\displaystyle v(x)=\ln(x)~\implies~v'(x)=\frac1x$

    Your integral is then:

    $\displaystyle \int x\cdot \ln(x) dx = \int u'(x) \cdot v(x) dx = u(x) \cdot v(x) - \int U(x) \cdot v'(x) dx = \frac12 x^2 \cdot \ln(x) - \int \frac12 x dx$

    Finally you have:

    $\displaystyle \int_1^2 x\cdot \ln(x) dx = \left[\frac12 x^2 \cdot \ln(x) - \frac14 x^2 \right]_1^2 = 2\ln(2) - \frac34 = \ln(4) - \frac34$

    Compare the last term with the given term to determine the values aof a and b.
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Yehia View Post
    Find the values of a and b if:

    ∫2,1 xlnx dx = lna -(b/a)

    btw, "∫2,1" means upper limit is 2, lower limit is 1

    Help is very appreciated!!! Thanks!

    $\displaystyle \int_{1}^{2} x\ ln x dx $

    du = x , u = x^2/2

    v = ln x , dv = 1/x

    $\displaystyle \int_{1}^{2} x\ln x \cdot dx = \frac{x^2 \cdot \ln x }{2}\mid_{1}^{2} - \int_{1}^{2} \frac{x^2}{2x} dx $

    $\displaystyle \int_{1}^{2} x\ln x \cdot dx = 2\ln 2 - \ln 1 - \left( \frac{x^2}{4} \mid^2_1\right) $

    $\displaystyle \int_{1}^{2} x\ln x \cdot dx = 2\ln 2 - 0 - \left( 1-\frac{1}{4}\right) $

    $\displaystyle \int_{1}^{2} x\ln x \cdot dx = 2\ln 2 - \frac{3}{4} $

    $\displaystyle \int_{1}^{2} x\ln x \cdot dx = \ln 2^2 - \frac{3}{4} $

    a=4 , b=-3
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