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Thread: Definite Integrals for Integration by Parts

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    Definite Integrals for Integration by Parts

    I'm having a bit of difficulty with this one. Can anyone help?
    Many thanks.

    Q. $\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$ logex dx

    Attempt: u = logex => du = $\displaystyle \frac{1}{x}$ dx
    dv = $\displaystyle \frac{1}{x^2}$ => dv = x-2 => v = $\displaystyle \int$x-2 => $\displaystyle \frac{-x^-^2}{2}$
    $\displaystyle \int$u dv = uv - $\displaystyle \int$v du => $\displaystyle \int$$\displaystyle \frac{1}{x^2}$logex dx => logex$\displaystyle (\frac{-x^2}{2})$ + $\displaystyle \int$$\displaystyle \frac{x^-^2}{2}$$\displaystyle (\frac{1}{x})$ => $\displaystyle \frac{-x^-^2log_ex}{2}$ + $\displaystyle \int$$\displaystyle \frac{x^-^3}{2}$ => $\displaystyle \frac{-x^-^2log_ex}{2}$ - $\displaystyle \frac{x^-^2}{4}$
    $\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$logex dx = $\displaystyle [\frac{-x^-^2log_ex}{2} - \frac{x^-^2}{4}]^2_1$ => $\displaystyle [\frac{-2^-^2log_e2}{2} - \frac{2^-^2}{4}] - [\frac{-1^-^2log_e1}{2} - \frac{1^-^2}{4}$ => $\displaystyle \frac{-log_e2}{8} - \frac{1}{16} + 0 + \frac{1}{4}$ => $\displaystyle \frac{-log_e2}{8} - \frac{5}{16}$

    Ans.:
    (From text book): $\displaystyle \frac{1}{2}$(1 - ln2)
    Last edited by GrigOrig99; Apr 19th 2012 at 01:42 PM.
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    Re: Definite Integrals for Integration by Parts

    Check $\displaystyle \int x^{-2}dx$
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    Re: Definite Integrals for Integration by Parts

    I've made the change I think you're referring to but I'm still off the mark...

    u = logex => du = $\displaystyle \frac{1}{x}$ dx
    dv = $\displaystyle \frac{1}{x^2}$ => => v = $\displaystyle \int$$\displaystyle \frac{1}{x^2}$ => ln(x2)
    $\displaystyle \int$u dv = uv - $\displaystyle \int$v du => $\displaystyle \int$$\displaystyle \frac{1}{x^2}$logex dx => logex(ln(x2)) -$\displaystyle \int$ln(x2)$\displaystyle (\frac{1}{x})$ => ln(x2)logex - (x2ln(x2) - x2)$\displaystyle (\frac{1}{x})$
    $\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$logex dx = [ln(x2)logex - (x2ln(x2) - x2)$\displaystyle (\frac{1}{x})$]$\displaystyle ^2_1$ => [ln(4)loge2 - (4ln(4) - 4)$\displaystyle (\frac{1}{2})$] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4)$\displaystyle (\frac{1}{2})$ + 0 + 1

    Ans.:
    (From text book): $\displaystyle \frac{1}{2}$(1 - ln2)
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    Re: Definite Integrals for Integration by Parts

    Quote Originally Posted by GrigOrig99 View Post
    I've made the change I think you're referring to but I'm still off the mark...

    u = logex => du = $\displaystyle \frac{1}{x}$ dx
    dv = $\displaystyle \frac{1}{x^2}$ => => v = $\displaystyle \int$$\displaystyle \frac{1}{x^2}$ => ln(x2) no
    $\displaystyle \int$u dv = uv - $\displaystyle \int$v du => $\displaystyle \int$$\displaystyle \frac{1}{x^2}$logex dx => logex(ln(x2)) -$\displaystyle \int$ln(x2)$\displaystyle (\frac{1}{x})$ => ln(x2)logex - (x2ln(x2) - x2)$\displaystyle (\frac{1}{x})$
    $\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$logex dx = [ln(x2)logex - (x2ln(x2) - x2)$\displaystyle (\frac{1}{x})$]$\displaystyle ^2_1$ => [ln(4)loge2 - (4ln(4) - 4)$\displaystyle (\frac{1}{2})$] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4)$\displaystyle (\frac{1}{2})$ + 0 + 1

    Ans.:
    (From text book): $\displaystyle \frac{1}{2}$(1 - ln2)
    $\displaystyle \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C$
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    Re: Definite Integrals for Integration by Parts

    Quote Originally Posted by GrigOrig99 View Post
    Ans.: [/B](From text book): $\displaystyle \frac{1}{2}$(1 - ln2)
    Take a look at this page.

    And this.
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    Re: Definite Integrals for Integration by Parts

    Thanks for that link. Really useful! The only question I have left is with the expression of the answer. The book has it as 1/2(1 - ln2) but the link you provided shows a final answer of 1/2(1 - log2). Is there no difference?
    Thanks again, guys.
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    Re: Definite Integrals for Integration by Parts

    Quote Originally Posted by GrigOrig99 View Post
    Thanks for that link. Really useful! The only question I have left is with the expression of the answer. The book has it as 1/2(1 - ln2) but the link you provided shows a final answer of 1/2(1 - log2). Is there no difference?
    Thanks again, guys.
    none ... in "big boy" math, log is understood base e
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