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Thread: Improper Integral - Ex 8

  1. #1
    MHF Contributor Jason76's Avatar
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    Improper Integral - Ex 8

    $\displaystyle \int_{-2}^{14} \dfrac{7}{(x + 2)^{1/4}} dx$

    $\displaystyle \int_{b}^{14} \dfrac{7}{(x + 2)^{1/4}} dx$

    $\displaystyle \int_{b}^{14} 7(x + 2)^{-1/4}$

    $\displaystyle u = x + 2$

    $\displaystyle du = dx$

    $\displaystyle \int_{b}^{14} 7(u)^{-1/4}$

    $\displaystyle = 7(\dfrac{u)^{3/4}}{3/4}$ evaluated at b and 14

    $\displaystyle = 7((\dfrac{4}{3})u^{3/4}$ evaluated at b and 14

    $\displaystyle = (\dfrac{28}{3})(u^{3/4}$ evaluated at b and 14

    $\displaystyle = (\dfrac{28}{3})(((x + 2))^{3/4}$ evaluated at b and 14

    $\displaystyle [(\dfrac{28}{3})(((14 + 2))^{3/4}] - [(\dfrac{28}{3})(((b + 2))^{3/4}]$

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(((14 + 2))^{3/4}] - [(\dfrac{28}{3})((((\infty) + 2))^{3/4}]$ ??

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(((16))^{3/4}] - [(\dfrac{28}{3})((((\infty) + 2))^{3/4}]$

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(8)] - [(\dfrac{28}{3})((((\infty) + 2))^{3/4}]$

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{224}{3})] - [\infty]$ ??
    Last edited by Jason76; Feb 26th 2015 at 06:13 PM.
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  2. #2
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    Re: Improper Integral - Ex 8

    You have again failed to change your limits.
    You are trying to evaluate a limit to infinity when the question asks for the limit to -2.
    The symbol $\infty$ cannot be used as a term in an equation - it isn't defined there.
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  3. #3
    MHF Contributor Jason76's Avatar
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    Re: Improper Integral - Ex 8

    $\displaystyle \int_{-2}^{14} \dfrac{7}{(x + 2)^{1/4}} dx$

    $\displaystyle \int_{b}^{14} \dfrac{7}{(x + 2)^{1/4}} dx$

    $\displaystyle \int_{b}^{14} 7(x + 2)^{-1/4}$

    $\displaystyle u = x + 2$

    $\displaystyle du = dx$

    $\displaystyle \int_{b}^{14} 7(u)^{-1/4}$

    $\displaystyle = 7(\dfrac{u)^{3/4}}{3/4}$ evaluated at b and 14

    $\displaystyle = 7((\dfrac{4}{3})u^{3/4}$ evaluated at b and 14

    $\displaystyle = (\dfrac{28}{3})(u^{3/4}$ evaluated at b and 14

    $\displaystyle = (\dfrac{28}{3})(((x + 2))^{3/4}$ evaluated at b and 14

    $\displaystyle [(\dfrac{28}{3})(((14 + 2))^{3/4}] - [(\dfrac{28}{3})(((b + 2))^{3/4}]$

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(((14 + 2))^{3/4}] - [(\dfrac{28}{3})(((-2 + 2))^{3/4}]$ ??

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(((16))^{3/4}] - [(\dfrac{28}{3})((((-2) + 2))^{3/4}]$

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(8)] - [(\dfrac{28}{3})(((-2 + 2))^{3/4}]$

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{28}{3})(8)] - [(\dfrac{28}{3})(((0))^{3/4}]$

    $\displaystyle \lim b \rightarrow \infty [(\dfrac{224}{3})] - [0 ] = \dfrac{224}{3}$

    Improper Integral - Ex 8-noname5.jpg - correct
    Last edited by Jason76; Feb 27th 2015 at 03:32 AM.
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  4. #4
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    Re: Improper Integral - Ex 8

    Did you not understand what Archie said? Your integral is from -2 to 4. Yes, this is an improper integral because the integrand is undefined at -2 so replacing -2 with a parameter, b, is appropriate. You can make the substitution u= x+ 2 so that du= dx. When x= b, u= b+2 and when x=4, u= 2. So your integral becomes [tex]\int_{b+2}^2 \frac{7}{u^{1/4}} du= \int_{b+2}^2 7u^{-1/4} du/tex]. You did not change the limits of integration and, once again, do not have the "du".

    The anti-derivative of $\displaystyle 7u^{-1/4}$ is $\displaystyle 7(\frac{4}{3})u^{3/4}$ and that is to be evaluated between b+2 and 2: $\displaystyle \frac{28}{3}\left(2^{3/4}- (b+2)^{3/4}\right)$. Then take the limit as b goes to -2 which is the same as b+ 2 going to 0. There is no "$\displaystyle \infty$" involved in this problem.
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