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Thread: Some integration questions

  1. #1
    Member roshanhero's Avatar
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    Some integration questions

    1.$\displaystyle \int a^xe^xdx$
    I tried to solve it by using integration by parts but couldnot get the required answer.
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  2. #2
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    Quote Originally Posted by roshanhero View Post
    1.$\displaystyle \int a^xe^xdx$
    I tried to solve it by using integration by parts but couldnot get the required answer.
    Try writing $\displaystyle a^x = e^{x \ln a}$
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    Quote Originally Posted by roshanhero View Post
    1.$\displaystyle \int a^xe^xdx$
    I tried to solve it by using integration by parts but couldnot get the required answer.
    Assuming $\displaystyle a>0$
    Actually it must be positive since the base must be positive.
    $\displaystyle \int a^{x} e^{x} dx = C1 + a^{x} e^{x} - ln(a) \int a^{x} e^{x} dx $
    then
    $\displaystyle (1+ln(a)) \int a^{x} e^{x} dx = a^{x} e^{x} + C1$
    $\displaystyle \int a^{x} e^{x} dx = \frac{a^{x} e^{x}}{1+ln(a)} + C$
    where $\displaystyle C = \frac{C1}{1+ln(a)} $
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    Quote Originally Posted by roshanhero View Post
    1.$\displaystyle \int a^xe^xdx$
    I tried to solve it by using integration by parts but couldnot get the required answer.
    Let $\displaystyle y = (ae)^x$ then

    $\displaystyle \ln y = x \ln (ae)$

    $\displaystyle \frac{1}{y} \frac{dy}{dx} = \ln (ae)$

    $\displaystyle dx = \frac{dy}{y \ln (ae)} $

    and

    $\displaystyle \int y ~dx = \int y \frac{dy}{y \ln (ae)}$

    $\displaystyle \int y ~dx = \int \frac{dy}{\ln (ae)}$
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  5. #5
    Member roshanhero's Avatar
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    How can we write$\displaystyle a^x = e^{x \ln a}$
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    Quote Originally Posted by roshanhero View Post
    How can we write$\displaystyle a^x = e^{x \ln a}$
    see Natural exponents and logarithms
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    Quote Originally Posted by roshanhero View Post
    How can we write$\displaystyle a^x = e^{x \ln a}$
    Its the definition of $\displaystyle a^x$ .
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  8. #8
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    $\displaystyle \int a^xe^xdx=\int e^{xln a}e^x dx
    =\int e^{x(log a+1)} dx$
    I hope that upto now all is right. Now I can use the formula of integration of e^ax.
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  9. #9
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    Quote Originally Posted by roshanhero View Post
    $\displaystyle \int a^xe^xdx=\int e^{xln a}e^x dx
    =\int e^{x(log a+1)} dx$
    I hope that upto now all is right. Now I can use the formula of integration of e^ax.
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  10. #10
    Member roshanhero's Avatar
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    I am sorry that I still cannot prove $\displaystyle a^x = e^{x \ln a}$ even with the link given by you guys.
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  11. #11
    Member roshanhero's Avatar
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    Allright guys I proved it but in a messy way,tell me whether it is right or wrong
    $\displaystyle
    Let,a^x=y
    $
    $\displaystyle
    ln a^x=lny
    $
    $\displaystyle
    a^x=e^{lny}
    $
    $\displaystyle
    a^x=e^{lna^x}
    $
    $\displaystyle
    a^x=e^{xlna}
    $
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    Quote Originally Posted by roshanhero View Post
    Allright guys I proved it but in a messy way,tell me whether it is right or wrong
    $\displaystyle
    Let,a^x=y
    $
    $\displaystyle
    ln a^x=lny
    $
    $\displaystyle
    a^x=e^{lny}
    $
    $\displaystyle
    a^x=e^{lna^x}
    $
    $\displaystyle
    a^x=e^{xlna}
    $
    Yeh that's fine, it's probably less messy to just say

    $\displaystyle e^{x \ln{a}} = \left( e^{\ln a} \right)^x = a^x$
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