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Math Help - the integral of a^u where u is a function of x

  1. #1
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    the integral of a^u where u is a function of x

    The integral of a^u where u is a function of x is given as

    1/ln(a) a^u + C

    but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0

    so f'g=0 and for fg' we have

    ua^(u-1)/ln(a) which does not look equal to a^u (but maybe it is).

    I'm assuming u is a function of x but I'm setting it as u = x so that u' =1

    taking the derivative of the integral should give back the original integral function correct? what am I doing wrong?




    thanks!
    Last edited by kingsolomonsgrave; July 19th 2013 at 03:28 AM.
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  2. #2
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    Re: the integral of a^u where u is a function of x

    Quote Originally Posted by kingsolomonsgrave View Post
    The integral of a^u where u is a function of x is given as
    1/ln(a) a^u + C

    but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0
    so f'g=0 and for fg' we have
    You are missing the point. Forget the product rule.

    If a>0~\&~f(x)=a^x then f'(x)=a^x\ln(a).
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    Re: the integral of a^u where u is a function of x

    I see!

    The integral of a^x ln(a) is a^x therefore the integral of a^x is a^x times 1/ln(a)
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    Re: the integral of a^u where u is a function of x

    Quote Originally Posted by kingsolomonsgrave View Post
    I see!

    The integral of a^x ln(a) is a^x therefore the integral of a^x is a^x times 1/ln(a)
    Note that this rule only holds for u = x. Any other function of x requires the chain rule.

    -Dan
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    Re: the integral of a^u where u is a function of x

    Quote Originally Posted by kingsolomonsgrave View Post
    The integral of a^u where u is a function of x is given as

    1/ln(a) a^u + C
    The integral with respect to what variable? The integral of e^u with respect to u is (1/ln(a))a^u+ C:
    \int a^u du= \frac{1}{ln(a)} a^u+ C

    The integral with respect to x will depend upon exactly what function of x u is.

    but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0
    Why use the product rule? Here 1/ln(a) is a constant and the rule (Cf(x))'= Cf'(x) is much simpler.

    so f'g=0 and for fg' we have

    ua^(u-1)/ln(a) which does not look equal to a^u (but maybe it is).
    Wrong rule. The derivative of u^a, with respect to u, is au^{a- 1} but that only applies when the variable is in the base and the exponent is constant. The derivative of a^x is ln(a) a^x.

    I'm assuming u is a function of x but I'm setting it as u = x so that u' =1

    taking the derivative of the integral should give back the original integral function correct? what am I doing wrong?




    thanks!
    Thanks from topsquark
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    Re: the integral of a^u where u is a function of x

    Quote Originally Posted by kingsolomonsgrave View Post
    The integral of a^u where u is a function of x is given as

    1/ln(a) a^u + C

    but if I take the derivative of that using the product rule we have C=0 because it's a constant and the derivative of 1/ln(a) = 0

    so f'g=0 and for fg' we have

    ua^(u-1)/ln(a) which does not look equal to a^u (but maybe it is).

    I'm assuming u is a function of x but I'm setting it as u = x so that u' =1

    taking the derivative of the integral should give back the original integral function correct? what am I doing wrong?




    thanks!
    \displaystyle \begin{align*} \int{a^x\,dx} &= \int{ e^{\ln{ \left( a^x \right) } } \, dx} \\ &= \int{ e^{x\ln{(a)}}\,dx} \\ &= \frac{1}{\ln{(a)}}\,e^{x\ln{(a)}} + C \\ &= \frac{1}{\ln{(a)}} \, e^{\ln{ \left( a^x \right) } } + C \\ &= \frac{1}{\ln{(a)}}\,a^x + C \end{align*}
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