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Thread: integral

  1. #1
    Junior Member
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    integral

    how do you integrate √(2^x) i know this equals 2^.5x . i know the general integral formula a^x =(a^x) /ln a but when i apply that here i get 2^.5x /ln(2) which is not right. i am not sure what i am doing wrong
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  2. #2
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    Re: integral

    Try a u-substitution:  u=\tfrac{1}{2}x
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  3. #3
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    Re: integral

    Check your work below. Better yet, check my work

    \sqrt{2^x}=(2^x)^{1/2} = 2^{\frac{1}{2}x}

    u=\tfrac{1}{2}x and du=\tfrac{1}{2}dx \implies dx = 2du

    \int 2^{\frac{1}{2}x} dx = \int 2^u (2du) = \frac{2^u}{\ln(2)}\cdot 2 +c =\frac{2^{u+1}}{\ln(2)}+c= \frac{2^{\frac{1}{2}x+1}}{\ln(2)}+c
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