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Math Help - Definite Integral

  1. #1
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    Definite Integral

    \int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx
    break it up 6\int_{x=1}^{x=2} (e^{3x}dx)-\int_{x=1}^{x=2} (\frac{1}{x}dx)
    2\int (e^udu-\ln{x}+c

    is this right so far, not sure what to do after.
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  2. #2
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    Quote Originally Posted by Jim Marnell View Post
    \int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx
    break it up 6\int_{x=1}^{x=2} (e^{3x}dx)-\int_{x=1}^{x=2} (\frac{1}{x}dx)
    2\int (e^udu-\ln{x}+c

    is this right so far, not sure what to do after.
    Well, yes, but this is a definite integral. So after you find the antiderivative F(x), you need to apply the fundamental theorem of calculus by taking F(2)-F(1).

    As for finding the antiderivative, what is \int e^x\,dx? This is a basic integration rule, and replacing the x with a u won't make any difference.
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  3. #3
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    yea i dont know why i put u in for the x, but, so now i would just insert the 2
    (x) values in and subtract them
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  4. #4
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    Quote Originally Posted by Jim Marnell View Post
    yea i dont know why i put u in for the x, but, so now i would just insert the 2
    (x) values in and subtract them
    No, no, your substitution was fine. I was just pointing out the integration rule:

    \int e^x\,dx=e^x+C.

    So,

    \int_1^2\left(6e^{3x}-\frac1x\right)dx

    =6\int_1^2e^{3x}\,dx-\int_1^2\frac1x\,dx

    =2\int_{x=1}^{x=2}e^u\,du-\bigg[\ln\lvert x\rvert\bigg]_1^2

    =2\bigg[e^u\bigg]_{x=1}^{x=2}-\ln 2

    =2\bigg[e^{3x}\bigg]_1^2-\ln 2

    =2(e^6-e^3)-\ln 2

    =2e^3(e^3-1)-\ln 2
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  5. #5
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    Thanks!, thats what i ended up getting so hopefully i understand this concept now
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