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Math Help - definite integral problem

  1. #1
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    definite integral problem

    can someone help me solve

    integral from 0 to 4 of

    (x)/ (square root of x^2+9) dx
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  2. #2
    Junior Member piglet's Avatar
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    Quote Originally Posted by cummings15 View Post
    can someone help me solve

    integral from 0 to 4 of

    (x)/ (square root of x^2+9) dx
    Have you made any attempt at it yourself? This is a pretty straight-forward problem and im sure you've covered problems similar to this in school or uni

    If ur really stuck, Hint: substitute u = x^2+9 and hence get a substitution for dx
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  3. #3
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    Quote Originally Posted by cummings15 View Post
    can someone help me solve

    integral from 0 to 4 of

    (x)/ (square root of x^2+9) dx
    use the method of substitution ...

    u = x^2+9
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by cummings15 View Post
    can someone help me solve

    integral from 0 to 4 of

    (x)/ (square root of x^2+9) dx
    Just note that x is (almost) the derivatived of x^2+9. Therefore you'd better substitute z := x^2+9 right away, which gives you that dz = 2xdx.

    Plugging this into your integral allows you to convert it to the form \tfrac{1}{2}\int \frac{1}{\sqrt{z}}\, dz.
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  5. #5
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    so far i got

    u=x^2+9
    du=2xdx
    (1/2)du=xdx

    so i pluged it in and got

    (1/2) integral of (du/u)

    what do i do with the 0 to 4 part?
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by cummings15 View Post
    so far i got

    u=x^2+9
    du=2xdx
    (1/2)du=xdx

    so i pluged it in and got

    (1/2) integral of (du/u)
    No, that's not quite correct, you should have \tfrac{1}{2}\int\frac{1}{\sqrt{u}}\, du

    what do i do with the 0 to 4 part?
    Finding the anti-derivative is the main problem, you have two choices as regards the limits of the integral: you can transform the limits along with the integrand:

    \int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(9)}\frac{du}{\sqrt{u}}  \, du = \Big[\sqrt{u}\Big]_{u=9}^{25} = \sqrt{25}-\sqrt{9}=2

    or you can do a back-substitution before you apply the old limits:
    \int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(4)}\frac{du}{\sqrt{u}}  \, du = \Big[\sqrt{u}\Big]_{u=u(0)}^{u(4)} = \Big[\sqrt{x^2+9}\Big]_{x=0}^4=\ldots = 2
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