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Math Help - Integration with substitution I think, but how?

  1. #1
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    Integration with substitution I think, but how?

    Hi guys, just found this site today and was hoping somebody here would have the time to help me out on what's probably a relatively simple problem...

    I'm asked to evaluate the integral x/sqrt(1+4x) from 0 to 3. This seems innocent enough, and seeing as we've worked with substitutions recently I suspect I'm intended to head down that path.

    However.. I can't get anything out of substituting u=1+4x; du ends up just being a constant (4) which isn't helping me any. I wonder if there's some trig substitution I'm just not seeing here.

    Also, where can I get a guide on using the math tags?

    thanks!
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  2. #2
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    Quote Originally Posted by storpotaten View Post
    Hi guys, just found this site today and was hoping somebody here would have the time to help me out on what's probably a relatively simple problem...

    I'm asked to evaluate the integral x/sqrt(1+4x) from 0 to 3. This seems innocent enough, and seeing as we've worked with substitutions recently I suspect I'm intended to head down that path.

    However.. I can't get anything out of substituting u=1+4x; du ends up just being a constant (4) which isn't helping me any. I wonder if there's some trig substitution I'm just not seeing here.

    Also, where can I get a guide on using the math tags?

    thanks!

    Your sub will work....

    \int_{0}^{3}\frac{x}{\sqrt{1+4x}}dx

    u=1+4x \iff \frac{u-1}{4}=x \to du=4dx

    using this we get

    \int_{0}^{3}\frac{x}{\sqrt{1+4x}}dx = \int_{1}^{13}\frac{\frac{u-1}{4}}{\sqrt{u}}\frac{du}{4}=\frac{1}{16}\int_{1}^  {13}u^{1/2}-u^{-1/2}du
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  3. #3
    Behold, the power of SARDINES!
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    Math tags

    You can hoover you cursor over anyone's code to see it.

    I use this wiki site

    http://en.wikipedia.org/wiki/Help Formula
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  4. #4
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    Hello, storpotaten!

    Welcome aboard!


    \int^3_0\frac{x}{\sqrt{1+4x}}\,dx
    With the square root of a linear expression, we can let u equal the entire radical . . .


    Let u \,=\,\sqrt{1+4x}\quad\Rightarrow\quad x\:=\:\frac{1}{4}(u^2-1) \quad\Rightarrow\quad dx \:=\:\frac{1}{2}u\,du


    Substitute: . \int\frac{\frac{1}{4}(u^2-1)}{u}\left(\frac{1}{2}u\,du\right) \;=\;\frac{1}{8}\int(u^2-1)\,du

    . . . . . = \;\frac{1}{8}\left(\frac{u^3}{3} - u\right) + C \;=\;\frac{1}{24}u(u^2-3) + C


    Back-substitute: . \frac{1}{24}\sqrt{1+4x}\left(\left[\sqrt{1+4x}\right]^2-3\right) + C \;=\;\frac{1}{24}\sqrt{1+4x}\,(1 + 4x - 3) +C

    . . . . = \;\frac{1}{24}\sqrt{1+4x}\,(4x-2) + C \;=\;\frac{1}{12}\sqrt{1+4x}\,(2x-1) + C


    Now evaluate the definite integral . . .

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    You guys are the best -- thanks!
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