anonymous
  • anonymous
Another integral
Calculus1
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1435507797590:dw|
anonymous
  • anonymous
\[\int\limits_{1}^{9} 1/\sqrt{x}(1+\sqrt{x})^2 dx\]
anonymous
  • anonymous
u=(1+ square root of x) du= 1/2(square root of x)

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idku
  • idku
\(\large \color{black}{\displaystyle \int_{1}^{9}\frac{1}{\sqrt{x}(1+\sqrt{x})^2}dx}\) I can hardly imagine what this function looks like if you want to find the area under the curve of it, but it is a continuous function over (0,\(\infty\)). Integral from 1 to 9 is applicable so why not? u-substitution: \(\large \color{black}{\displaystyle u=1+\sqrt{x}}\) what would sqrt(x) be? \(\large \color{black}{\displaystyle \sqrt{x}=(u-1)}\) what is going to replace the dx? \(\large \color{black}{\displaystyle du=\frac{1}{2\sqrt{x}}dx \\[1.6em] ~~\rightarrow~~\left(2\sqrt{x}\right)du=dx\\[1.3em]~~\rightarrow~~2(u-1)du=dx}\) what are the new limits of integration? \(\large \color{black}{\displaystyle x=1~~~~\rightarrow ~~~~u=1+\sqrt{1}=2}\) \(\large \color{black}{\displaystyle x=9~~~~\rightarrow ~~~~u=1+\sqrt{9}=4}\) OUR NEW INTEGRAL WITH u ((do NOT sub back the x)) \(\large \color{black}{\displaystyle \int_{2}^{4}\frac{1}{(u-1)(u)^2}~~2(u-1)du}\) u-1 cancelled. (((You can tell someone probably made this problem up by making a sub of a square root instead of a whole variable))) \(\large \color{black}{\displaystyle \int_{2}^{4}2u^{-2}du~=~2~\frac{u^{-2+1}}{(-2+1)}~|^{4}_{2} ~=~\frac{-2}{u}~|^{4}_{2}~\\[1.2em] \displaystyle =\frac{-2}{2}-\frac{-2}{4}=-1+(1/2)=-1/2}\)

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