Find the exact area of the region bounded byhttp://edugen.wiley.com/edugen/share...=1241573763553 , http://edugen.wiley.com/edugen/share...=1241573763553

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- May 5th 2009, 05:40 PMZosoPageArea of a region
Find the exact area of the region bounded byhttp://edugen.wiley.com/edugen/share...=1241573763553 , http://edugen.wiley.com/edugen/share...=1241573763553

- May 5th 2009, 06:03 PMmr fantastic
- May 6th 2009, 10:51 AMZosoPage
Sorry again.

Here's the problem without any red x's.

Find the exact area of the region bounded by $\displaystyle y=\frac{x^3}{\sqrt{4-x^2}}, y=0, x=0, x=\sqrt{2}$ - May 6th 2009, 11:05 AMskeeter
- May 6th 2009, 12:19 PMZosoPage
- May 6th 2009, 12:39 PMskeeter
$\displaystyle u^2 = 4-x^2$

$\displaystyle x^2 = 4-u^2$

$\displaystyle 2x \, dx = -2u \, du$

$\displaystyle x \, dx = -u \, du$

$\displaystyle \int \frac{x^3}{\sqrt{4-x^2}} \, dx = \int \frac{x^2}{\sqrt{4-x^2}} \cdot x \, dx $

substitute ...

$\displaystyle \int \frac{4-u^2}{u} \cdot (-u) \, du$

$\displaystyle -\int 4 - u^2 \, du$

lower limit ... $\displaystyle x = 0$ , $\displaystyle u = 2$

upper limit ... $\displaystyle x = \sqrt{2}$ , $\displaystyle u = \sqrt{2}$

$\displaystyle -\int_2^{\sqrt{2}} 4 - u^2 \, du = \int_{\sqrt{2}}^2 4 - u^2 \, du$