# Area of a region

• May 5th 2009, 06:40 PM
ZosoPage
Area of a region
• May 5th 2009, 07:03 PM
mr fantastic
Quote:

Originally Posted by ZosoPage

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• May 6th 2009, 11:51 AM
ZosoPage
Sorry again.

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

Find the exact area of the region bounded by $y=\frac{x^3}{\sqrt{4-x^2}}, y=0, x=0, x=\sqrt{2}$
• May 6th 2009, 12:05 PM
skeeter
Quote:

Originally Posted by ZosoPage
Sorry again.

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

Find the exact area of the region bounded by $y=\frac{x^3}{\sqrt{4-x^2}}, y=0, x=0, x=\sqrt{2}$

try the substitution ...

$u^2 = 4 - x^2$

you'll end up with an easier definite integral ...

$\int_{\sqrt{2}}^2 4 - u^2 \, du
$
• May 6th 2009, 01:19 PM
ZosoPage
Quote:

Originally Posted by skeeter
try the substitution ...

$u^2 = 4 - x^2$

you'll end up with an easier definite integral ...

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

I'm confused. How did you use that substitution to come up with that integral? I need to take this integral but I don't know how:

$
\int_{0}^{\sqrt{2}}\frac{x^3}{\sqrt{4-x^2}}dx
$
• May 6th 2009, 01:39 PM
skeeter
Quote:

Originally Posted by ZosoPage
I'm confused. How did you use that substitution to come up with that integral? I need to take this integral but I don't know how:

$
\int_{0}^{\sqrt{2}}\frac{x^3}{\sqrt{4-x^2}}dx
$

$u^2 = 4-x^2$

$x^2 = 4-u^2$

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

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

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

substitute ...

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

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

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

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

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