# Double Integration

• Nov 29th 2007, 11:54 AM
stuwy
Double Integration
∫R (x^1/2 - y^1/2) dydx
where y=x^2, y= x^1/4, x=0, x=1
im struggling to solve this!! i came up with an answer of 1.5 but i dont think i have done it correct!...could anyone help please?
• Nov 29th 2007, 01:33 PM
galactus
$\displaystyle \int_{0}^{1}\int_{x^{\frac{1}{4}}}^{x^{2}}(\sqrt{x }-\sqrt{y})dydx$

First, integrate wrt y:

$\displaystyle \int_{x^{\frac{1}{4}}}^{x^{2}}{\sqrt{x}-\sqrt{y}}dy=y\sqrt{x}-\frac{2y^{\frac{3}{2}}}{3}$

Now use the limits of integration for y. By the FTC:

$\displaystyle \frac{-2x^{3}}{3}+x^{\frac{5}{2}}-x^{\frac{3}{4}}+\frac{2x^{\frac{3}{8}}}{3}$

Now, integrate this wrt x:

$\displaystyle \int_{0}^{1}{\left[\frac{-2x^{3}}{3}+x^{\frac{5}{2}}-x^{\frac{3}{4}}+\frac{2x^{\frac{3}{8}}}{3}\right]}dx$

We get:

$\displaystyle \frac{x^{4}}{6}+\frac{2x^{\frac{7}{2}}}{7}-\frac{4x^{\frac{7}{4}}}{7}+\frac{16x^{\frac{11}{8} }}{33}$

Now, use the limits of integration for x, 0 and 1:

We get $\displaystyle \frac{154}{5}$