- #1

Bucky

- 82

- 0

## Homework Statement

Evaluate the following repeated integral

[tex] \int^2_1 \int^4_2 \sqrt{x-y} dx dy [/tex]

## Homework Equations

## The Attempt at a Solution

[tex] \int^2_1 \int^4_2 x^{1/2} - y^{1/2} dx dy [/tex]

[tex] \int^2_1 [ \frac {2x^{3/2}}{3} - xy^{1/2} ]^4_2 dy [/tex]

[tex] \int^2_1 [ \frac {2(4)^{3/2}}{3} - (4)y^{1/2} ] - [ \frac {2(2)^{3/2}}{3} - (2)y^{1/2} ] dy [/tex]

[tex] \int^2_1 [ \frac {2\sqrt{64}}{3} - (4)y^{1/2} ] - [ \frac {2\sqrt{8}}{3} - (2)\sqrt{y} ] dy [/tex]

[tex] \int^2_1 [ \frac {16}{3} - (4)\sqrt{y}] - [ \frac {2(2)^{3/2}}{3} - (4)\sqrt{2} ] dy [/tex]

[tex] \int^2_1 [ \frac {16 - 4 \sqrt{2}}{3} - 2 y^{1/2}] dy [/tex]

[tex] \frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ - 2 y^{1/2}] dy [/tex]

(can i take it out since it's a constant?)

[tex] \frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ - 2 y^{3/2}]^2_1 [/tex]

[tex] \frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ \frac{- 2 y^{3/2}}{\frac{3}{2}}]^2_1 [/tex]

[tex] \frac {16 - 4 \sqrt{2}}{3} \int^2_1 [ \frac{- 4 y^{3/2}}{3}]^2_1 [/tex]

...and so on. basically it doesn't work out. answer I'm meant to get is

[tex]\frac{4(9\sqrt{3} - 4\sqrt{2} - 1)}{15}[/tex]

can anyone see what I've done wrong? Invairably it's sloppy algebra -.-