Question

integrate (y^2)sqrt(y^3+x^3) dy dx, 0<=x<=2x, 0 <= y <=4

Answer 1

\[\int\limits_{0}^{4}\int\limits_{0}^{2x} y ^{2}\sqrt{y ^{3}+x^{3}} dy dx\]

Answer 2

what's going on? lol

Answer 3

double integral

Answer 4

\[v=y^3, dv=3y^2dy\]\[\text{When y=2x, v=}8x^3\]
\[\int_0^{2x}y^2\sqrt{y^3+x^3}dy\]\[=\frac13\int_0^{8x^3}\sqrt{v+x^3}dv\]\[=\frac13\left[\frac23(v+x^3)^{\frac32}\right]_0^{8x^3}\]\[=\frac29\left((9x^3)^{\frac32}-(x^3)^{\frac32}\right)\]\[=\frac29\left(27(x^3)^{\frac32}-(x^3)^{\frac32}\right)\]\[=\frac{52}9x^{4.5}\]

Answer 5

the outer-integral i bet you can do it yourself

Answer 6

(i've only considered the inner integral)

Answer 7

If it's not big enough to view the details, enlarge the browser by Ctrl+mouse_scroll or any other means

Answer 8

:O calc iv oh boy I barely passed on a needle and a thread.

Answer 9

I think this one should be...