calculate the integral.

Question

∫₀²∫₀^{√2x-x^2} xdydx

Answer 1

Do the "inner" integral first. Notice that integrating x with respect to y behaves as if x is a constant.

∫₀^{√2x-x^2} xdy = [xy]₀^{√2x-x^2} = √2x^2-x^3

The do the "ourter" integral.

∫₀² (√2x^2-x^3) dx = (8√2)/3 - 4 ≈ -0.228764