# First, switch the order of integration and evaluate I = integral 9 0 integral 3 square root 9...

## Question:

First, switch the order of integration and evaluate {eq}I=\int_0^9 \int_{\sqrt{9}}^3 \cos(x^3) \ dxdy {/eq}

## Double Integral:

We will reverse the order of integration where we will solve the problem by reversing the limits as the limits of y will be varibale and the limits of x will be constant.

To solve the problem we will reverse the order of integration:

{eq}\int_{0}^{9}\int_{\sqrt{y}}^{3}\cos x^{3}dxdy {/eq}

Now reverse the order we will get:

{eq}0\leq y\leq x^{2}\\ 0\leq x\leq 3 {/eq}

The integral will be:

{eq}=\int_{0}^{3}\int_{0}^{x^{2}}\cos x^{3}dydx\\ =\frac{1}{3}\left [ \cos 27-1 \right ] {/eq}