Question

In Exercises \(51-54,\) evaluate the iterated integral. (Note that it is necessary to switch the order of integration.) $$ \int_{0}^{2} \int_{x}^{2} x \sqrt{1+y^{3}} d y d x $$

Short answer

To solve this problem, we have to change the order of integration, then take the integral with respect to x then y using basic integration techniques and substitution. The final solution is obtained after evaluating these integrals.

Step-by-step solution

Change the Order of Integration

First we change the order of integration. To do this, we have to understand the area over which we are integrating. The original constraints were \(0 \leq x \leq 2\) and \(x \leq y \leq 2\). Switching these around, we get \(0 \leq y \leq 2\) and \(0 \leq x \leq y\). So, our integral now becomes: \[ \int_{0}^{2} \int_{0}^{y} x \sqrt{1+y^{3}} dx dy \]

Evaluate the Inner Integral With Respect to x

Now, we can easily integrate the inner integral with respect to x since it is now only involving x. \(x\) will be integrated as \(\frac{1}{2}x^{2}\), while \(\sqrt{1+y^{3}}\) is a constant with respect to x and therefore will remain unchanged. So we will have: \[ \int_{0}^{2} [0.5*y^{2} * \sqrt{1+y^{3}} |_{0}^{y}] dy \], which simplifies to \[ \int_{0}^{2} 0.5*y^{2} * \sqrt{1+y^{3}} dy \]

Evaluate the Outer Integral With Respect to y

Now we can integrate with respect to y. This is a slightly more complex integral, but with substitution (\(u=1+y^{3}, du = 3y^{2}dy\)), it can be reduced to a simpler form. The solution to this integral yields the final answer.