Question

Use a symbolic integration utility to evaluate the double integral. $$ \int_{0}^{2} \int_{x^{2}}^{2 x}\left(x^{3}+3 y^{2}\right) d y d x $$

Short answer

-8.5714

Step-by-step solution

Compute the inner integral with respect to y.

The first step is to compute the inner integral with respect to \(y\), which is \(\int_{x^{2}}^{2x} (x^3 + 3y^2) dy\). We integrate the function \(x^3 + 3y^2\) with respect to \(y\). Note that \(x^3\) can be treated as a constant while integrating with respect to \(y\). This gives \(x^3y + y^3\). Now evaluate this from \(y = x^2\) to \(y = 2x\).

Substitute the limits for the inner integral

Replacing \(y\) by \(2x\), we get: \(2x^{4} + 8x^{3}\). Replacing \(y\) by \(x^{2}\), we get: \(x^{5} + x^{6}\). Subtract the two results to get: \(2x^{4} + 8x^{3} - x^{5} - x^{6}\).

Compute the outer integral with respect to x

Now, we have simplified the problem to a single variable integral. The final step is to compute the outer integral from 0 to 2 on the result from Step 2, which is \(\int_{0}^{2} (2x^4 + 8x^3 - x^5 - x^6) dx\).

Substitute the limits for the outer integral

After integrating \(2x^4 + 8x^3 - x^5 - x^6\) with respect to \(x\), we get \(0.4x^5 + 1.6x^4 - 0.1667x^6 - 0.1429x^7\). Evaluate from \(x=0\) to \(x=2\) to get the final result.