Question

Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{0}^{2} x \sqrt[3]{4+x^{2}} d x $$

Short answer

The solution to the definite integral \(\int_{0}^{2} x \sqrt[3]{4+x^{2}} dx\) is \(\frac{3}{4} [8^{4/3} - 4^{4/3}]\).

Step-by-step solution

Identify the Correct Substitution

Look for a function and its derivative within the integral. Here, we see that if we choose \(u=4+x^{2}\), its derivative \(du=2x dx\) is also present. So make substitution \(u=4+x^{2}\). Then, calculate \(du\) by taking the derivative of \(u\) with respect to \(x\) which yields \(du= 2x dx\). Now, let's solve for \(dx\) which is \(dx=du/(2x)\).

Substitute and Simplify

Replace \(4+x^{2}\) with \(u\) and \(dx\) with \(du/(2x)\) in the integral: \(\int_{0}^{2} x \sqrt[3]{u} (du/2x)\). The \(x\) in the numerator and denominator cancels out and the integral becomes \(\frac{1}{2}\int_{u=4}^{u=8} \sqrt[3]{u} du\). Notice the change of the limits of integration because we are now integrating with respect to \(u\).

Perform the Integration

Now perform the integration. The integral of \(\sqrt[3]{u}\) or \(u^{1/3}\) is \(\frac{3}{4} u^{4/3}\). This yields \(\frac{3}{4}\int_{u=4}^{u=8} u^{4/3} du\).

Evaluate the Definite Integral

Finally, substitute the limits of integration into the result obtained in step 3. The result is \(\frac{3}{4} [8^{4/3} - 4^{4/3}]\). Simplifying gives the final result.