Question

Evaluate the integral using the properties of even and odd functions as an aid. $$ \int_{-2}^{2} x^{2}\left(x^{2}+1\right) d x $$

Short answer

The value of the integral is \(2 \left[\frac{32}{5} + \frac{8}{3}\right]\).

Step-by-step solution

Identify the function type

Check if the given function \(x^{2}(x^{2} + 1)\) is even or odd. We know a function is even if \(f(-x) = f(x)\) for all x in the function's domain. In this case, replacing x by -x, we get \((-x)^{2} ((-x)^{2} + 1) = x^{2}(x^{2} + 1)\). So, the function is even.

Apply the property of the definite integral of an even function

As it's an even function, the integral between -a and a simplifies to twice the integral from 0 to a. Hence, we have\[\int_{-2}^{2} x^{2} (x^{2} + 1) \, dx = 2\int_{0}^{2} x^{2}(x^{2} + 1) \, dx\]

Compute the integral

Finally, compute the integral \[2\int_{0}^{2} x^{2}(x^{2} + 1) \, dx\]By expanding and applying the power rule for integration (the integral of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\)), we get\[2 \left[\frac{x^5}{5} + \frac{x^3}{3} \right]_0^2\]

Evaluate the integral at the limits

Evaluate the integral at limits 0 and 2. \[= 2 \left[\frac{2^5}{5} + \frac{2^3}{3} \right] - 2 \left[\frac{0^5}{5} + \frac{0^3}{3} \right] = 2 \left[\frac{32}{5} + \frac{8}{3}\right]\]