Question

Evaluate: \(\displaystyle\int_{-\pi/2}^{\pi/2} (x^3 + x\cos x)\, dx\)

Hint:

Both x³ and x·cos x are odd functions, so their integral over symmetric limits is 0.

Answer 1

Step 1: Split the integral into two parts:
$\displaystyle\int_{-\pi/2}^{\pi/2}(x^3+x\cos x)\,dx = \int_{-\pi/2}^{\pi/2}x^3\,dx + \int_{-\pi/2}^{\pi/2}x\cos x\,dx$

Step 2: Evaluate the first part directly:
$\displaystyle\int_{-\pi/2}^{\pi/2}x^3\,dx = \left[\frac{x^4}{4}\right]_{-\pi/2}^{\pi/2} = \frac{(\pi/2)^4}{4} - \frac{(-\pi/2)^4}{4} = 0$, since $(\pi/2)^4=(-\pi/2)^4$.

Step 3: Evaluate the second part using integration by parts, then substitute limits:
$\int x\cos x\,dx = x\sin x + \cos x$. Evaluating from $-\pi/2$ to $\pi/2$: at $\pi/2$, this is $\frac{\pi}{2}(1)+0=\frac{\pi}{2}$; at $-\pi/2$, this is $-\frac{\pi}{2}(-1)+0=\frac{\pi}{2}$. Subtracting gives $\frac{\pi}{2}-\frac{\pi}{2}=0$.

Final Answer:
Both parts vanish, so the total is 0. \[ \boxed{0} \]