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: Key Formula or Approach:
For a function integrated over symmetric limits \([-a,a]\): if the function is odd, the integral is 0; if it is even, the integral is \(2\int_0^a\). Check each term for odd/even behaviour first.

Step 2: Checking \(x^3\):
\((-x)^3 = -x^3\), so \(x^3\) is an odd function. Its integral over \([-\pi/2,\pi/2]\) is 0.

Step 3: Checking \(x\cos x\):
\(x\) is odd and \(\cos x\) is even, and odd \(\times\) even = odd. So \((-x)\cos(-x) = -x\cos x\), confirming \(x\cos x\) is odd, and its integral over the symmetric interval is also 0.

Step 4: Adding the pieces:
\[ \int_{-\pi/2}^{\pi/2}(x^3+x\cos x)\,dx = 0 + 0 = 0 \]

Final Answer:
The integral evaluates to 0. \[ \boxed{0} \]