The value of \(\int_{-\pi/2}^{\pi/2}\displaystyle (x^5+x^3\cos x)dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Over a symmetric interval \([-a, a]\), the integral of an odd function is 0.
Let \(f(x) = x^5 + x^3\cos x\). Since \(\cos(-x) = \cos x\),
\(f(-x) = (-x)^5 + (-x)^3\cos x\)
\(= -x^5 - x^3\cos x = -f(x)\)
So f is odd, and
\( \displaystyle \int_{-\pi/2}^{\pi/2} (x^5 + x^3\cos x)\,dx = 0 \)