Question

The integral $ \int_{-1}^{\frac{3}{2}} \left( \pi^2 x \sin(\pi x) \right) dx $ is equal to:

• A. \( 2 + 3\pi \)
• B. \( 3 + 2\pi \)
• C. \( 1 + 3\pi \)
• D. \( 4 + \pi \)

Hint:

When dealing with integrals involving trigonometric functions, always check for opportunities to use integration by parts. Ensure you compute boundary terms carefully.

Answer 1

The Correct Option is C

Let, \( I = \pi^2 \int_{-1}^{1} \left| x \sin \pi x \right| dx \)

\( = \pi^2 \left( \int_{-1}^{0} (-x \sin \pi x) dx + \int_{0}^{1} (x \sin \pi x) dx \right) \)

We consider the integral \( \int x \sin \pi x dx \).

Using integration by parts, we get \( \int x \sin \pi x dx = -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \).

Applying the limits:

\( \int_{0}^{1} x \sin \pi x dx = \left[ -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \right]_0^1 = \left( -\frac{1}{\pi} \cos \pi + \frac{1}{\pi^2} \sin \pi \right) - (0) = \frac{1}{\pi} \)

\( \int_{-1}^{0} (-x \sin \pi x) dx = \left[ \frac{x}{\pi} \cos \pi x - \frac{1}{\pi^2} \sin \pi x \right]_{-1}^0 = (0) - \left( \frac{-1}{\pi} \cos (-\pi) - \frac{1}{\pi^2} \sin (-\pi) \right) = - \left( -\frac{1}{\pi} (-1) \right) = -\frac{1}{\pi} \)

Therefore,

\( I = \pi^2 \left( -\frac{1}{\pi} + \frac{1}{\pi} \right) = 0 \)