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

1. Analyze the integrand
The integrand is \( | \pi^2 x \sin(\pi x) | \). The absolute value will make the integral slightly tricky. We need to determine where \( \pi^2 x \sin(\pi x) \) is positive and negative in the interval \( [-1, 3/2] \).

• \(\pi^2\) is always positive.
• For \( x > 0 \), \( x \) is positive; for \( x < 0 \), \( x \) is negative.
• \(\sin(\pi x)\) is positive for \( 0 < x < 1 \), negative for \( -1 < x < 0 \), and again positive for \( 1 < x < 2 \), and so on.

2. Split the integral into intervals based on the sign of \( \pi^2 x \sin(\pi x) \)

• Interval 1: \( -1 \leq x \leq 0 \):
  • For \( x \) negative and \( \sin(\pi x) \) negative, \( \pi^2 x \sin(\pi x) \) is positive.
  • Therefore, \( | \pi^2 x \sin(\pi x) | = \pi^2 x \sin(\pi x) \) in this interval.
  • We need to evaluate \( \int_{-1}^{0} \pi^2 x \sin(\pi x) \, dx \).
• Interval 2: \( 0 \leq x \leq 1 \):
  • For \( x \) positive and \( \sin(\pi x) \) positive, \( \pi^2 x \sin(\pi x) \) is positive.
  • Therefore, \( | \pi^2 x \sin(\pi x) | = \pi^2 x \sin(\pi x) \) in this interval.
  • We need to evaluate \( \int_{0}^{1} \pi^2 x \sin(\pi x) \, dx \).
• Interval 3: \( 1 \leq x \leq \frac{3}{2} \):
  • For \( x \) positive and \( \sin(\pi x) \) negative, \( \pi^2 x \sin(\pi x) \) is negative.
  • Therefore, \( | \pi^2 x \sin(\pi x) | = - \pi^2 x \sin(\pi x) \) in this interval.
  • We need to evaluate \( \int_{1}^{\frac{3}{2}} - \pi^2 x \sin(\pi x) \, dx \).

3. Evaluate the integrals
We'll use integration by parts. Let \( u = x \) and \( dv = \sin(\pi x) \, dx \). Then \( du = dx \) and \( v = -\frac{\cos(\pi x)}{\pi} \). The integral of \( x \sin(\pi x) \, dx = -x \frac{\cos(\pi x)}{\pi} + \int \frac{\cos(\pi x)}{\pi} \, dx = -x \frac{\cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} + C \). Now, let's evaluate each interval:

• Interval 1: \[ \int_{-1}^{0} \pi^2 x \sin(\pi x) \, dx = \pi^2 \left[ -x \frac{\cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{-1}^{0} \] \[ = \pi^2 \left[ (0) - \left( -(-1) \frac{\cos(-\pi)}{\pi} + \frac{\sin(-\pi)}{\pi^2} \right) \right] \] \[ = \pi^2 \left[ - \left( \frac{-\cos(\pi)}{\pi} + 0 \right) \right] = \pi^2 \left[ - \left( \frac{-(-1)}{\pi} \right) \right] = \pi^2 \left( \frac{1}{\pi} \right) = \pi \]
• Interval 2: \[ \int_{0}^{1} \pi^2 x \sin(\pi x) \, dx = \pi^2 \left[ -x \frac{\cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{0}^{1} \] \[ = \pi^2 \left[ \left( -1 \frac{\cos(\pi)}{\pi} + \frac{\sin(\pi)}{\pi^2} \right) - (0) \right] = \pi^2 \left[ \frac{1}{\pi} + 0 \right] = \pi \]
• Interval 3: \[ \int_{1}^{\frac{3}{2}} -\pi^2 x \sin(\pi x) \, dx = -\pi^2 \left[ -x \frac{\cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{\frac{3}{2}} \] \[ = -\pi^2 \left[ \left( - \frac{3}{2} \frac{\cos(\frac{3\pi}{2})}{\pi} + \frac{\sin(\frac{3\pi}{2})}{\pi^2} \right) - \left( -1 \frac{\cos(\pi)}{\pi} + \frac{\sin(\pi)}{\pi^2} \right) \right] \] \[ = -\pi^2 \left[ \left( 0 - \frac{1}{\pi^2} \right) - \left( \frac{1}{\pi} + 0 \right) \right] = -\pi^2 \left[ -\frac{1}{\pi^2} - \frac{1}{\pi} \right] = -\pi^2 \left[ - \frac{1 + \pi}{\pi^2} \right] = 1 + \pi \]

4. Sum the results
Total Integral = \( \pi + \pi + (1 + \pi) = 3\pi + 1 \).
Answer: The integral is equal to \( 1 + 3\pi \). So the answer is option 3.