Question

Evaluate the integral: $ \int_{-1}^{3/2} | \pi^2 x \sin(\pi x) | \, dx $

• A. \(4\pi + 1\)
• B. \(3\pi + 1\)
• C. \(5\pi + 1\)
• D. \(6\pi + 1\)

Hint:

When dealing with absolute value integrals involving trigonometric functions, break the integral into intervals where the function inside the absolute value changes its sign.

Answer 1

The Correct Option is B

We need to evaluate the integral: \[ I = \int_{-1}^{3/2} | \pi^2 x \sin(\pi x) | \, dx \] Since \( \sin(\pi x) \) changes its sign at integer multiples of 1, we need to break the integral into intervals where the absolute value expression will simplify.
Step 1: Simplifying the Absolute Value Expression
The absolute value \( |\sin(\pi x)| \) is split depending on the sign of \( \sin(\pi x) \).
We can express this integral as: \[ I = \int_{-1}^{0} \pi^2 x \sin(\pi x) \, dx + \int_{0}^{3/2} \pi^2 x |\sin(\pi x)| \, dx \]
Step 2: Integral from \(-1\) to 0
In the interval \([-1, 0]\), \( \sin(\pi x) \) is negative, so \( |\sin(\pi x)| = -\sin(\pi x) \).
The integral becomes: \[ I_1 = \int_{-1}^{0} -\pi^2 x \sin(\pi x) \, dx \]
Step 3: Integral from 0 to 3/2
In the interval \([0, 3/2]\), \( \sin(\pi x) \) remains positive for \( x \in (0,1) \), and negative for \( x \in (1,3/2) \).
We will evaluate these parts separately.
Step 4: Compute Each Part Evaluating each part will give us: \[ I = 3\pi + 1 \] Thus, the correct answer is \( 3\pi + 1 \).