Question

If \[ \int_{0}^{x} t^2 \sin(x - t)\,dt = x^2, \] then the sum of values of \( x \), where \( x \in [0,100] \), is:

• A. \( 300\pi \)
• B. \( 272\pi \)
• C. \( 200\pi \)
• D. \( 240\pi \)

Hint:

Integral equations involving variable limits often simplify by differentiating using Leibniz’s rule.

Answer 1

The Correct Option is D

Step 1: Differentiate both sides with respect to \( x \).
Using Leibniz’s rule, \[ \frac{d}{dx}\int_{0}^{x} t^2 \sin(x - t)\,dt = \int_{0}^{x} t^2 \cos(x - t)\,dt \]
Step 2: Differentiate again.
\[ \frac{d^2}{dx^2}\int_{0}^{x} t^2 \sin(x - t)\,dt = -\int_{0}^{x} t^2 \sin(x - t)\,dt + x^2 \] Using the given equation, this becomes \[ - x^2 + x^2 = 0 \]
Step 3: Solve the resulting differential equation.
This leads to \[ x^2 = 2x\sin x \Rightarrow x = 0 \quad \text{or} \quad x = 2\sin x \]
Step 4: Find solutions in the given interval.
The non-zero solutions occur at \[ x = 2n\pi,\quad n = 1,2,\dots \] within \( [0,100] \).
Step 5: Compute the sum.
The largest multiple of \( 2\pi \) less than 100 is \( 76\pi \). Sum of all such solutions gives \[ \boxed{240\pi} \]