Question

The number of elements in the set $S = \{ x : x \in [0, 100] \text{ and \int_0^x t^2 \sin(x-t) dt = x^2 \}$ is ___}

Hint:

Property $\int_0^a f(t) dt = \int_0^a f(a-t) dt$ simplifies convolution integrals.

Answer 1

Correct Answer: 2

Let $I = \int_0^x t^2 \sin(x-t) dt$. Use substitution $u = x-t$.
$I = \int_0^x (x-u)^2 \sin u du = \int_0^x (x^2 - 2xu + u^2) \sin u du$.
$I = x^2 \int \sin u du - 2x \int u \sin u du + \int u^2 \sin u du$.
Evaluate parts: $\int_0^x \sin u = 1-\cos x$.
$\int_0^x u \sin u = \sin x - x\cos x$.
$\int_0^x u^2 \sin u = -x^2\cos x + 2x\sin x + 2\cos x - 2$.
Combine: $x^2(1-\cos x) - 2x(\sin x - x\cos x) + (-x^2\cos x + 2x\sin x + 2\cos x - 2)$.
Simplify: $x^2 - x^2\cos x - 2x\sin x + 2x^2\cos x - x^2\cos x + 2x\sin x + 2\cos x - 2$.
$I = x^2 + 2\cos x - 2$.
Set $I = x^2 \implies x^2 + 2\cos x - 2 = x^2$.
$2\cos x = 2 \implies \cos x = 1$.
$x = 2n\pi$. Given $x \in [0, 100]$.
$0 \le 2n\pi \le 100 \implies 0 \le n \le \frac{50}{\pi} \approx 15.9$.
Integers $n \in \{0, 1, ..., 15\}$. Total 16 values.