Question:

Given the function $f(x) = \int_0^x t \sin(x-t) dt$. **Step 1: Simplify $f(x)$** To simplify the integral, perform a substitution. Let $u = x-t$. Then $t = x-u$, and $dt = -du$. The limits of integration change as follows: When $t=0$, $u=x$. When $t=x$, $u=0$. Substituting these into the integral for $f(x)$: $$f(x) = \int_x^0 (x-u)\sin u (-du)$$ Reversing the limits of integration changes the sign of the integral: $$f(x) = \int_0^x (x-u)\sin u \,du$$ Distribute $x$ and $u$: $$f(x) = x\int_0^x \sin u \,du - \int_0^x u\sin u \,du$$ Evaluate the first integral: $$\int_0^x \sin u \,du = [-\cos u]_0^x = -\cos x - (-\cos 0) = 1 - \cos x$$ Evaluate the second integral using integration by parts, $\int v\,dw = vw - \int w\,dv$. Let $v=u$ and $dw=\sin u\,du$. Then $dv=du$ and $w=-\cos u$. $$\int_0^x u\sin u \,du = [-u\cos u]_0^x - \int_0^x (-\cos u)\,du$$ $$= (-x\cos x - 0\cdot\cos 0) + \int_0^x \cos u\,du$$ $$= -x\cos x + [\sin u]_0^x$$ $$= -x\cos x + (\sin x - \sin 0)$$ $$= -x\cos x + \sin x$$ Substitute these results back into the expression for $f(x)$: $$f(x) = x(1-\cos x) - (-x\cos x + \sin x)$$ $$f(x) = x - x\cos x + x\cos x - \sin x$$ $$f(x) = x - \sin x$$ **Step 2: Find $f'(x)$** Differentiate $f(x)$ with respect to $x$: $$f'(x) = \frac{d}{dx}(x - \sin x)$$ $$f'(x) = 1 - \cos x$$ **Step 3: Find $f''(x)$** Differentiate $f'(x)$ with respect to $x$: $$f''(x) = \frac{d}{dx}(1 - \cos x)$$ $$f''(x) = 0 - (-\sin x)$$ $$f''(x) = \sin x$$