1. The problem is to explain the integration by parts formula: $$\int_a^b u(x)v'(x)\,dx = [u(x)v(x)]_a^b - \int_a^b u'(x)v(x)\,dx.$$\n\n2. Integration by parts is derived from the product rule of differentiation which states: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$\n\n3. Integrate both sides of the product rule over the interval $[a,b]$: $$\int_a^b \frac{d}{dx}[u(x)v(x)]\,dx = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx.$$\n\n4. The integral of a derivative over $[a,b]$ simplifies to the evaluation of the function at the boundaries: $$[u(x)v(x)]_a^b = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx.$$\n\n5. Rearrange to isolate the integral of $u(x)v'(x)$ on the left: $$\int_a^b u(x)v'(x)\,dx = [u(x)v(x)]_a^b - \int_a^b u'(x)v(x)\,dx.$$\n\n6. This formula allows substitution when integrating products of functions, converting the integral into something easier to evaluate.