Funtegration

$\begin{aligned} I & = \int_0^{10} g(x) f'(x) \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = g(x) f(x) \bigg|_0^{10} - \int_0^{10} f(x) g'(x) \ dx \\ & = g(10)f(10)-g(0)f(0) - 2015 \\ & = 99^2 - 88^2 - 2015 \\ & = \boxed{42} \end{aligned}$

Taking the definite integral $\int_{0}^{10} f(x)g'(x) dx = 2015$ , a simple application of integration-by-parts yields:

$\int_{0}^{10} f(x)g'(x) dx = f(x)g(x) - \int_{0}^{10} f'(x)g(x) dx = 2015 \Rightarrow \int_{0}^{10} f'(x)g(x) dx = f(x)g(x)|_{0}^{10} - 2015;$

or $\int_{0}^{10} f'(x)g(x) dx = [f(10)g(10) - f(0)g(0)] - 2015 = 99^2 - 88^2 - 2015 = \boxed{42}.$