Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 390: 46

$\int_{0}^{\pi/2}(2~cos~x-5x)~dx= 2-\frac{5\pi^2}{8}$

We can evaluate the integral using properties of integrals: $\int_{0}^{\pi/2}(2~cos~x-5x)~dx$ $= \int_{0}^{\pi/2}2~cos~x~dx-\int_{0}^{\pi/2}5x~dx$ $= 2\int_{0}^{\pi/2}cos~x~dx-5\int_{0}^{\pi/2}x~dx$ $= 2(1)-5[\frac{(\pi/2)^2-0^2}{2}]$ $= 2-5(\frac{\pi^2}{8})$ $= 2-\frac{5\pi^2}{8}$