Single Variable Calculus, Chapter 5, 5.2, Section 5.2, Problem 46

Evaluate $\displaystyle \int^{\pi/2}_0 (2 \cos x - 5x) dx$ using the properties of integrals and the fact that $\displaystyle \int^{\pi/2}_0 \cos x dx = 1$ and $\displaystyle \int^a_b x dx = \frac{b^2 - a^2}{2}$

Applying the properties of integrals

$
\begin{equation}
\begin{aligned}
\int^a_b [f(x) - g(x)] dx &= \int^a_b f(x)dx - \int^a_b g(x) dx \\
\\
\int^{\pi/2}_0 ( 2 \cos x - 5x) dx &= \int^{\pi/2}_0 2 \cos x dx - \int^{\pi/2}_0 5x dx
\end{aligned}
\end{equation}
$


Then apply $\displaystyle \int^a_b c f(x) dx = c \int^a_b f(x) dx$, where $c$ is any constant then
$\displaystyle \int^{\pi/2}_0 2\cos x dx - \int^{\pi/2}_0 2 5 x dx = 2 \int^{\pi/2}_0 2 \cos x dx - 5 \int^{\pi/2}_0 2 x dx$
Since $\displaystyle \int^{\pi/2}_0 2 \cos x dx 1$ and $\displaystyle \int^b_a x dx = \frac{b^2 - a^2}{2}$, we have

$
\begin{equation}
\begin{aligned}
2 \int^{\pi/2}_0 2 \cos dx - 5 \int^{\pi/2}_0 2 x dx &= 2 (1) - 5 \left[ \frac{\left(\frac{\pi}{2} \right)^2 - (0)^2}{2} \right]\\
\\
2 \int^{\pi/2}_0 2 \cos dx - 5 \int^{\pi/2}_0 2 x dx &= 2 2 - 5 \left( \frac{\frac{\pi^2}{4}}{2}\right)\\
\\
2 \int^{\pi/2}_0 2 \cos dx - 5 \int^{\pi/2}_0 2 x dx &= 2 2 - \frac{5\pi^2}{8}
\end{aligned}
\end{equation}
$