Answer: \(\displaystyle 2\mathrm{e}^{\pi/8}-\frac{11}{5}\).

Answer: \(\displaystyle 2\mathrm{e}^{\pi/8}-\frac{11}{5}\).

The curve \(y=\cos5x\) cuts the \(x\)-axis where

\(\cos5x=0\).

The first positive solution is

\(5x=\frac{\pi}{2}\),

so

\(x=\frac{\pi}{10}\).

Thus \(B\) has coordinates \(\left(\frac{\pi}{10},0\right)\).

The shaded area is the area under \(y=\mathrm{e}^{x/2}\) from \(x=0\) to \(x=\frac{\pi}{4}\), minus the area under \(y=\cos5x\) from \(x=0\) to \(x=\frac{\pi}{10}\):

Now

\(\displaystyle \int \mathrm{e}^{x/2}\,dx=2\mathrm{e}^{x/2}\),

and

\(\displaystyle \int \cos5x\,dx=\frac{1}{5}\sin5x\).

Therefore

This gives

Since \(\sin\frac{\pi}{2}=1\),

\(\displaystyle \text{area}=2\mathrm{e}^{\pi/8}-2-\frac15=2\mathrm{e}^{\pi/8}-\frac{11}{5}\).