The area of the figure bounded by the lines $x = 0$, $x =\pi /2$, $f \left(x\right) = \sin x$ and $g\left(x\right) = \cos x$ is

We can divide the interval $[0, \pi /2]$ into the parts, $[0, \pi /4]$, where $\sin x\le \cos x$, and $[\mathrm{\pi }/4, \mathrm{\pi }/2]$, where $\sin x \ge \cos x$. so the required area is
$\begin{array}{l}{\int }_0^{\pi /4} (\cos x-\sin x)\mathrm{d}x+{\int }_{\pi /4}^{\pi /2} (\sin x-\cos x)\mathrm{d}x\\ =[\sin x+\cos x{]}_0^{\pi /4}+[-\cos x-\sin x{]}_{\pi /4}^{\pi /2}\\ =(\sqrt{2}-1)+(-1+\sqrt{2})=2(\sqrt{2}-1)\end{array}$