JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Area Under Curves

On $[-\pi ,\pi ]$ the graphs of $\sin x$ and $\cos x$ cross at $x=-\frac{3\pi }{4}$ and $x=\frac{\pi }{4}$, so $f(x)=\sin x$ on $\left[-\pi ,-\frac{3\pi }{4}\right]$, $f(x)=\cos x$ on $\left[-\frac{3\pi }{4},\frac{\pi }{4}\right]$ and $f(x)=\sin x$ on $\left[\frac{\pi }{4},\pi \right]$. The enclosed area is the sum of the moduli of the signed integrals over the sign-intervals: ${\int }_{-\pi }^{-3\pi /4}|\sin x|dx=1-\frac{1}{\sqrt{2}}$, ${\int }_{-3\pi /4}^{-\pi /2}|\cos x|dx=1-\frac{1}{\sqrt{2}}$, ${\int }_{-\pi /2}^{\pi /4}\cos xdx=1+\frac{1}{\sqrt{2}}$ and ${\int }_{\pi /4}^{\pi }\sin xdx=1+\frac{1}{\sqrt{2}}$. On adding, the $\frac{1}{\sqrt{2}}$ terms cancel and the area equals $4$.