IIT-JEE 1994Question 1
Show that $\displaystyle\int_{0}^{n\pi+v}|\sin x|\,dx=2n+1-\cos v$ where $n$ is a positive integer and $0\le v<\pi$.
Definite Integration
Answerproved
Solution
Concept. $|\sin x|$ has period $\pi$, and each arch has area $2$. Split $[0,n\pi+v]$ into $n$ full arches and the piece $[n\pi,n\pi+v]$. Formulas used. (1) $\int_0^\pi|\sin x|\,dx=2$, so $\int_0^{n\pi}|\sin x|\,dx=2n$(2) $\int_{n\pi}^{n\pi+v}|\sin x|\,dx=\int_0^v\sin x\,dx$ (period $\pi$, and $\sin x\ge0$ on $[0,v]$)Step 1 - Picture.Step 2 - Full arches. By (1), $2n$. Step 3 - Last piece. By (2), $\int_0^v\sin x\,dx=1-\cos v$. Step 4 - Add. $2n+1-\cos v$. Hence proved. Answer: proved