Statement-1: $\int\limits\limits_0^{n \pi+v}|\sin x| d x=2 n+1-\cos v$ where $n \in N$ and $0 \leq v<\pi$.
Statement-2: If $f(x)$ is a periodic function with period $T$, then
(i) $\int\limits_0^{n T} f(x) d x=n \int\limits_0^T f(x) d x$, where $n \in N$ and
(ii) $\int\limits_{n T}^{n t+a} f(x) d x=\int\limits_0^a f(x) d x$, where $n \in N$
Answer & explanation
Correct answer: option 1
Clearly, statement-2 is true.
Now,
$\int\limits_0^{n \pi+v}|\sin x| d x$
$=\int\limits_0^{n \pi}|\sin x| d x+\int\limits_{n \pi}^{n \pi+v}|\sin x| d x$
$=n \int\limits_0^\pi|\sin x| d x+\int\limits_0^v|\sin x| d x$
$=n \int\limits_0^\pi \sin x d x+\int\limits_0^v \sin x d x$
$[∵ 0<\sin x<1$ for $x \in(0, \pi)]$
$=2 n+1-\cos v=(2 n+1)-\cos v$
So, statement- 1 is also true and statement-2 is a correct explanation for statement-1.