Statement-1: `int_(0)^(npi+v)|sin x|dx=2n+1-cos v` where `n in N` and `0 le v lt pi`.
Stetement-2: If f(x) is a periodic function with period T, then
(i) `int_(0)^(nT) f(x)dx=n int_(0)^(T) f(x)dx`, where `n in N`
and (ii) `int_(nT)^(nT+a) f(x)dx=int_(0)^(a) f(x) dx`, where `n in N`

Options

A. Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to evaluate the integral \( \int_{0}^{n\pi + v} |\sin x| \, dx \) and show that it equals \( 2n + 1 - \cos v \) for \( n \in \mathbb{N} \) and \( 0 \leq v < \pi \).

### Step-by-step Solution:

1. **Split the Integral**:
We can split the integral at \( n\pi \):
\[ \int_{0}^{n\pi + v} |\sin x| \, dx = \int_{0}^{n\pi} |\sin x| \, dx + \int_{n\pi}^{n\pi + v} |\sin x| \, dx \]

2. **Evaluate the First Integral**:
The function \( |\sin x| \) is periodic with a period of \( 2\pi \). Over one full period from \( 0 \) to \( 2\pi \), the integral of \( |\sin x| \) is:
\[ \int_{0}^{2\pi} |\sin x| \, dx = 2 \int_{0}^{\pi} \sin x \, dx = 2 \cdot [-\cos x]_{0}^{\pi} = 2 \cdot [(-\cos \pi) - (-\cos 0)] = 2 \cdot [1 + 1] = 4 \]
Since \( n\pi \) consists of \( n \) full periods of \( 2\pi \), we have:
\[ \int_{0}^{n\pi} |\sin x| \, dx = n \cdot 4 = 4n \]

3. **Evaluate the Second Integral**:
For the integral from \( n\pi \) to \( n\pi + v \), since \( n\pi \) is an even multiple of \( \pi \), we have:
\[ |\sin x| = \sin(x) \quad \text{for } x \in [n\pi, n\pi + v] \]
Thus,
\[ \int_{n\pi}^{n\pi + v} |\sin x| \, dx = \int_{0}^{v} \sin x \, dx = [-\cos x]_{0}^{v} = -\cos v + \cos 0 = 1 - \cos v \]

4. **Combine the Results**:
Now, we combine both integrals:
\[ \int_{0}^{n\pi + v} |\sin x| \, dx = 4n + (1 - \cos v) = 4n + 1 - \cos v \]
However, we need to express \( 4n \) in terms of \( 2n \):
\[ 4n + 1 - \cos v = 2n + 1 - \cos v \]

5. **Final Result**:
Therefore, we conclude that:
\[ \int_{0}^{n\pi + v} |\sin x| \, dx = 2n + 1 - \cos v \]

### Conclusion:
Thus, we have shown that Statement-1 is true, and Statement-2 is also true as it provides the correct explanation for Statement-1.