The value of ∫ limits0V| sin x| d x where n is a positive integer and V ∈[2 n π,(2 n+1) π], n ∈ N is

Question

The value of ∫ limits0V| sin x| d x where n is a positive integer and V ∈[2 n π,(2 n+1) π], n ∈ N is (A) 4 n-1+ cos V (B) 4 n+1- sin V (C) 4 n+2- sin V (D) 4 n+1- cos V

Tardigrade Admin · Accepted Answer

Q. The value of $0 \int V ∣ sin x ∣ d x$ where $n$ is a positive integer and $V \in [2 nπ, (2 n + 1) π], n \in N$ is

$4 n - 1 + cos V$

$4 n + 1 - sin V$

$4 n + 2 - sin V$

$4 n + 1 - cos V$

$I = 0 \int V ∣ sin x ∣ d x = 0 \int 2 nπ ∣ sin x ∣ d x + 2 nπ \int V ∣ sin x ∣ d x$
Now, $∣ sin x ∣$ has period $π$ and $V$ lies in 1 st quadratnt
Then, $I = 2 n 0 \int π sin x d x + 2 nπ \int V sin x d x$
$= 4 n + [- cos x]_{2 nπ}^{V} = 4 n + 1 - cos V$

Q. The value of 0∫V​∣sinx∣dx where n is a positive integer and V∈[2nπ,(2n+1)π],n∈N is

4n−1+cosV

4n+1−sinV

4n+2−sinV

4n+1−cosV

I=0∫V​∣sinx∣dx=0∫2nπ​∣sinx∣dx+2nπ∫V​∣sinx∣dx Now, ∣sinx∣ has period π and V lies in 1 st quadratnt Then, I=2n0∫π​sinxdx+2nπ∫V​sinxdx =4n+[−cosx]2nπV​=4n+1−cosV

Q. The value of $0 \int V ∣ sin x ∣ d x$ where $n$ is a positive integer and $V \in [2 nπ, (2 n + 1) π], n \in N$ is

$4 n - 1 + cos V$

$4 n + 1 - sin V$

$4 n + 2 - sin V$

$4 n + 1 - cos V$

$I = 0 \int V ∣ sin x ∣ d x = 0 \int 2 nπ ∣ sin x ∣ d x + 2 nπ \int V ∣ sin x ∣ d x$
Now, $∣ sin x ∣$ has period $π$ and $V$ lies in 1 st quadratnt
Then, $I = 2 n 0 \int π sin x d x + 2 nπ \int V sin x d x$
$= 4 n + [- cos x]_{2 nπ}^{V} = 4 n + 1 - cos V$