Question

Let $I=\underset{0}{\overset{n\pi +\lambda }{\int }}|\sin x|dx$, where $n\in N,0\le \lambda <\pi$. Then the value of $I$ is

Answer 1

Given: $I=\underset{0}{\overset{n\pi +\lambda }{\int }}|\sin x|dxn\in N,0\le \lambda <\pi$
$\Rightarrow I=\underset{0}{\overset{\lambda }{\int }}|\sin x|dx+\underset{\lambda }{\overset{n\pi +\lambda }{\int }}|\sin x|dx$
Using the property:
$\underset{a}{\overset{a+nT}{\int }}f\left(x\right)dx=n\underset{0}{\overset{T}{\int }}f\left(x\right)dx$
Where $T$ is the period of $f\left(x\right)$
For $|\sin x|$, the period is $\pi$
Now,
$\Rightarrow I=\underset{0}{\overset{\lambda }{\int }}|\sin x|dx+n\underset{0}{\overset{\pi }{\int }}|\sin x|dx=[-\cos x{]}_0^{\lambda }+n\underset{0}{\overset{\pi }{\int }}\sin xdx=-(\cos \lambda -1)+n(1+1)\therefore I=2n+1-\cos \lambda$