Question

Consider a function $f:R\to R$ is periodic function with period $\pi$ and is defined in $[0,\pi ]$ as $f\left(x\right)=\{\begin{array}{c}1-\cos x,0\le x<\frac{\pi }{2}\\ 2-\frac{2x}{\pi },\frac{\pi }{2}\le x\le \pi \end{array}$ Then the area(in $\text{sq.units}$) bounded by $y=f\left(x\right)$ and the $x-$axis from $x=-n\pi$ to $x=n\pi (n\in Z^{+}),$ is

• A. $\frac{n(\pi +4)}{4}$
• B. $\frac{n(3\pi -4)}{2}$
• C. $\frac{n(3\pi -2)}{4}$
• D. $\frac{n(\pi +2)}{2}$

Answer 1

The correct option is B $\frac{n(3\pi -4)}{2}$

The bounded region between $0$ to $\pi$ is shown :
So , Required area $=2n\underset{0}{\overset{\pi /2}{\int }}(1-\cos x)dx+2n(\frac{1}{2}\cdot 1\cdot \frac{\pi }{2})$
$[\because f(x\left)\text{has period}\pi \right]$
$=\frac{n(3\pi -4)}{2}$