Question

If $g\left(x\right)=\underset{0}{\overset{x}{\int }}(|\sin t|+|\cos t|)dt$, then the value of $g(x+\frac{n\pi }{2})$,( where $n\in N$) is

Answer 1

We have,
$g\left(x\right)=\underset{0}{\overset{x}{\int }}(|\sin t|+|\cos t|)dt$
Now,
$g(x+\frac{n\pi }{2})=\underset{0}{\overset{x+\frac{n\pi }{2}}{\int }}(|\sin t|+|\cos t|)dt=\underset{0}{\overset{x}{\int }}(|\sin t|+|\cos t|)dt+\underset{x}{\overset{x+\frac{n\pi }{2}}{\int }}(|\sin t|+|\cos t|)dt$
Using the property:
$\underset{a}{\overset{a+nT}{\int }}f\left(x\right)dx=\underset{0}{\overset{nT}{\int }}f\left(x\right)dx$
$\Rightarrow g(x+\frac{n\pi }{2})=g\left(x\right)+\underset{0}{\overset{\frac{n\pi }{2}}{\int }}(|\sin t|+|\cos t|)dt$
As $(|\sin t|+|\cos t|)$ has a period $\frac{\pi }{2}$
$\Rightarrow g(x+\frac{n\pi }{2})=g\left(x\right)+g\left(\frac{n\pi }{2}\right)\Rightarrow g(x+\frac{n\pi }{2})=g\left(x\right)+\underset{0}{\overset{\frac{n\pi }{2}}{\int }}(|\sin x|+|\cos x|)dx\Rightarrow g(x+\frac{n\pi }{2})=g\left(x\right)+n\underset{0}{\overset{\frac{\pi }{2}}{\int }}(\sin x+\cos x)dx\Rightarrow g(x+\frac{n\pi }{2})=g\left(x\right)+n{[\sin x-\cos x]}_0^{\frac{\pi }{2}}\therefore g(x+\frac{n\pi }{2})=g\left(x\right)+2n$