The value of $\int\limits_{0}^{1}{\left| \sin (2\pi x) \right|dx}$ is equal to
A. 0
B. $\dfrac{-1}{\pi }$
C. $\dfrac{1}{\pi }$
D. $\dfrac{2}{\pi }$

Hint: According to definition of absolute value function $f(x)=\left| x \right|$ we can write
$f(x)=x$ when $x\ge 0$
$f(x)=-x$ when $x<0$
And also we can use$\int{\sin (x)dx=-\cos (x)}$

Complete step-by-step answer:
Given integral is $\int\limits_{0}^{1}{\left| \sin (2\pi x) \right|dx}$
For \[0\le x\le \dfrac{1}{2}\] , \[\sin (2\pi x)>0\]
For \[\dfrac{1}{2}\le x\le 1\], \[\sin (2\pi x)<0\]
Hence we can write integral as
$\int\limits_{0}^{1}{\sin (2\pi x)dx=\int\limits_{0}^{{}^{1}/{}_{2}}{\sin (2\pi x)dx}}+\int\limits_{{}^{1}/{}_{2}}^{1}{-\sin (2\pi x)dx}$
Now calculate $\int{\sin (2\pi x)dx}$
Let $2\pi x=t$
On differentiating
$2\pi dx=dt$
$dx=\dfrac{dt}{2\pi }$
Hence integral is
$\Rightarrow \int{\sin (2\pi x)dx}=\dfrac{1}{2\pi }\int{\sin (t)dt}$
$\Rightarrow \int{\sin (2\pi x)dx}=\dfrac{-\cos (t)}{2\pi }$
On substituting $2\pi x=t$
$\Rightarrow \int{\sin (2\pi x)dx}=\dfrac{-\cos (2\pi x)}{2\pi }$
As we have $\int\limits_{0}^{1}{\sin (2\pi x)dx=\int\limits_{0}^{{}^{1}/{}_{2}}{\sin (2\pi x)dx}}+\int\limits_{{}^{1}/{}_{2}}^{1}{-\sin (2\pi x)dx}$
After substituting value of $\int{\sin (2\pi x)dx}$
\[\Rightarrow \int\limits_{0}^{1}{\sin (2\pi x)dx=\left[ \dfrac{-\cos (2\pi x)}{2\pi } \right]_{0}^{{}^{1}/{}_{2}}}+\left[ \dfrac{\cos (2\pi x)}{2\pi } \right]_{{}^{1}/{}_{2}}^{1}\]
\[\Rightarrow \int\limits_{0}^{1}{\sin (2\pi x)dx=\left[ \dfrac{-\cos (\pi )}{2\pi }+\dfrac{\cos (0)}{2\pi } \right]}+\left[ \dfrac{\cos (2\pi )}{2\pi }-\dfrac{\cos (\pi )}{2\pi } \right]\]
\[\Rightarrow \int\limits_{0}^{1}{\sin (2\pi x)dx=\left[ \dfrac{-(-1)}{2\pi }+\dfrac{1}{2\pi } \right]}+\left[ \dfrac{1}{2\pi }-\dfrac{(-1)}{2\pi } \right]\].
\[\Rightarrow \int\limits_{0}^{1}{\sin (2\pi x)dx=\left[ \dfrac{1}{2\pi }+\dfrac{1}{2\pi } \right]}+\left[ \dfrac{1}{2\pi }+\dfrac{1}{2\pi } \right]\]
\[\Rightarrow \int\limits_{0}^{1}{\sin (2\pi x)dx=\dfrac{2}{2\pi }+}\dfrac{2}{2\pi }\]
\[\Rightarrow \int\limits_{0}^{1}{\sin (2\pi x)dx=\dfrac{4}{2\pi }}\]
\[\Rightarrow \int\limits_{0}^{1}{\sin (2\pi x)dx=\dfrac{2}{\pi }}\]
Hence option D is correct.

Note: In above question we need to classify range of x very carefully.
We can understand it from graph of sin(x) as below:


As from 0 to $\pi $ , sin(x) is positive and from $\pi $ to $2\pi $
That’s we classify range as below:
For \[0\le x\le \dfrac{1}{2}\] , \[\sin (2\pi x)>0\]