Question

Find the value of \[\int_0^{2n\pi } {\left( {\left| {\sin x} \right| - \left| {\dfrac{1}{2}\sin x} \right|} \right)dx} \].
A \[n\]
B \[2n\]
C \[2n\pi \]
D None of these.

Answer 1

Hint: To find value of definite integral first solve the integral. After simplification of integral substitute limits of the integral and perform subtraction.



Formula Used:\[\int {\sin xdx} = - \cos x + c\]



Complete step by step solution:The given definite integral is \[\int_0^{2n\pi } {\left( {\left| {\sin x} \right| - \left| {\dfrac{1}{2}\sin x} \right|} \right)dx} \].
First simplify the integral as follows,
\[\int_0^{2n\pi } {\left( {\left| {\sin x} \right| - \dfrac{1}{2}\left| {\sin x} \right|} \right)dx} \]
Further simplify the integral by taking common terms out.
\[\begin{array}{l}\int_0^{2n\pi } {\left( {\left| {\sin x} \right|\left( {1 - \dfrac{1}{2}} \right)} \right)dx} \\ \Rightarrow \int_0^{2n\pi } {\left( {\dfrac{1}{2}\left| {\sin x} \right|} \right)dx} \end{array}\]
Take constant out of integral.
\[ \Rightarrow \dfrac{1}{2}\int_0^{2n\pi } {\left| {\sin x} \right|dx} \]
Find integration with respect to x.
\[\dfrac{1}{2} \times 4n\int\limits_0^{\dfrac{\pi }{2}} {\sin xdx} \]
Now the integral becomes as follows.
\[ \Rightarrow 2n\int\limits_0^{\dfrac{\pi }{2}} {\sin xdx} \]
Now find integration
\[ \Rightarrow 2n\left[ { - \cos x} \right]_0^{\dfrac{\pi }{2}}\]
Substitute the limit in the expression.
\[\begin{array}{l}2n\left( { - \cos \dfrac{\pi }{2} - \left( { - \cos 0} \right)} \right)\\ \Rightarrow 2n\left( {0 + 1} \right)\\ \Rightarrow 2n\end{array}\]
Hence, the value of \[\int_0^{2n\pi } {\left( {\left| {\sin x} \right| - \left| {\dfrac{1}{2}\sin x} \right|} \right)dx} \]is \[2n\].



Option ‘B’ is correct



Note: The common mistake happen by student is taking integration of \[\int_0^{2n\pi } {\left| {\sin x} \right|dx} \]as \[\left[ { - \cos x} \right]_0^{2n\pi }\]which is wrong.