Question

Evaluate the given integral.

\(\int\limits_0^{\frac{{3\pi }}{2}} {\left| {\sin x} \right|dx} \)

Short answer

The value of the given integral is 3.

Step-by-step solution

Definition of function

The definition of the function is as follows:

If

\(x \in \left( {0,\pi } \right),\sin x \succ 0 \Rightarrow \left| {\sin x} \right| = \sin x\)

And if

\(x \in \left( {\pi .\frac{{3\pi }}{2}} \right),\sin x \prec 0 \Rightarrow \left| {\sin x} \right| = - \sin x\)

Rewrite the integral

\(\int\limits_0^{\frac{{3\pi }}{2}} {\left| {\sin x} \right|dx = \int\limits_0^\pi {\left| {\sin x} \right|dx + \int\limits_\pi ^{\frac{{3\pi }}{2}} {\left| {\sin x} \right|dx} } } \)

Evaluate the integral

\(\begin{aligned}{c}\int\limits_0^{\frac{{3\pi }}{2}} {\left| {\sin x} \right|dx} &= \int\limits_0^\pi {\left| {\sin x} \right|dx + \int\limits_\pi ^{\frac{{3\pi }}{2}} {\left| {\sin x} \right|dx} } \\ &= \int\limits_0^\pi {\sin xdx} - \int\limits_\pi ^{\frac{{3\pi }}{2}} {\sin xdx} \\ &= - \left. {\cos x} \right)_0^\pi + \left. {\cos x} \right)_\pi ^{\frac{{3\pi }}{2}}\\ &= - \cos \pi + \cos 0 + \cos \frac{{3\pi }}{2} - \cos \pi \\ &= 1 + 1 + 0 + 1\\ &= 3\end{aligned}\)

Therefore, the value of the integral is 3.