Question

Evaluate : ${\int }_0^{\mathrm{\pi }}\left|\cos x\right|dx$

• A. $1/2$
• B. $-2$
• C. $1$
• D. $-1$
• E. $2$

Answer 1

The correct option is E

$2$

Step 1:Consider the given equation as :

$I={\int }_0^{\mathrm{\pi }}\left|\cos x\right|dx$ …… (1)

We have, $f\left(x\right)=\left|\cos x\right|$

In the first quadrant $\cos x$ is positive that is from $0to\frac{\mathrm{\pi }}{2}$ and in the second quadrant is $\cos x$is negative that is from $\frac{\mathrm{\pi }}{2}to\mathrm{\pi }$, so that we can say that :

$f\left(x\right)=\left\{\begin{array}{ll}\cos x,& 0\le x\le \frac{\mathrm{\pi }}{2}\\ -\cos x,& \frac{\mathrm{\pi }}{2}\le x\le \mathrm{\pi }\end{array}\right.$

Step 2: Rewrite the Equation $\left(1\right)$ as,

$\begin{array}{rcl}I& =& {\int }_0^{\mathrm{\pi }}\left|\cos x\right|dx\\ I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\cos xdx+{\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}\left(-\cos x\right)dx\\ I& =& {\left[\sin x\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[-\sin x\right]}_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}\\ I& =& \left(\sin \frac{\mathrm{\pi }}{2}-\sin 0\right)-\left(\sin \mathrm{\pi }-\sin \frac{\mathrm{\pi }}{2}\right)\\ I& =& 1-0-0+1\\ I& =& 2\end{array}$

Therefore, the correct answer is Option (E).