Question

The area between x-axis and curve y = cos x when 0 ≤ x ≤ 2 π is
(a) 0
(b) 2
(c) 3
(d) 4

Answer 1

(d) 4




$$\begin{gathered} \mathrm{Required}\mathrm{shaded}\mathrm{area}, \\ A={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cos xdx+{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(-\cos x\right)dx+{\int }_{3\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{2\mathrm{\pi }}\cos xdx \\ ={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cos xdx-{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cos xdx+{\int }_{3\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{2\mathrm{\pi }}\cos xdx \\ ={\left[\sin x\right]}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left[\sin x\right]}_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+{\left[\sin x\right]}_{3\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{2\mathrm{\pi }} \\ ={\left[\sin x\right]}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left[\sin x\right]}_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+{\left[\sin x\right]}_{3\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{2\mathrm{\pi }} \\ =\left(1-0\right)-\left(-1-1\right)+\left[0-\left(-1\right)\right] \\ =1+2+1 \\ =4\mathrm{sq}\mathrm{units} \end{gathered}$$