Q63 / 180 · KCET 2018 · Mathematics
The area of the region bounded by the curve \( y=\cos x \) between \( x=0 \) and \( x=\pi \) is
- A\( 1 \) sq. unit
- B\( 4 \) sq. unit
- Correct answerC\( 2 \) sq. unit
- D\( 3 \) sq. unit
Solution
Given curve, \( y=\cos x \) between \( x=0 \) and \( x=\pi \)
So, required area is given by
\[ \begin{array}{l} \int_{0}^{\pi}|\cos x| d x=\int_{0}^{\pi / 2}(\cos x) d x+\int_{\pi / 2}^{\pi}(-\cos x) d x \\ \int_{0}^{\pi}|\cos x| d x=\left.(\sin x)\right|_{0} ^{\pi / 2}+.\left.(-\sin x)\right|_{\Pi / 2} ^{\Pi} \\ =1+1=\text { 2sq.units. } \end{array} \]
So, required area is given by
\[ \begin{array}{l} \int_{0}^{\pi}|\cos x| d x=\int_{0}^{\pi / 2}(\cos x) d x+\int_{\pi / 2}^{\pi}(-\cos x) d x \\ \int_{0}^{\pi}|\cos x| d x=\left.(\sin x)\right|_{0} ^{\pi / 2}+.\left.(-\sin x)\right|_{\Pi / 2} ^{\Pi} \\ =1+1=\text { 2sq.units. } \end{array} \]
Hence, the correct answer is option: 'C'.