Q120 / 180 · KCET 2018 · Mathematics
\( \int_{-2}^{2}|x \cos \pi x| d x \) is equal to
- Correct answerA\( \frac{8}{\pi} \)
- B\( \frac{4}{\pi} \)
- C\( \frac{2}{\pi} \)
- D\( \frac{1}{\pi} \)
Solution
Given that,
\[ \begin{array}{l} \int_{-2}^{2}|x \cos \Pi x| d x=2 \int_{0}^{2}|x \cos x| d x \\ =2\left[\int_{0}^{1 / 2} x \cos \pi x-\int_{1 / 2}^{3 / 2}(x \cos \Pi x) d x+\int_{3 / 2}^{2}(x \cos \Pi x) d x\right] \\ =2\left[\frac{1}{2 \Pi}-\frac{1}{\Pi^{2}}+\frac{4}{2 \Pi}+\frac{3}{2 \Pi}+\frac{1}{\Pi^{2}}\right] \\ =2\left(\frac{8}{2 \Pi}\right)=\frac{8}{\Pi} \end{array} \]
\[ \begin{array}{l} \int_{-2}^{2}|x \cos \Pi x| d x=2 \int_{0}^{2}|x \cos x| d x \\ =2\left[\int_{0}^{1 / 2} x \cos \pi x-\int_{1 / 2}^{3 / 2}(x \cos \Pi x) d x+\int_{3 / 2}^{2}(x \cos \Pi x) d x\right] \\ =2\left[\frac{1}{2 \Pi}-\frac{1}{\Pi^{2}}+\frac{4}{2 \Pi}+\frac{3}{2 \Pi}+\frac{1}{\Pi^{2}}\right] \\ =2\left(\frac{8}{2 \Pi}\right)=\frac{8}{\Pi} \end{array} \]
Hence, the correct answer is option: 'A'.