Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{-2}^2|x \cos \pi x| d x$ is equal to
✓
$\frac{8}{\pi}$
B
$\frac{4}{\pi}$
C
$\frac{2}{\pi}$
D
$\frac{1}{\pi}$
✓
Answer
Correct option: A.
$\frac{8}{\pi}$
(A) Let $I =\int_{-2}^2|x \cos \pi x| d x$ $=2 \int_0^2|x \cos \pi x| d x$ ... $\left[\begin{array}{l}\because \int_{- a }^{ a } f (x) d x=2 \int_0^{ a } f (x) d x \\ \text { if } f (x) \text { is an even function }\end{array}\right]$ $=2\left[\int_0^{1 / 2} x \cos \pi x d x-\int_{1 / 2}^{3 / 2} x \cos \pi x d x\right.$ $\left.+\int_{3 / 2}^2 x \cos \pi x d x\right]$ Since $\int x \cos \pi x=\frac{x \sin \pi x}{\pi}+\frac{\cos \pi x}{\pi^2}$ $\therefore \quad I=2\left[\left(\frac{1}{2 \pi}-\frac{1}{\pi^2}\right)-\left(\frac{-3}{2 \pi}-\frac{1}{2 \pi}\right)+\left(\frac{1}{\pi^2}+\frac{3}{2 \pi}\right)\right]$ $\begin{array}{l}=2\left[\frac{1}{2 \pi}-\frac{1}{\pi^2}+\frac{3}{2 \pi}+\frac{1}{2 \pi}+\frac{1}{\pi^2}+\frac{3}{2 \pi}\right] \\ =2\left(\frac{8}{2 \pi}\right)=\frac{8}{\pi}\end{array}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.