Question bank/JEE Main/Mathematics

The area of the region A = in square units, is

A$\frac{4}{3}$B$\frac{2}{3}$C$\frac{1}{3}$D2

Show answer and solution▾

$= \int_{- 1}^{0}\mspace{2mu}\left( - x^{2} + 1 \right)dx + \int_{- 1}^{1}\mspace{2mu}\left( x^{2} + 1 \right)dx$
$= \left. \ \left( - \frac{x^{3}}{3} + x \right) \right|_{- 1}^{0} + \left. \ \left( \frac{x^{3}}{3} + x \right) \right|_{0}^{1} = 0 - \left\lbrack \frac{1}{3} - 1 \right) + \frac{1}{3} + 1 = 2$

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx