Question

The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .

विकल्प

•  4

•  2

• −2

• 0

Answer 1

4

\[\text{We have}, \]
\[I = \int_{- 2}^2 \left| 1 - x^{{}^2} \right| d x\]
\[\left| 1 - x^{2} \right| = \begin{cases}- \left( 1 - x^{2} \right)&,& - 2 < x < - 1 \\ \left( 1 - x^{2} \right)&,& - 1 < x < 1\\ - \left( 1 - x^{2} \right)&,& 1 < x < 2\end{cases}\]
\[ \therefore I = \int_{- 2}^{- 1} \left| 1 - x^{2} \right| d x + \int_{- 1}^1 \left| 1 - x^{2} \right| d x + \int_1^2 \left| 1 - x^{2} \right| d x\]
\[ = \int_{- 2}^{- 1} - \left( 1 - x^{2} \right) d x + \int_{- 1}^1 \left( 1 - x^{2} \right) d x + \int_1^2 - \left( 1 - x^{2} \right) d x\]
\[ = - \int_{- 2}^{- 1} \left( 1 - x^{2} \right) d x + \int_{- 1}^1 \left( 1 - x^{2} \right) d x - \int_1^2 \left( 1 - x^{2} \right) d x\]
\[ = - \left[ x - \frac{x^3}{3} \right]_{- 2}^{- 1} + \left[ x - \frac{x^3}{3} \right]_{- 1}^1 - \left[ x - \frac{x^3}{3} \right]_1^2 \]
\[ = - \left[ - 1 + \frac{1}{3} + 2 - \frac{8}{3} \right] + \left[ 1 - \frac{1}{3} + 1 - \frac{1}{3} \right] - \left[ 2 - \frac{8}{3} - 1 + \frac{1}{3} \right]\]
\[ = - \left[ 1 - \frac{7}{3} \right] + \left[ 2 - \frac{2}{3} \right] - \left[ 1 - \frac{7}{3} \right]\]
\[ = - 1 + \frac{7}{3} + 2 - \frac{2}{3} - 1 + \frac{7}{3}\]
\[ = 4\]