The value of the integral `int_(-2)^2|1-x^2|dx` is `4` b. `2` c. `-2` d. `0`

Explanation

`4`

We have,


`I=int_{-2}^{2}|1-x^{2}| d x `

`|1-x^{2}|= {(-(1-x^{2}) , ,quad-2 < x < -1),((1-x^{2}) , ,quad-1 < x < 1),(-(1-x^{2}) , ,quad 1 < x < 2):} `

`therefore I=int_{-2}^{-1}|1-x^{2}| d x+int_{1-1}^{1}|1-x^{2}| d x+int_{1}^{2}|1-x^{2}| d x `

`=int_{-2}^{-1}-(1-x^{2}) d x+int_{-1}^{1}(1-x^{2}) d x+int_{1}^{2}-(1-x^{2}) d x `

`=-int_{-2}^{-1}(1-x^{2}) d x+int_{-1}^{1}(1-x^{2}) d x-int_{1}^{2}(1-x^{2}) d x `

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