MCQ
$\int_{ - 2}^2 {|1 - {x^2}|\,dx = } $
- A$2$
- ✓$4$
- C$6$
- D$8$
$ + \int_1^2 {|1 - {x^2}|dx} $
$= - \int_{ - 2}^{ - 1} {(1 - {x^2})\,dx + \int_{ - 1}^1 {(1 - {x^2})\,dx - \int_1^2 {(1 - {x^2})\,dx} } } $
$= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4.$
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$\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$
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