Question
$\int\limits^1_{-1}|1-\text{x}|\text{dx}$ is equal to:
- -2
- 2
- 0
- 4
Solution:
$\int\limits^1_{-1}|1-\text{x}|\text{dx}$
$=\int\limits^0_{-1}(1-\text{x})\text{dx}+\int\limits^1_0(1-\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^0_{-1}+\Big[\text{x}-\frac{\text{x}^2}{2}\Big]^1_0$
$=0+1+\frac{1}{2}+1-\frac{1}{2}-0$
$=2$
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$f(x)=\sin ^{-1}\left(\frac{3 x^{2}+x-1}{(x-1)^{2}}\right)+\cos ^{-1}\left(\frac{x-1}{x+1}\right)$ is :