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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$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$
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$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$
has a non-trivial solution, then the value of $\theta$ is :