Question
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ is equal to:
- 1
- 2
- -1
- -2
Solution:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$
$=-\int\limits^0_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}+\int\limits^\frac{\pi}{2}_0\sin\text{x}\text{ dx}$
$=-\big[-\cos\text{x}\big]^0_{-\frac{\pi}{2}}+\big[-\cos\text{x}\big]^\frac{\pi}{2}_0$
$=1-0-0+1$
$=2$
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$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.
$( S 1): A \cap B =(1, \infty)-N$ and
$( S 2): A \cup B=(1, \infty)$