Question
Solve: $\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
- $\frac{1}{2}$
- $1$
- $2$
- $\frac{3}{2}$
Solution:
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1+\sin2\text{x}\text{dx}}=$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{xdx}}$
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$\int\limits_{0}^{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$
$=[-\cos\text{x}+\sin\text{x}]\frac{\frac{\pi}{2}}{{0}}$
$=\Big[-\Big(\cos\frac{\pi}{2}-\cos0\Big)+\Big(\sin\frac{\pi}{2}-\sin0\Big)\Big]$
$=\big[-\big(0-1\big)+\big(1-0\big)\big]$
$=1+1=2$
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| $X$ | $\alpha$ | $1$ | $0$ | $-3$ |
| $P(X)$ | $\frac{1}{3}$ | $K$ | $\frac{1}{6}$ | $\frac{1}{4}$ |
be $\mu$ and $\sigma$, respectively. If $\sigma-\mu=2$, then $\sigma+\mu$ is equal to....................