$\int^\frac{2}{3}_{0}\frac{\text{dx}}{4+9\text{x}^{2}}\text{equals}$
$\frac{\pi}{6}$
$\frac{\pi}{12}$
$\frac{\pi}{24}$
$\frac{\pi}{4}$
$\ \text{put}\ 3\text{x}=\text{t}\Rightarrow3\text{dx}=\text{dt}$
$\therefore\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}=\frac{1}{3}\int\frac{\text{dt}}{(2)^{2}+\text{t}^{2}}$
$=\frac{1}{3}\bigg[\frac{1}{2}\tan^{-1}\frac{t}{2}\bigg]$
$=\frac{1}{6}\tan^{-1}\bigg(\frac{3\text{x}}{2}\bigg)$
$=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain $\int\limits_{0}^{\frac{2}{3}}\frac{\text{dx}}{4+9\text{x}^{2}}=\text{F}\bigg(\frac{2}{3}\bigg)-\text{F}(0)$ $=\frac{1}{6}\tan^{-1}\bigg(\frac{3}{2}.\frac{2}{3}\bigg)-\frac{1}{6}\tan^{-1}0$ $=\frac{1}{6}\tan^{-1}1-0$ $=\frac{1}{6}\times\frac{\pi}{4}$ $=\frac{\pi}{24}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X)$ | $K^2$ | $2K$ | $K$ | $2K$ | $5K^2$ |
Then $\mathrm{P}(\mathrm{X}> 2)$ is equal to
| $\text{X}:$ | $2$ | $3$ | $4$ | $5$ |
| $\text{P}(\text{X}):$ | $\frac{5}{\text{k}}$ | $\frac{7}{\text{k}}$ | $\frac{9}{\text{k}}$ | $\frac{11}{\text{k}}$ |
The value of k is: