Question
$\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$ is equal to:
-
$\frac{\pi}{4}$
-
$\frac{\pi}{2}$
-
$\pi$
-
$1$
$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\pi$
$1$
Solution:
We have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_01(-\cos\text{x})\text{dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\int\limits^{\frac{\pi}{2}}_0\cos\text{x dx}$
$=-\big[\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=-\big[0-0\big]+\big[1-0\big]$
$=1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\left| x \right| + \left[ x \right],}&{ - 1 \leq x < 1} \\
{x + \left| x \right|,}&{1 \leq x < 2} \\
{x + \left| x \right|,}&{2 \leq x \leq 3}
\end{array}} \right.$
where $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is discontinuous at:
The value of objective function is maximum under linear constraints