MCQ
$\int\limits_0^\infty {} x^{2n + 1}·{e^{ - {x^2}}}\, dx$ is equal to $(n \in N).$
- A$n !$
- B$2 (n !)$
- ✓$\frac{{n\,\,!}}{2}$
- D$\frac{{(n + 1)!}}{2}\,$
put $x^2 = t \Rightarrow x\, dx = - dt/2$
$=\frac{1}{2}\,\int\limits_0^\infty {\,{t^n}\,\,{e^{ - t}}\,\,dt\,} $
$=\frac{1}{2}\,\left[ {\,\,{t^n}\,\left. {{e^{ - t}}} \right]_0^\infty \,\,\,\, + \,n\int\limits_0^\infty {\,{t^{n - 1}}\,\,{e^{ - t}}\,\,dt\,} } \right]$
$= \frac{1}{2}\,\left[ {\,\,0\,\, + \,\,n\int\limits_0^\infty {\,{t^{n - 1}}\,\,{e^{ - t}}\,\,dt\,} } \right]$
Hence $I = \frac{{n\,!\,}}{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.