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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
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
    $\frac{{(n + 1)!}}{2}\,$
✓

Answer

Correct option: C.
$\frac{{n\,\,!}}{2}$
c
$I = \int\limits_0^\infty  {{{({x^2})}^n}\,.\,x\,{e^{ - {x^2}}}\,\,dx\,} $

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 $I = \frac{{n\,!\,}}{2}$

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