MCQ
$I = \int_{\,0}^{\,1} {\,x{{(1 - x)}^n}dx} =$
- A$\frac{1}{{n + 1}}$
- B$\frac{1}{{n + 2}}$
- ✓$\frac{1}{{n + 1}} - \frac{1}{{n + 2}}$
- D$\frac{1}{{n + 1}} + \frac{1}{{n + 2}}$
$ - I = \int_0^1 { - x{{(1 - x)}^n}dx = \int_0^1 {(1 - x - 1){{(1 - x)}^n}dx} } $
$ = \int_0^1 {{{(1 - x)}^{n + 1}}dx - \int_0^1 {{{(1 - x)}^n}dx} } $
$ = \left[ {\frac{{{{(1 - x)}^{n + 2}}}}{{ - (n + 2)}}} \right]_0^1 - \left[ {\frac{{{{(1 - x)}^{n + 1}}}}{{ - (n + 1)}}} \right]_0^1 = \frac{1}{{n + 2}} - \frac{1}{{n + 1}}$
$ \Rightarrow I = \frac{1}{{n + 1}} - \frac{1}{{n + 2}}.$
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