Question

If I (m, n) = $\int _ { 0 } ^ { 1 } \mathrm { t } ^ { \mathrm { m } }$ (1 + t)ⁿ dt ; m, n Ī R, then I (m, n) is -

• A. $\frac { \mathrm { n } } { 1 + \mathrm { m } }$ I [(m + 1), (n – 1)]
• B. $\frac { \mathrm { m } } { \mathrm { n } + 1 }$ I [(m + 1), (n – 1)]
• C. – I [(m + 1, n – 1)]
• D. – I [(m + 1, n – 1)]

Answer 1

Answer: (C)

– I $(m + 1, n – 1)$

Answer 2

I (m, n) = $\int _ { 0 } ^ { 1 } \mathrm { t } ^ { \mathrm { m } }$(1 + t)ⁿ dt =

[Integrating by Parts]

= – t^{m +1} dt

= – I [(m + 1), (n – 1)].