Question

Evaluate ${\int }_0^1{(tx+1-x)}^{n}dx,$ where n is a positive integer and t is a parameter independent of x. Hence ${\int }_0^1{x}^k{(1-x)}^{n-k}dx=\frac{P}{\left[{}^n{C}_k(n+1)\right]}fork=0,1,......n$, then $P=$

• A. 2
• B. 1
• C. 3
• D. None of these

Answer 1

Answer: (B)

1

Let $I={\int }_0^1{(tx+1-x)}^{n}dx={\int }_0^1{\{(t-1)x+1\}}^{n}dx={\left[\frac{(t-1){x+1}^{n+1}}{(n+1)(t-1)}\right]}_0^1$
$=\frac{1}{n+1}\left(\frac{t^{n+1}-1}{t-1}\right)=\frac{1}{n+1}(1+t+t^2+...t^n)$ ...(1)
Again ${\int }_0^1{(tx+1-x)}^{n}dx={\int }_0^1{[(1-x)+tx]}^{n}dx$

$={\int }_0^1[{}^n{C}_0{(1-x)}^n{+}^n{C}_1{(1-x)}^{n-1}\left(tx\right){+}^n{C}_2{(1-x)}^{n-2}{\left(tx\right)}^2+...{+}^n{C}_n{\left(tx\right)}^n]dx$
$={\int }_0^1\sum _{r=1}^n{}^n{C}_r{(1-x)}^{n-r}{\left(tx\right)}^{r}dx=\sum _{r=1}^n{}^n{C}_r\left[{\int }_0^1{(1-x)}^{n-r}{x}^{r}dx\right]t^r$ ...(2)
From (1) and (2)
$\sum _{r=1}^n{}^n{C}_r\left[{\int }_0^1{(1-x)}^{n-r}.x^{r}dx\right]t^r=\frac{1}{n+1}(1+t+...t^n)$
On equating coefficient of $t^k$ on both sides , we get
${}^n{C}_k\left[{\int }_0^1{(1-x)}^{n-r},x^{r}dx\right]\Rightarrow {\int }_0^1{(1-x)}^{n-k}{x}^{h}dx=\frac{1}{{(n+1)}^n{C}_k}$