Definite Integration Question 86
Question 86
- Evaluate $\int_{0}^{1}(t x+1-x)^{n} d x$,
where $n$ is a positive integer and $t$ is a parameter independent of $x$. Hence, show that
$$ \int_{0}^{1} x^{k}(1-x)^{n-k} d x=\frac{1}{{ }^{n} C_{k}(n+1)}, \text { for } k=0,1, \ldots, n \text {. } $$
$(1981,4 M)$
Integer Answer Type Questions
Show Answer
Answer:
Correct Answer: 88. $\frac{t^{n+1}-1}{(t-1)(n+1)}$
Solution:
- Let $I=\int_{0}^{1}(t x+1-x)^{n} d x=\int_{0}^{1}{(t-1) x+1}^{n} d x$
$$ \begin{aligned} & =\frac{((t-1) x+1)^{n+1}}{(n+1)(t-1)}=\frac{1}{n+1} \frac{t^{n+1}-1}{t-1} \ & =\frac{1}{n+1}\left(1+t+t^{2}+\ldots+t^{n}\right) \end{aligned} $$
Again, $\quad I=\int_{0}^{1}(t x+1-x)^{n} d x=\int_{0}^{1}[(1-x)+t x]^{n} d x$
$$ =\int_{0}^{1}\left[{ }^{n} C_{0}(1-x)^{n}+{ }^{n} C_{1}(1-x)^{n-1}(t x)\right. $$
$$ \left.{ }^{n} C_{2}(1-x)^{n-2}(t x)^{2}+\ldots+{ }^{n} C_{n}(t x)^{n}+\right] d x $$
$=\int_{0}^{1} \sum_{r=0}^{n}{ }^{n} C_{r}(1-x)^{n-r}(t x)^{r} d x$
$=\sum_{r=0}^{n}{ }^{n} C_{r} \int_{0}^{1}(1-x)^{n-r} \cdot x^{r} d x t^{r}$
From Eqs. (i) and (ii), we get
$$ \sum_{r=0}^{n}{ }^{n} C_{r} \quad \int_{0}^{1}(1-x)^{n-r} \cdot x^{r} d x \quad t^{r}=\frac{1}{n+1}\left(1+t+\ldots+t^{n}\right) $$
On equating coefficient of $t^{k}$ on both sides, we get
$$ \begin{aligned} { }^{n} C_{k} \quad \int_{0}^{1}(1-x)^{n-k} \cdot x^{k} d x & =\frac{1}{n+1} \ \Rightarrow \quad \int_{0}^{1}(1-x)^{n-k} x^{k} d x & =\frac{1}{(n+1)^{n} C_{k}} \end{aligned} $$
