Definite Integration Ques 88
- 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
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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
$$ \begin{aligned} { }^{n} C _k \quad \int _0^{1}(1-x)^{n-k} \cdot x^{k} d x & =\frac{1}{n+1} \\ \Rightarrow \quad \int _0^{1}(1-x)^{n-k} x^{k} d x & =\frac{1}{(n+1)^{n} C _k} \end{aligned} $$
BETA
