SATHEE: Indefinite Integration 1 Question 87

Indefinite Integration 1 Question 87

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

$\int_{0}^{1} x^{k} (1 - x)^{n - k} d x = \frac{1}{^{n} C_{k} (n + 1)}, for k = 0, 1, \dots, n .$

$(1981, 4 M)$

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Answer:

Correct Answer: 88. $\frac{t^{n + 1} - 1}{(t - 1) (n + 1)}$

Solution:

Let $I = \int_{0}^{1} (t x + 1 - x)^{n} d x = \int_{0}^{1} {(t - 1) x + 1}^{n} d x$

$\begin{aligned} = \frac{((t - 1) x + 1)^{n + 1}}{(n + 1) (t - 1)} = \frac{1}{n + 1} \frac{t^{n + 1} - 1}{t - 1} \\ = \frac{1}{n + 1} (1 + t + t^{2} + \dots + t^{n}) \end{aligned}$

Again, $I = \int_{0}^{1} (t x + 1 - x)^{n} d x = \int_{0}^{1} [(1 - x) + t x]^{n} d x$

$= \int_{0}^{1} [^{n} C_{0} (1 - x)^{n} +^{n} C_{1} (1 - x)^{n - 1} (t x)$

$^{n} C_{2} (1 - x)^{n - 2} (t x)^{2} + \dots +^{n} C_{n} (t x)^{n} +] d x$

$= \int_{0}^{1} \sum_{r = 0}^{n}^{n} C_{r} (1 - x)^{n - r} (t x)^{r} d x$

$= \sum_{r = 0}^{n}^{n} C_{r} \int_{0}^{1} (1 - x)^{n - r} \cdot x^{r} d x t^{r}$

From Eqs. (i) and (ii), we get

$\sum_{r = 0}^{n}^{n} C_{r} \int_{0}^{1} (1 - x)^{n - r} \cdot x^{r} d x t^{r} = \frac{1}{n + 1} (1 + t + \dots + t^{n})$

On equating coefficient of $t^{k}$ on both sides, we get

$\begin{aligned} ^{n} C_{k} \int_{0}^{1} (1 - x)^{n - k} \cdot x^{k} d x & = \frac{1}{n + 1} \\ \Rightarrow \int_{0}^{1} (1 - x)^{n - k} x^{k} d x & = \frac{1}{(n + 1)^{n} C_{k}} \end{aligned}$

Indefinite Integration 1 Question 87

88. Evaluate $\int_{0}^{1} (t x + 1 - x)^{n} d x$ ,

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Indefinite Integration 1 Question 87

88. Evaluate ∫01(tx+1−x)ndx,

Integer Answer Type Questions

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88. Evaluate $\int_{0}^{1} (t x + 1 - x)^{n} d x$ ,