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Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 50 maths textbook solution.

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Answer:-  \frac{2}{(n+1)(n+2)(n+3)}

Hints:-  You must know the rules of integration with its limits.

Given:- Prove  \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x

             Hence evaluate \int_{0}^{1} x^{2}(1-x)^{n} d x

Solution : Let I=\int_{0}^{a} f(x) d x                                          ....(1)

\begin{aligned} &\text { Put } x=a-t \\ &\therefore d x=-d t \end{aligned}

When, x=0,t=a-0=a


           \begin{aligned} I=\int_{0}^{a} f(x) d x &=\int_{a}^{0} f(a-t)(-d t) \\ &=-\int_{a}^{0} f(a-t) d t \\ &=\int_{0}^{a} f(a-x) d x \end{aligned}                                \left[\because \int_{a}^{b} f(x) \cdot d x=-\int_{b}^{a} f(x) \cdot d x\right]

         \therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x

Now, \int_{0}^{1} x^{2}(1-x)^{n} d x

          \begin{aligned} I &=\int_{0}^{1}\left(1-x^{2}\right)[1-(1-x)]^{n} d x \\ &=\int_{0}^{1}\left(1-2 x+x^{2}\right) x^{n} d x \\ &=\int_{0}^{1}\left(x^{n}-2 x^{n+1}+x^{n+2}\right) d x \end{aligned}

             \begin{aligned} &=\left[\frac{x^{n+1}}{n+1}-2 \cdot \frac{x^{n+2}}{n+2}+\frac{x^{n+3}}{n+3}\right]_{0}^{1} \\ &=\left[\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}\right] \end{aligned}

             \begin{aligned} &=\frac{(n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)}{(n+1)(n+2)(n+3)} \\ &=\frac{n^{2}+5 n+6-2 n^{2}-8 n-6+n^{2}+3 n+2}{(n+1)(n+2)(n+3)} \\ &=\frac{2}{(n+1)(n+2)(n+3)} \end{aligned}

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