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

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

Answers (1)

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

           x=a,t=a-a=0

           \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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