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  • Explain solution rd sharma class 12 chapter 19 Definite Integrals exercise 19.5, question 13

Explain solution rd sharma class 12 chapter 19 Definite Integrals exercise 19.5, question 13

Answers (1)

\frac{27}{2}

Hint:

To solve the given statement, we have to use the formula of addition limits.

Given:

\int_{4}^{1}\left ( x^{2}-x \right )dx

Solution:

\int_{a}^{b}\left ( fx \right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]

Where, h=\frac{b-a}{n}

Here,

a=1,b=4,f(x)=x^{2}-x\\ h=\frac{4-1}{n}=\frac{3}{n}

Thus, we have

\begin{aligned} &I=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ &=\lim _{h \rightarrow 0} h\left[\left(1^{2}-1\right)+\left((1+h)^{2}-(1+h)\right)+\left((1+2 h)^{2}-(1+h)\right)\right] \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} h\left[0+\left(h+h^{2}\right)+\left(2 h+(2 h)^{2}\right)+\ldots\right] \\ &=\lim _{h \rightarrow 0} h\left[h+\left(1+2+3+\ldots(n-1)+h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots(n-1)^{2}\right)\right)\right] \\ \end{aligned}

\begin{aligned}&=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\frac{3}{n}\left(\frac{n(n-1)}{2}\right)+\frac{9}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right]\left[\begin{array}{c} \mathrm{Q} h=\frac{3}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{9}{2 n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)+\frac{9}{2 n^{3}} \cdot n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ &=\lim _{n \rightarrow \infty} \frac{9}{2}\left(1-\frac{1}{n}\right)+\frac{9}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \end{aligned}

=\frac{9}{2}\left ( 1-0 \right )+\frac{9}{2}\left ( 1-0 \right )(2-0)\\ =\frac{9}{2}+9\\ =\frac{9+18}{2}\\ =\frac{27}{2}

 

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