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Need solution for RD sharma maths class 12 chapter 19 Definite Integrals exercise 19.5 question 9

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


\int_{1}^{2}\left(x^{2}\right) d x


We have,

\int_{a}^{b}(f x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1) h)]

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


\begin{aligned} &a=1, b=2, f(x)=x^{2} \\ &h=\frac{2-1}{n}=\frac{1}{n} \end{aligned}

Thus, we have

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

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

=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{2}{n}\left(\frac{n(n-1)}{2}\right)+\frac{1}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \mathbf{Q} h=\frac{1}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right]

=\lim _{n \rightarrow \infty} 1+\frac{n^{2}}{n^{2}}\left(1-\frac{1}{n}\right)+\frac{1}{6 n^{3}} n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)

\begin{aligned} &=1+1+\frac{2}{6} \\ &=\frac{6+6+2}{6}=\frac{14}{6} \\ &=\frac{7}{3} \end{aligned}


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