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

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


\int_{0}^{2}\left ( x^{2}+4 \right )dx


We have,

\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}


a=0,b=2,f(x)=x^{2}-x\\ h=\frac{2-0}{n}=\frac{2}{n}

Thus, we have

\int_{0}^{2}\left ( x^{2}+4 \right )dx

\begin{aligned} I &=\lim _{h \rightarrow 0} h[f(0)+f(h)+f(2 h)+\ldots f(0+(n-1)) h] \\ &=\lim _{h \rightarrow 0} h\left[4\left(h^{2}+4\right)+\left((2 h)^{2}+4\right)+\ldots+\left((n-1)^{2} h+4\right)\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[4 n+\frac{4}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)\right] \quad\left[\begin{array}{l} \text { Q } h=\frac{2}{n} \text { if } \\ h \rightarrow 0, n \rightarrow \infty \end{array}\right] \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} 8+\frac{4}{3 n^{3}} . n^{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) \\ \end{aligned}

\begin{aligned} &=8+\frac{4}{3}(1-0)(2-0) \\ &=8+\frac{4}{3} \times 1 \times 2 \\ &=8+\frac{8}{3} \\ &=\frac{24+8}{3} \\ &=\frac{32}{3} \end{aligned}

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