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

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

Answers (1)

\frac{32}{3}

Hint:

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

Given:

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

Solution:

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}

Here,

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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