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  • Provide solution for RD sharma maths class 12 chapter19 Definite Integrals exercise 19.5 question 6

Provide solution for RD sharma maths class 12 chapter19 Definite Integrals exercise 19.5 question 6

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best_answer

14

Hint:

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

Given:

\int_{1}^{3}(2 x+3) d x

Solution:

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}

Here,

\begin{aligned} &a=1, b=3, f(x)=2 x+3 \\ &h=\frac{3-1}{n}=\frac{2}{n} \end{aligned}

Thus, we have

\begin{aligned} I &=\int_{1}^{3}(2 x+3) 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[2+3+\{2(1+h)+3\}+\ldots 2(1+(n-1)+h)] \\ &=\lim _{h \rightarrow 0} h[5+(5+2 h)+(5+4 h)+\ldots 5+2(n-1) h] \\ &=\lim _{h \rightarrow 0} h[5 n+2 h(1+2+3+\ldots(n-1))] \end{aligned}

=\lim _{n \rightarrow \infty} \frac{2}{n}\left[5 n+\frac{4}{n} \frac{n(n-1)}{2}\right] \left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]

\begin{aligned} &=\lim _{n \rightarrow \infty}\left[10+\frac{4}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=\lim _{n \rightarrow \infty} 10+4\left(1-\frac{1}{n}\right) \\ &=10+4=14 \end{aligned}

 

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