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  • please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19 point 5 question 3 maths textbook solution

please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19 point 5 question 3 maths textbook solution

Answers (1)

8

Hint:

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

Given:

\int_{1}^{3}(3x-2) 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)=3x-1 \\ &h \rightarrow \frac{2}{n} \end{aligned}

Thus, we have

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

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

\begin{aligned} &=\lim _{n \rightarrow \infty}\left[2+\frac{6}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right)\right] \\ &=6+2=8 \end{aligned}

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