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  • Explain solution rd sharma class 12 chapter 19 Definite Integrals exercise 19.5, question 14

Explain solution rd sharma class 12 chapter 19 Definite Integrals exercise 19.5, question 14

Answers (1)

\frac{7}{2}

Hint:

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

Given:

\int_{1}^{0}\left ( 3x^{2}-5x \right )dx

Solution:

\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=1,f(x)=3x^{2}+5x\\ h=\frac{1-0}{n}=\frac{1}{n}

Thus, we have

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

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

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{3}{n^{2}}\left(\frac{n(n-1)(2 n-1)}{6}\right)+\frac{5}{n}\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{3}{n^{3}} \cdot \frac{n^{3}}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{5}{2 n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ \end{aligned}

\begin{aligned} &=\lim _{n \rightarrow \infty} \frac{1}{2}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)+\frac{5}{2}\left(1-\frac{1}{n}\right) \\ \end{aligned}

\begin{aligned} &=\frac{1}{2}(1-0)(2-0)+\frac{5}{2}(1-0) \\ &=1+\frac{5}{2} \\ &=\frac{2+5}{2} \\ &=\frac{7}{2} \end{aligned}

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