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  • please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19 point 5 question 1 maths textbook solution

please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19 point 5 question 1 maths textbook solution

Answers (2)

\frac{33}{2}

Hint:

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

Given:

\begin{aligned} &\int_{0}^{3}(x+4) d x \\ \end{aligned}

Solution:
We have,

\begin{aligned} &\int_{0}^{3}(x+4) d x \\ \end{aligned}

\begin{aligned}&\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)] \end{aligned}

Where, h=\frac{b-a}{n}

Here,

\begin{aligned} &a=0, b=3, f(x)=(x+4) \\ &h=\frac{3}{n} \Rightarrow n h=3 \end{aligned}

Thus, we have

\begin{aligned} I &=\int_{0}^{3}(x+4) d x \\ \end{aligned}

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

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

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

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

   \begin{aligned} &=\lim _{n \rightarrow \infty} \frac{3}{n}\left[4 n+\frac{3}{n} \frac{\left(n^{2}-1\right)}{2}\right]\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{3}{n} \\ n \rightarrow \infty \end{array}\right] \\ \end{aligned}

    \begin{aligned} &=\lim _{n \rightarrow \infty} 12+\frac{9}{2}\left(1-\frac{1}{n}\right) \\ &=12+\frac{9}{2}=\frac{33}{2} \end{aligned}

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\frac{33}{2}

Hint:

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

Given:

\begin{aligned} &\int_{0}^{3}(x+4) d x \\ \end{aligned}

Solution:
We have,

\begin{aligned} &\int_{0}^{3}(x+4) d x \\ \end{aligned}

\begin{aligned}&\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)] \end{aligned}

Where, h=\frac{b-a}{n}

Here,

\begin{aligned} &a=0, b=3, f(x)=(x+4) \\ &h=\frac{3}{n} \Rightarrow n h=3 \end{aligned}

Thus, we have

\begin{aligned} I &=\int_{0}^{3}(x+4) d x \\ \end{aligned}

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

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

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

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

   \begin{aligned} &=\lim _{n \rightarrow \infty} \frac{3}{n}\left[4 n+\frac{3}{n} \frac{\left(n^{2}-1\right)}{2}\right]&&&&&&&&&&&&&\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{3}{n} \\ n \rightarrow \infty \end{array}\right] \\ \end{aligned}

    \begin{aligned} &=\lim _{n \rightarrow \infty} 12+\frac{9}{2}\left(1-\frac{1}{n}\right) \\ &=12+\frac{9}{2}=\frac{33}{2} \end{aligned}

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Info Expert 29

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