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Provide solution for RD sharma maths class 12 chapter19 Definite Integrals exercise 19.5 question 7

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


\int_{3}^{5}(2-x) d x


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)]


\begin{aligned} &a=3, b=5, f(x)=2-x \\ &h=\frac{2}{n} \end{aligned}

Thus, we have

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

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

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

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