Exercise 9.2 - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $9.2$
Question 1.
Evaluate the following integrals as the limits of sums:
(i) $\int_0^1(5 x+4) d x$
Solution:
We use the formula
$
\begin{aligned}
& \int_a^b f(x) d x=\lim _{n \rightarrow \infty} \Delta x \sum_{r=1}^n f(a+r \Delta x) \\
& \text { Here, } a=0 ; b=1 ; f(x)=5 x+4 \\
& \Delta x=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n} \\
& f(a+r \Delta x)=f\left(0+r \cdot \frac{1}{n}\right)=f\left(\frac{r}{n}\right) \\
&=5\left(\frac{r}{n}\right)+4=\frac{5 r}{n}+4
\end{aligned}
$
$
\text {} \begin{aligned}
\Rightarrow \int_0^1(5 x+4) d x & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{5 r}{n}+4\right) \\
& =\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{5}{n} \sum_{r=1}^n r+\sum_{r=1}^n 4\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{5}{n^2} \sum_{r=1}^n r+\frac{1}{n} \sum_{r=1}^n 4\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{5}{n^2} \frac{n(n+1)}{2}+\frac{1}{n}(4 n)\right]=\lim _{n \rightarrow \infty}\left[\frac{5}{2}\left(1+\frac{1}{n}\right)+4\right]
\end{aligned}
$
Applying limit we get,
$
=\frac{5}{2}+4=\frac{13}{2} \text {. }
$
(ii) $\int_1^2\left(4 x^2-1\right) d x$
Solution:
We use the formula

$\begin{aligned}
& \int_a^b f(x) d x=\lim _{n \rightarrow \infty} \Delta x \sum_{r=1}^n f(a+r \Delta x) \\
& \text { Here, } a=1 ; b=2 ; f(x)=4 x^2-1 \\
& \Delta x=\frac{b-a}{n}=\frac{2-1}{n}=\frac{1}{n} \\
& f(a+r \Delta x)=f\left(1+r \cdot \frac{1}{n}\right)=f\left(1+\frac{r}{n}\right) \\
& =4\left(1+\frac{r}{n}\right)^2-1 \\
& =4\left[1+\frac{r^2}{n^2}+\frac{2 r}{n}\right]-1 \\
& =4+\frac{4 r^2}{n^2}+\frac{8 r}{n}-1=\frac{4 r^2}{n^2}+\frac{8 r}{n}+3 \\
& \text { (1) } \Rightarrow \int_1^2\left(4 x^2-1\right) d x=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{4 r^2}{n^2}+\frac{8 r}{n}+3\right) \\
& =\lim _{n \rightarrow \infty}\left[\frac{4}{n^3} \sum_{r=1}^n r^2+\frac{8}{n^2} \sum_{r=1}^n r+\frac{1}{n} \sum_{r=1}^n 3\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{4}{n^3} \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{8}{n^2} \cdot \frac{n(n+1)}{2}+\frac{1}{n}(3 n)\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{2}{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)+4\left(1+\frac{1}{n}\right)+3\right] \\
& =\frac{2}{3}(1)(2)+4(1)+3=\frac{4}{3}+4+3 \\
& =\frac{4}{3}+7=\frac{25}{3} \text {. } \\
&
\end{aligned}$