Exercise 2.11 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.11$
Evaluate the following integrals as the limit of the sum:
Question 1.
$
\int_0^1(x+4) d x
$
Solution:
We have $\quad \int_a^b f(x) d x=\lim _{\substack{n \rightarrow \infty \\ h \rightarrow 0}} \sum_{r=1}^n h f(a+r h)$
Here, $a=0, b=1, h=\frac{b-a}{n}=\frac{1}{n} f(x)=x+4$
$
\begin{aligned}
& f(a+r h)=f\left(o+\frac{r}{n}\right)=f\left(\frac{r}{n}\right)=\frac{r}{n}+4 \\
& \text { So } \int_0^1(x+4) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{r}{n}+4\right) \\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(\frac{r}{n^2}+\frac{4}{n}\right) \\
& =\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sum_{r=1}^n r+\frac{1}{n} \sum_{r=1}^n 4\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \frac{n(n+1)}{2}+\frac{1}{n} 4 n\right] \\
& =4+\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right) \\
& =4+\frac{1}{2}=\frac{9}{2} \\
&
\end{aligned}
$
Question 2.
$
\int_1^3 x d x
$

Solution:
$
\begin{aligned}
\int_1^3 x d x \quad \text { Here } f(x)=x, a & =1, b=3 h=\frac{b-a}{n}=\frac{2}{n} \\
f(a+r h) & =f\left(1+\frac{2 r}{n}\right)=1+\frac{2 r}{n} \\
\text { So } \quad \int_1^3 x d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{2}{n}\left(1+\frac{2 r}{n}\right) \\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(\frac{2}{n}+\frac{4 r}{n^2}\right) \\
& =\lim _{n \rightarrow \infty}\left[\frac{1}{n} \sum_{r=1}^n 2+\frac{4}{n^2} \sum_{r=1}^n r\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{1}{n}(2 n)+\frac{4}{n^2} \frac{n(n+1)}{2}\right] \\
& =\lim _{n \rightarrow \infty}\left[2+2\left(1+\frac{1}{n}\right)\right] \\
& =2+2 \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=2+2=4
\end{aligned}
$
Question 3.
$
\int_1^3(2 x+3) d x
$
Solution:
$
\int_1^3(2 x+3) d x \quad \int_a^b f(x) d x=\lim _{\substack{n \rightarrow \infty \\ h \rightarrow 0}} \sum_{r=1}^n h f(a+r h)
$
Here $a=1, b=3, h=\frac{3-1}{n}=\frac{2}{n} \quad f(x)=2 x+3$
$
f(a+r h)=f\left(1+\frac{2 r}{n}\right)
$

$\begin{aligned}
& =2\left(1+\frac{2 r}{n}\right)+3=\left(5+\frac{4 r}{n}\right) \\
\int_1^3 f(x) d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{2}{n}\left(5+\frac{4 r}{n}\right) \\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(\frac{10}{n}+\frac{8 r}{n^2}\right) \\
& =\lim _{n \rightarrow \infty}\left[\frac{10}{n} \sum_{r=1}^n 1+\frac{8}{n^2} \sum_{r=1}^n r\right] \\
& =\lim _{n \rightarrow \infty}\left[\frac{10}{n} n+\frac{8}{n^2} \frac{n(n+1)}{2}\right] \\
& =10+4 \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=10+4=14
\end{aligned}$

Question 4.
$\int_0^1 x^2 d x$
Solution:
$
\begin{array}{rl}
\int_0^1 x^2 d x \quad \text { Here } a=0, b=1 & h=\frac{1}{n} f(x)=x^2 \\
\text { Now } f(a+r h) & =f\left(0+\frac{r}{n}\right)=\frac{r^2}{n^2} \\
\therefore \int_0^1 x^2 d x & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{r^2}{n^2}\right) \\
& =\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n r^2 \\
& =\lim _{n \rightarrow \infty} \frac{1}{n^3} \frac{n(n+1)(2 n+1)}{6} \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}=\frac{2}{6}=\frac{1}{3}
\end{array}
$