Exercise 9.1 - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Applications of Integration
$\operatorname{Ex} 9.1$
Question 1.
Find an approximate value of $\int_1^{1.5} x d x$ by applying the left-end rule with the partition $\{1.1,1.2$, $1.3,1.4,1.5\}$
Solution:
Here $\mathrm{a}=1, \mathrm{~b}=1.5, \mathrm{n}=5, \mathrm{f}(\mathrm{x})=\mathrm{x}$
So, the width of each subinterval is
$
\begin{aligned}
& h=\Delta x=\frac{b-a}{n}=\frac{1.5-1}{5}=\frac{0.5}{5}=0.1 \\
& x_0=1 ; x_1=1.1 ; x_2=1.2 ; x_3=1.3 ; x_4=1.4 ; x_5=1.5
\end{aligned}
$
The left hand rule for Riemann sum,
$
\begin{aligned}
\mathrm{S} & \left.=\left[\mathrm{f}\left(\mathrm{x}_0\right)+\mathrm{f}\left(\mathrm{x}_1\right)\right)+\mathrm{f}\left(\mathrm{x}_2\right)+\mathrm{f}\left(\mathrm{x}_3\right)+\mathrm{f}\left(\mathrm{x}_4\right)\right] \Delta \mathrm{x} \\
& =[\mathrm{f}(1)+\mathrm{f}(1.1)+\mathrm{f}(1.2)+\mathrm{f}(1.3)+\mathrm{f}(1.4)](0.1) \\
& =[1+1.1+1.2+1.3+1.4](0.1) \\
& =[6](0.1) \\
& =0.6 .
\end{aligned}
$
Question 2.
Find an approximate value of $\int_1^{1.5} x^2 d x$ by applying the right-end rule with the partition $\{1.1,1.2$, $1.3,1.4,1.5\}$
Solution:
Here $\mathrm{a}=1$;
$
\begin{aligned}
& \mathrm{b}=1.5 \\
& \mathrm{n}=5 ; \\
& \mathrm{f}(\mathrm{x})=\mathrm{x}^2
\end{aligned}
$
So, the width of each subinterval is
$
\begin{aligned}
& h=\Delta x=\frac{b-a}{n}=\frac{1.5-1}{5}=\frac{0.5}{5}=0.1 \\
& x_0=1 ; x_1=1.1 ; x_2=1.2 ; x_3=1.3 ; x_4=1.4 ; x_5=1.5
\end{aligned}
$

The Right hand rule for Riemann sum,
$
\begin{aligned}
& \mathrm{S}=\left[\mathrm{f}\left(\mathrm{x}_1\right)+\mathrm{f}\left(\mathrm{x}_2\right)+\mathrm{f}\left(\mathrm{x}_3\right)+\mathrm{f}\left(\mathrm{x}_4\right)+\mathrm{f}\left(\mathrm{x}_5\right)\right] \Delta \mathrm{x} \\
& =[f(1.1)+f(1.2)+f(1.3)+f(1.4)+f(1.5)](0.1) \\
& =[1.21+1.44+1.69+1.96+2.25](0.1) \\
& =[8.55](0.1) \\
& =0.855 \text {. } \\
&
\end{aligned}
$
Question 3.
Find an approximate value of $\int_1^{1.5}(2-x) d x$ by applying the mid-point rule with the partition $\{1.1,1.2,1.3,1.4,1.5\}$
Solution:
Here $\mathrm{a}=1$
$
\begin{aligned}
& \mathrm{b}=1.5 \\
& \mathrm{n}=5 \\
& \mathrm{f}(\mathrm{x})=2-\mathrm{x}
\end{aligned}
$
So, the width of each subinterval is
$
\begin{aligned}
& h=\Delta x=\frac{b-a}{n}=\frac{1.5-1}{5}=\frac{0.5}{5}=0.1 \\
& x_0=1 ; x_1=1.1 ; x_2=1.2 ; x_3=1.3 ; x_4=1.4 ; x_5=1.5
\end{aligned}
$
The mid-point rule for Riemann sum,
$
\begin{aligned}
& \mathrm{S}=\left[f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+f\left(\frac{x_3+x_4}{2}\right)+f\left(\frac{x_4+x_5}{2}\right)\right] \Delta x \\
& =\left[f\left(\frac{1+1.1}{2}\right)+f\left(\frac{1.1+1.2}{2}\right)+f\left(\frac{1.2+1.3}{2}\right)+f\left(\frac{1.3+1.4}{2}\right)+f\left(\frac{1.4+1.5}{2}\right)\right] \\
& =[\mathrm{f}(1.05)+\mathrm{f}(1.15)+\mathrm{f}(1.25)+\mathrm{f}(1.35)+\mathrm{f}(1.45)](0.1) \\
& =[0.95+0.85+0.75+0.65+0.55](0.1) \\
& =[3.75](0.1) \\
& =0.375 \text {. } \\
&
\end{aligned}
$