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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Evaluate as the limit of sums:

Solution:
Let $f(x)=2 x^2+5 ; a=1 ; b=3$ and $n h=3-1=2$
$
\begin{aligned}
& \therefore \int_1^3\left(2 x^2+5\right) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots f(a+(n-1)) h] \\
& \text { Here, } \mathrm{f}(\mathrm{a})=\mathrm{f}(1)=2(1)^2+5 \\
& \mathrm{f}(\mathrm{a}+\mathrm{h})=\mathrm{f}(1+\mathrm{h})=2(1+\mathrm{h})^2+5 \\
& \mathrm{f}(\mathrm{a}+2 \mathrm{~h})=\mathrm{f}(1+2 \mathrm{~h})=2(1+2 \mathrm{~h})^2+5 \\
& \mathrm{f}\left[\mathrm{a}+(\mathrm{n}-1) \mathrm{h}=\mathrm{f}[1+(\mathrm{n}-1) \mathrm{h}]=2[1+(\mathrm{n}-1) \mathrm{h}]^2+5\right.
\end{aligned}
$

$\begin{aligned}
& \therefore \int_1^3\left(2 x^2+5\right) d x=\lim _{h \rightarrow 0} h\left[\begin{array}{r}
2(1)^2+5+2(1+h)^2+5+2(1+2 h)^2+5+ \\
\ldots .+2[1+(n-1) h]^2+5
\end{array}\right] \\
& =\lim _{h \rightarrow 0} h\left[\begin{array}{r}
\left.2(1)^2+5+2\left(1+h^2+2 h\right)+5+2\left(1+4 h^2+4 h\right)+5+\right] \\
\ldots .+2\left[1+(n-1)^2 h^2+2(n-1) h\right]+5
\end{array}\right] \\
& =\lim _{h \rightarrow 0} h\left[\begin{array}{r}
(2+5)+\left(2+2 h^2+4 h+5\right)+\left(2+8 h^2+8 h+5\right)+ \\
\left.\ldots . .+\left[2+2(n-1)^2 h^2+4(n-1)\right] h+5\right]
\end{array}\right] \\
& =\lim _{h \rightarrow 0} h\left[\begin{array}{r}
(2+2+2+\ldots \text { to } n \text { terms })+\left(2 h^2+8 h^2+\ldots .2(n-1)^2 h^2\right) \\
+[4 h+8 h+. .4(n-1) h]+(5+5+\ldots \text { to } n \text { terms })
\end{array}\right] \\
& =\lim _{h \rightarrow 0} h\left[2 n+2 h^2\left[1^2+2^2+\ldots(n-1)^2\right]+4 h(1+2+\ldots(n-1))+5 n\right] \\
& =\lim _{h \rightarrow 0} h\left[(2 n+5 n)+\frac{2 h^2(n-1) n(2 n-1)}{6}+4 h \frac{n(n-1)}{2}\right] \\
& =\lim _{h \rightarrow 0} h\left[7 n+\frac{h^2(n-1) n(2 n-1)}{3}+2 n h(n-1)\right] \\
& =\lim _{h \rightarrow 0}\left[7 n h+\frac{n h(n h-h)(2 n h-h)}{3}+2 n h(n h-h)\right] \\
& =\left[7(2)+\frac{2(2-0)(2(2)-0)}{3}+2(2)(2-0)\right] \\
& =\left[14+\frac{2(2)(4)}{3}+8\right] \\
& =\left[22+\frac{16}{3}\right]=\left[\frac{66+16}{3}\right]=\frac{82}{3} \text {. } \\
&
\end{aligned}$

Question 2.
Evaluates as the limit of sums: $\int_1^2\left(x^2-1\right) d x$
Solution:
Let $f(x)=x^2-1$ for $1 \leq x \leq 2$
We divide the interval $[1,2]$ into $n$ equal sub-intervals each of length $h$.
We have $\mathrm{a}=1, \mathrm{~b}=2$
$
\begin{gathered}
h=\frac{b-a}{n} \Rightarrow h=\frac{2-1}{n} \Rightarrow n h=1 \\
\int_1^2\left(x^2-1\right) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots . f(a+(n-1)) h]
\end{gathered}
$
Here, $f(a)=f(1)=(1)^2-1=0$
$
\begin{aligned}
& f(a+h)=f(1+h)=(1+h)^2-1=1+h^2+2 h-1=h^2+2 h \\
& f(a+2 h)=f(1+2 h)=(1+2 h)^2-1=1+4 h^2+4 h-1=4 h^2+4 h \\
& \mathrm{f}[\mathrm{a}+(\mathrm{n}-1) \mathrm{h}]=\mathrm{f}[1+(\mathrm{n}-1) \mathrm{h}]=[1+(\mathrm{n}-1) \mathrm{h}]^2-1 \\
& =1+(\mathrm{n}-1)^2 \mathrm{~h}^2+2(\mathrm{n}-1) \mathrm{h}-1=(\mathrm{n}-1)^2 \mathrm{~h}^2+2(\mathrm{n}-1) \mathrm{h} \\
& \therefore \int_1^2\left(x^2-1\right) d x=\lim _{h \rightarrow 0} h\left[0+\left(h^2+2 h\right)+\left(4 h^2+4 h\right)+\ldots(n-1)^2 h^2+2(n-1) h\right] \\
& =\lim _{h \rightarrow 0} h\left[\left\{1+2^2+\ldots .(n-1)^2\right\} h^2+2 h\{1+2+\ldots .(n-1)\}\right] \\
& =\lim _{h \rightarrow 0} h\left[\frac{n(n-1)(2 n-1) h^2}{6}+2 h \frac{n(n-1)}{2}\right] \\
& =\lim _{h \rightarrow 0}\left[\frac{n h(n h-h)(2 n h-h)}{6}+n h(n h-h)\right] \\
& =\lim _{h \rightarrow 0}\left[\frac{1(1-h)\{2(1)-h\}}{6}+1(1-h)\right] \\
& =\left[\frac{1(1-0)(2-0\}}{6}+1(1-0)\right]=\frac{1(1)(2)}{6}+1 \\
& =\frac{1}{3}+1=\frac{4}{3} \text {. } \\
&
\end{aligned}
$

Question 3.
Evalute: $\int_0^2\left(x^2+x+2\right) d x$
Solution:
Let $f(x)=x^2+x+2$, for $0 \leq x \leq 2$ Here, $a=0, b=2$
We divide the closed interval $[0,2]$ into $\mathrm{n}$ subintervals each of length $h$.
$
\therefore \quad h=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n} \Rightarrow n h=2
$
By definition,

$\begin{aligned}
& \int_a^b f(x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots .+f(a+\overline{n-1} h)] \\
& f(a)=f(0)=2 \\
& f(a+h)=f(h)=h^2+h+2 \\
& f(a+2 h)=f(2 h)=(2 h)^2+2 h+2=4 h^2+2 h+2 \\
& \therefore \int_0^2\left(x^2+x+2\right) d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)+\ldots+f(0+\overline{n-1} h)] \\
& =\lim _{h \rightarrow 0} h\left[(2)+\left(h^2+h+2\right)+\left(4 h^2+2 h+2\right)+\right. \\
& \left.\ldots+\left\{(n-1)^2 h^2+(n-1) h+2\right\}\right] \\
& =\lim _{h \rightarrow 0} h\left[(2+2+\ldots \text { to } n \text { terms })+h^2\left(1+4+\ldots+\overline{(n-1)}^2\right)\right. \\
& +h(1+2+\ldots+\overline{(n-1)}] \\
& =\lim _{h \rightarrow 0} h\left[2 n+h^2 \frac{(n-1) n(2 n-1)}{6}+h \cdot \frac{(n-1) n}{2}\right] . \\
& =\lim _{h \rightarrow 0}\left[2 n h+\frac{(n h-h)(n h)(2 n h-h)}{6}+\frac{(n h-h)(n h)}{2}\right] \\
& =\lim _{h \rightarrow 0}\left[4+\frac{(2-h)(2)(4-h)}{6}+\frac{(2-h)(2)}{2}\right] \quad[\because n h=2] \\
& =4+\frac{(2)(2)(4)}{6}+\frac{(2)(2)}{3}=4+\frac{8}{3}+2=\frac{26}{3} \text {. } \\
&
\end{aligned}$