Integral Calculus I Exercise 2.11 - 12th Text Book Solutions

Dear Samacheer Kalvi Students, here are the answers for book back exercise 2.11 in Business Maths Chapter 2 Integral Calculus I.

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Important Formulas in Integral Calculus I

Text Book Solutions for Integral Calculus I Exercise 2.1

Text Book Solutions for Integral Calculus I Exercise 2.2

Text Book Solutions for Integral Calculus I Exercise 2.3

Text Book Solutions for Integral Calculus I Exercise 2.4

Text Book Solutions for Integral Calculus I Exercise 2.5

Text Book Solutions for Integral Calculus I Exercise 2.6

Text Book Solutions for Integral Calculus I Exercise 2.7

Text Book Solutions for Integral Calculus I Exercise 2.8

Text Book Solutions for Integral Calculus I Exercise 2.9

Text Book Solutions for Integral Calculus I Exercise 2.10

Text Book Solutions for Integral Calculus I Exercise 2.12 (MCQ)

Exercise 2.11

Evaluate the following integrals as the limit of the sum:

$1.\displaystyle \int_{_{0}}^{1}(x+4)dx$

$a=0; \:b=1\:; f(x)=x+4\\$
$h=\dfrac{b-a}{n}=\dfrac{1-0}{n}=\dfrac{1}{n}\\$
$f(a+rh)=f(0+r\times\frac{1}{n})\\$
$=f(\dfrac{r}{n})\\$
$f(x)=x+4\\$
$f(\dfrac{r}{n}) =\dfrac{r}{n}+4\\$
$\displaystyle \int_{_{a}}^{b} f(x)\:dx =\lim_{\substack {n \to \infty \\ h\to 0}} \sum\limits_{r=1}^{n}h f(a+rh)\\$
$=\displaystyle \int_{_{0}}^{1} (x+4)\:dx=\lim_ {n\to \infty}\sum\limits _{r=1}^{n}\dfrac{1}{n} (\dfrac{r}{n}+4)\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{r}{n^{2}}+\dfrac{4}{n}\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{r}{n^{2}}+\sum\limits_{r=1}^{n}\dfrac{4}{n}\\$
$we \:know \:that \:\sum_{}^{}n=\dfrac{n(n+1)}{2} \: and\: \sum\limits_{a}^{b}k=kb \\$
$=\lim_{n\to \infty}\dfrac{1}{n^{2}}\dfrac{n(n+1)}{2}+\dfrac{4}{n}\times n\\$
$=\lim_{n\to \infty}\dfrac{1}{n^{2}}\dfrac{(1+\frac{1}{h})}{2}+4\\$
$=\dfrac{(1+\frac{1}{\infty})}{2}+4\\$
$=\dfrac{(1+0)}{2}+4\\$
$=\dfrac{1}{2}+4\\$
$=\dfrac{9}{2}$

$2.\displaystyle \int_{_{1}}^{3}x\: dx$

$a=1; \:b=3\:; f(x)=x\\$
$h=\dfrac{b-a}{n}=\dfrac{3-1}{n}=\dfrac{2}{n}\\$
$f(a+rh)=f(1+r\times\frac{2}{n})\\$
$=f(1+\dfrac{2r}{n})\\$
$f(x)=x\\$
$f(1+\dfrac{2r}{n}) =1+\dfrac{2r}{n}\\$
$\displaystyle \int_{_{a}}^{b} f(x)\:dx =\lim_{\substack {n \to \infty \\ h\to 0}} \sum\limits_{r=1}^{n}h f(a+rh)\\$
$=\displaystyle \int_{_{1}}^{3} x\:dx=\lim_ {n\to \infty}\sum\limits _{r=1}^{n}\dfrac{2}{n} (1+\dfrac{2r}{n})\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{2}{n}+\dfrac{4r}{n^{2}}\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{2}{n} + \sum\limits_{r=1}^{n}\dfrac{4r}{n^{2}}\\$
$we \:know \:that \sum\limits_{a}^{b}k=kb \: and \:\sum_{}^{}n=\dfrac{n(n+1)}{2} \\$
$=\lim_{n\to \infty} \dfrac{2}{n}} (n) + \dfrac{4}{n^{2}} \left (\dfrac{n(n+1)}{2}\\ \right )\\$
$=\lim_{n\to \infty}2+\dfrac{2}{n^{2}} \: n\: . \: n(1+\dfrac{1}{n})\: (Taking\: n \:common) \\$
$=\lim_{n\to \infty}2+\dfrac{2}{n^{2}} \: n^{2}(1+\dfrac{1}{n})\\$
$=\lim_{n\to \infty}2+2(1+\dfrac{1}{n})\\$
$=2+2(1+\dfrac{1}{\infty})\\$
$=2+2(1+0)\\$
$=2+2\\$
$=4$

$3.\displaystyle \int_{_{1}}^{3}2x+3\: dx$

$a=1; \:b=3\:; f(x)=2x+3\\$
$h=\dfrac{b-a}{n}=\dfrac{3-1}{n}=\dfrac{2}{n}\\$
$f(a+rh)=f(1+r\times\frac{2}{n})\\$
$=f(1+\dfrac{2r}{n})\\$
$f(x)=2x+3\\$
$=2 \bigg[1+\dfrac{2r}{n} \bigg]+3\\$
$=2+\dfrac{4r}{n}+3\\$
$=5+\dfrac{4r}{n}\\$
$\displaystyle \int_{_{a}}^{b} f(x)\:dx =\lim_{\substack {n \to \infty \\ h\to 0}} \sum\limits_{r=1}^{n}h f(a+rh)\\$
$=\displaystyle \int_{_{1}}^{3} 2x+3\:dx=\lim_ {n\to \infty}\sum\limits _{r=1}^{n}\dfrac{2}{n} (5+\dfrac{4r}{n})\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{10}{n}+\dfrac{8r}{n^{2}}\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{10}{n} + \sum\limits_{r=1}^{n}\dfrac{8r}{n^{2}}\\$
$we \:know \:that \sum\limits_{a}^{b}k=kb \: and \:\sum_{}^{}n=\dfrac{n(n+1)}{2} \\$
$=\lim_{n\to \infty} {\dfrac{10}{n}} (n) + \dfrac{8}{n^{2}} \left (\dfrac{n(n+1)}{2} \right )\\$
$=\lim_{n\to \infty}10+\dfrac{4}{n^{2}} \: n\: . \: n(1+\dfrac{1}{n})\: (Taking\: n \:common) \\$
$=\lim_{n\to \infty}10+4(1+\dfrac{1}{n})\\$
$=10+4(1+\dfrac{1}{\infty})\\$
$=10+4(1+0)\\$
$=10+4\\$
$=14$

$4.\displaystyle \int_{_{0}}^{1}x^{2} \: dx$

$a=0; \:b=1\:; f(x)=x^{2}\\$
$h=\dfrac{b-a}{n}=\dfrac{1-0}{n}=\dfrac{1}{n}\\$
$f(a+rh)=f(0+r\times\dfrac{1}{n})\\$
$=f(\dfrac{r}{n})\\$
$f(x)=x^{2}\\$
$f(\dfrac{r}{n})=\bigg[\dfrac{r^{2}}{n^{2}} \bigg]\\$
$\displaystyle \int_{_{a}}^{b} f(x)\:dx =\lim_{\substack {n \to \infty \\ h\to 0}} \sum\limits_{r=1}^{n}h f(a+rh)\\$
$=\displaystyle \int_{_{1}}^{3} x^{2}\:dx=\lim_ {n\to \infty}\sum\limits _{r=1}^{n}\dfrac{1}{n} (\dfrac{r^{2}}{n^{2}})\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{r^{2}}{n^{3}}\\$
$=\lim_{n\to \infty}\sum\limits_{r=1}^{n}\dfrac{r^{2}}{n^{3}}\\$
$we \:know \:that \sum\limits_{a}^{b}n^{2}=\dfrac{n(n+1)(2n+1)}{6} \\$
$=\lim_{n\to \infty}\dfrac{1}{n^{3}}\left [\dfrac{n(n+1)(2n+1)}{6} \right ]\\$
$=\lim_{n\to \infty}\dfrac{1}{n^{3}}\left [\dfrac{n\:. n\: (1+\dfrac{1}{n})\: n. (2+\dfrac{1}{n})}{6} \right ]\\(Taking\: n \:common) \\$
$=\lim_{n\to \infty}\dfrac{1}{n^{3}}\: \dfrac{n^{3}(1+\dfrac{1}{n})(2+\dfrac{1}{n})}{6}\\$
$=\lim_{n\to \infty} \dfrac{(1+\dfrac{1}{n})(2+\dfrac{1}{n})}{6}\\$
$=\lim_{n\to \infty} \dfrac{(1+\dfrac{1}{\infty})(2+\dfrac{1}{\infty})}{6}\\$
$=\lim_{n\to \infty} \dfrac{(1+0)(2+0)}{6}\\$
$=\dfrac{2}{6}\\$
$=\dfrac{1}{3}$