Integral Calculus I - Exercise 2.1 Text Book Solutions

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Important Formulas in Integral Calculus I

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 2.11

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

TEXT BOOK SOLUTIONS EXERCISE 2.1

Integrate the following with respect to x.

$1. \: \int \sqrt{(3x+5)} dx\\$

$Let \: I \: = \int \sqrt{(3x+5)} dx\\$

$=\int (3x+5)^{\dfrac{1}{2}} dx\\$

$=\dfrac{(3x+5)^{\dfrac{1}{2}+1}}{3 (\dfrac{1}{2}+1)}+c\\$

$\dfrac{(3x+5)^{\dfrac{3}{2}}}{3 (\dfrac{3}{2})}+c \\$

$\dfrac{(3x+5)^{\dfrac{3}{2}}}{\dfrac{9}{2}}+c \\$

$I = \frac{2}{9}(3x+5)^{\frac{3}{2}}+c \\$

$2. \: \left (9x^{2}-\dfrac{4}{x^{2}} \right )^{2}\\$

$Let\: I = \: \int \left (9x^{2}-\dfrac{4}{x^{2}} \right )^{2} dx \\$

$= \int \left (81x^{4}-72+\dfrac{16}{x^{4}} \right ) dx \\$

$= \int \left (81x^{4}-72+ 16x^{-4} \right ) dx \\$

$= \dfrac{81x^{5}}{5}-72x+\dfrac{16x^{-3}}{-3}+c \\$

$= I = \dfrac{81x^{5}}{5}-72x-\dfrac{16}{3x^{3}}+c \\$

3. (3+x) (2-5x)

$Let \: I =\int (3+x)(2-5x) dx \\$

$=\int (6-15x+2x-5x^{2}) dx \\$

$=\int (6-13x-5x^{2}) dx \\$

$= 6x-\frac{13x^{2}}{2}-\frac{5x^{3}}{3}+c \\$

$4. \: \sqrt{x}(x^{2}-2x+3) dx \\$

$Let \: I = \int \sqrt{x}(x^{2}-2x+3) dx \\$

$= \int x^{\frac{1}{2}} (x^{2}-2x+3) dx \\$

$= \int \left (x^{\frac{1}{2}} x^{3}-x^{\frac{1}{2}} 2x +x^{\frac{1}{2}}3 \right )dx\\$

$=\int \left (x^{\frac{7}{2}}-2x^{\frac{3}{2}}+3x^{\frac{1}{2}} \right ) dx \\$

$= \dfrac{x^{\frac{7}{2}+1}}{\frac{7}{2}+1}- \dfrac{2x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+ \dfrac{3x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \\$

$=\dfrac{x^{\frac{9}{2}}}{\frac{9}{2}}- \dfrac{2x^{\frac{5}{2}}}{\frac{5}{2}}+\dfrac{3x^{\frac{3}{2}}}{\frac{3}{2}} +c \\$

$=\frac{2}{9}x^{\frac{9}{2}}-\frac{4}{5}x^{\frac{5}{2}}+3\times \frac{2}{3}x^{\frac{3}{2}}+c \\$

$I =\frac{2}{9}x^{\frac{9}{2}}-\frac{4}{5}x^{\frac{5}{2}}+ 2x^{\frac{3}{2}}+c \\$

$5. \dfrac{8x+13}{\sqrt{4x+7}}\\$

$Let \: I = \int \dfrac{8x+13}{\sqrt{4x+7}}\\$

$=\int \dfrac{8x+13+1-1}{\sqrt{4x+7}}\\$

$=\int \dfrac{8x+14}{\sqrt{4x+7}}\\$

$=\int \dfrac{2(4x+7)-1}{\sqrt{4x+7}}\\$

$=\int \dfrac{2(4x+7)}{\sqrt{4x+7}}-\int \dfrac{1}{\sqrt{4x+7}}\\$

$=\int {2(\sqrt{4x+7})}dx-\int \dfrac{1}{\sqrt{4x+7}}dx \\$

$=\int 2 (4x+7)^{\dfrac{1}{2}} dx -(4x+7)^{\dfrac{-1}{2}}dx \\$

$= \dfrac{2 (4x+7)^{\dfrac{1}{2}+1} }{4(\dfrac{1}{2}+1)}-\dfrac{(4x+7)^{\dfrac{-1}{2}+1}}{4(\dfrac{-1}{2}+1)} +c\\$

$= \dfrac{(4x+7)^{\dfrac{3}{2}} }{2(\dfrac{3}{2})}-\dfrac{(4x+7)^{\dfrac{1}{2}}}{4(\dfrac{1}{2})}+c \\$

$I = \dfrac{(4x+7)^{\dfrac{3}{2}} }{3}-\dfrac{(4x+7)^{\dfrac{1}{2}}}{{2}}+c \\$

$6. \dfrac{1}{\sqrt{(x+1)}+\sqrt{x-1}}\\$

$Let \: I =\int \dfrac{1}{\sqrt{(x+1)}+\sqrt{x-1}} dx \\$

$Let \dfrac{1}{\sqrt{(x+1)}+\sqrt{x-1}} \times \dfrac{{\sqrt{(x+1)}-\sqrt{x-1}}}{\sqrt{(x+1)}-\sqrt{x-1}}\\$

$=\dfrac{{\sqrt{(x+1)}-\sqrt{x-1}}}{(x+1)-(x-1)}\\$

$=\dfrac{{\sqrt{(x+1)}-\sqrt{x-1}}}{(x+1-x+1)}\\$

$=\dfrac{{\sqrt{(x+1)}-\sqrt{x-1}}}{2} \\$

$I =\int \dfrac{{\sqrt{(x+1)}-\sqrt{x-1}}}{2}\\$

$=\dfrac{1}{2} \int \sqrt{(x+1)}-\sqrt{x-1}\\$

$= \dfrac{1}{2} \int {(x+1)^{\dfrac{1}{2}}-(x-1)^{\dfrac{1}{2}}} dx\\$

$=\dfrac{1}{2} \left [\dfrac{(x+1)^{\dfrac{1}{2}+1}}{\dfrac{1}{2}+1}-\dfrac{(x-1)^{\dfrac{1}{2}+1}}{\dfrac{1}{2}+1} +c \right ] \\$

$= \dfrac{1}{2} \left [\dfrac{(x+1)^{\dfrac{3}{2}}}{\dfrac{3}{2}}-\dfrac{(x-1)^{\dfrac{3}{2}}}{\dfrac{3}{2}} \right ]+c \\$

$= \dfrac{1}{2} \left [\dfrac{2}{3}\times (x+1)^{\dfrac{3}{2}}-\dfrac{2}{3} \times (x-1)^{\dfrac{3}{2}}\right ] +c\\$

$=\dfrac{1}{2} \times \dfrac{2}{3}\left [(x+1)^{\dfrac{3}{2}} \:- \: \dfrac{2}{3}(x-1)^{\dfrac{3}{2}} \right ] +c\\$

$=\dfrac{1}{3}\left [(x+1)^{\dfrac{3}{2}}-(x-1)^{\dfrac{3}{2}} \right ]+c$

$7. If \: f'(x)=x+b\: ; \: f(1)=5\: and \:f(2)=13\: then \: find\: f(x).\\$

$Let \: \int f'(x) dx = f(x)\\$

$=\int (x+b)dx = f(x)\\$

$=\dfrac{x^{2}}{2}+bx +c = f(x) \cdots \cdots (1)\\$

$When \: x=1\\$

$f(x) = \dfrac{1}{2}+b+c =f(1)\\$

$\dfrac{1}{2}+b+c =5\\$

$b+c=5-\dfrac{1}{2}\\$

$b+c=\dfrac{9}{2}\\$

$2b+2c=9 \cdots \cdots (2)\\$

$When \: x=2 \\$

$In \: f(x) = \dfrac{4}{2} + 2b +c =f (2)\\$

$2 + 2b +c =13\\$

$2b+c=11\cdots \cdots (3)\\$

$Solving \: (2)\: and (3)\\$

$2b+2c=9$
$2b+c=11$
$\noindent\rule{2.5cm}{0.5pt}$
$c=-2 \\$

$In \: (3) \: 2b-2=11\\$
$2b=13\\$
$b=\dfrac{13}{2}\\$

$f(x) = \dfrac{x^{2}}{2}+\dfrac{13x}{2}-2 \\$

$=\dfrac{x^{2}}{2}+\dfrac{13x}{2}-2 \\$

$f(x) = \dfrac{x^{2}}{2}+\dfrac{13x}{2}-2\\$

$8. If \: f'(x)=8x^{3}-2x\: and \:f(2)=8\: then \: find\: f(x).\\$

$Let \: \int f'(x)=8x^{3}-2x\: dx= f(x) \\$

$\int f'(x)=\left (8x^{3}-2x \right )\: dx= f(x) \\$

$\dfrac{8x^{4}}{4}-\dfrac{2x^{2}}{2} +c = f(x) \\$

${2x^{4}}-x^{2}+c = f(x) \cdots \cdots (1)\\$

$When \: x=2\\$

$In f(x)= {2\times 2^{4}}-2^{2}= f(2) +c \\$
$2 \times 16 -4 +c =8 \\$
$32-4 +c = 8\\$
$28-8=-c\\$
$c =-20\\$

$In \: (1) \cdots \: {2\times x^{4}}-x^{2}= f(x) +20 \\$
$f(x)= 2x^{4}}-x^{2}-20$

Integral Calculus I – Exercise 2.1 Text Book Solutions 12th Business Maths

TEXT BOOK SOLUTIONS EXERCISE 2.1

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