7– Integrals Exercise 9

Question 1: [Tex]\int_{-1}^{1} (x+1) \,dx [/Tex]

Answer:

[Tex]Let \,I = \int_{-1}^{1} (x+1) \,dx [/Tex]

[Tex]\int(x+1)dx = \frac{x^2}{2}+x=F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I =F(1)-F(-1)[/Tex]

[Tex]=(\frac{1}{2}+1)-(\frac{1}{2}-1)[/Tex]

[Tex]=\frac{1}{2}+1-\frac{1}{2}+1[/Tex]

= 2

Question 2: [Tex]\int_{2}^{3} \frac{1}{x} \,dx [/Tex]

Answer:

Let [Tex]I=\int_{2}^{3} \frac{1}{x} \,dx [/Tex]

[Tex]\int \frac{1}{x} dx = log|x|=F(x)[/Tex]

Using second fundamental theorem of calculus

I = F(3) – F(2)

= log|3| – log|2|

= log(3/2)

Question 3: [Tex]\int_{1}^{2} (4x^3-5x^2+6x+9) \,dx [/Tex]

Answer:

[Tex]Let\,I = \int_{1}^{2} (4x^3-5x^2+6x+9) \,dx [/Tex]

[Tex]\int (4x^3-5x^2+6x+9) \,dx =4(\frac{x^4}{4})-5(\frac{x^3}{3})+6(\frac{x^2}{2})+9(x)[/Tex]

[Tex]=x^4-5\frac{x^3}{3}+3x^2+9x=F(x)[/Tex]

Using second fundamental theorem of calculus

I =F(2)-F(1)

[Tex]I=\{2^4-\frac{5.(2)^3}{3}+3(2)^2+9(2)\}-\{1^4-\frac{5.(1)^3}{3}+3(1)^2+9(1)\}[/Tex]

[Tex]=( \,16-\frac{40}{3}+12+18) \,-( \,1-\frac{5}{3}+3+9) \,[/Tex]

[Tex] =\,16-\frac{40}{3}+12+18 \,- \,1+\frac{5}{3}-3-9 \,[/Tex]

=[Tex]33-\frac{35}{3}[/Tex]

=[Tex]\frac{99-35}{3}[/Tex]

=[Tex]\frac{64}{3}[/Tex]

Question 4: [Tex]\int_{0}^{\frac{\pi}{4}} sin2x \,dx [/Tex]

Answer:

Let [Tex]I=\int_{0}^{\frac{\pi}{4}} sin2x \,dx [/Tex]

[Tex]\int sin2x \,dx = ( \,\frac{-cos2x}{2}) \,=F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I = F(\frac{\pi}{4})-F(0)[/Tex]

[Tex]=-\frac{1}{2}[ \,cos2( \,\frac{\pi}{4}) \,-cos0] \,[/Tex]

[Tex]=-\frac{1}{2}[ \,cos( \,\frac{\pi}{2}) \,-cos0] \,[/Tex]

[Tex]=-\frac{1}{2}[0-1][/Tex]

[Tex]=\frac{1}{2}[/Tex]

Question 5: [Tex]\int_{0}^{\frac{\pi}{2}} cos2x \,dx [/Tex]

Answer:

Let [Tex]I=\int_{0}^{\frac{\pi}{2}} cos2x \,dx [/Tex]

[Tex]\int cos2x \,dx = \frac{sin2x}{2} = F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I = F(\frac{\pi}{2})-F(0)[/Tex]

[Tex]=\frac{1}{2}[ \,sin2( \,\frac{\pi}{2}) \,-sin0] \,[/Tex]

[Tex]=\frac{1}{2}[ \,sin\pi \,-sin0] \,[/Tex]

[Tex]=\frac{1}{2}[0-0]=0[/Tex]

Question 6: [Tex]\int_{4}^{5} e^x\,dx [/Tex]

Answer:

Let [Tex]I=\int_{4}^{5} e^x\,dx [/Tex]

[Tex]\int e^x dx = e^x = F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I=F(5)-F(4)[/Tex]

[Tex]=e^5-e^4[/Tex]

[Tex]=e^4(e-1)[/Tex]

Question 7: [Tex]\int_{0}^{\frac{\pi}{4}} tanx \,dx [/Tex]

Answer:

Let [Tex]I=\int_{0}^{\frac{\pi}{4}} tanx \,dx [/Tex]

[Tex]\int tanx \,dx=-log|cosx|=F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I=F(\frac{\pi}{4})-F(0)[/Tex]

[Tex]=-log|cos\frac{\pi}{4}|+log|cos0|[/Tex]

[Tex]=-log|\frac{1}{\sqrt{2}}|+log|1|[/Tex]

[Tex]=-log(2)^{-\frac{1}{2}}[/Tex]

[Tex]=\frac{1}{2}log2[/Tex]

Question 8: [Tex]\int_{{\frac{\pi}{6}}}^{\frac{\pi}{4}} cosecx \,dx [/Tex]

Answer:

Let [Tex]I = \int_{{\frac{\pi}{6}}}^{\frac{\pi}{4}} cosecx \,dx [/Tex]

[Tex]\int cosecx dx= log|cosecx-cotx|=F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I=F(\frac{\pi}{4})-F(\frac{\pi}{6})[/Tex]

[Tex]=log|cosec\frac{\pi}{4}-cot\frac{\pi}{4}|-log|cosec\frac{\pi}{6}-cot\frac{\pi}{6}|[/Tex]

[Tex]=log|\sqrt{2}-1|-log|2-\sqrt{3}|[/Tex]

[Tex]=log( \,\frac{\sqrt{2}-1}{2-\sqrt{3}}) \,[/Tex]

Question 9: [Tex]\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}[/Tex]

Answer:

Let [Tex]I=\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}[/Tex]

[Tex]\int \frac{dx}{\sqrt{1-x^2}} = sin^{-1}x = F(x)[/Tex]

Using second fundamental theorem of calculus

I = F(1)-F(0)

[Tex]= sin^{-1}(1)- sin^{-1}(0)[/Tex]

[Tex]=\frac{\pi}{2}-0[/Tex]

[Tex]=\frac{\pi}{2}[/Tex]

Question 10: [Tex]\int_{0}^{1} \frac{dx}{1+x^2}[/Tex]

Answer:

[Tex]let\, I = \int_{0}^{1} \frac{dx}{1+x^2}\\ \int \frac{dx}{1+x^2}=tan^{-1}x=F(x)[/Tex]

Using second fundamental theorem of calculus

I = F(1)-F(0)

[Tex]= tan^{-1}(1)-tan^{-1}(0)[/Tex]

[Tex]=\frac{\pi}{4}[/Tex]

Question 11: [Tex]\int_{2}^{3} \frac{dx}{x^2-1}[/Tex]

Answer:

Let [Tex]I=\int_{2}^{3} \frac{dx}{x^2-1}[/Tex]

[Tex]\int\frac{dx}{x^2-1}=\frac{1}{2}log|\frac{x-1}{x+1}|=F(x)[/Tex]

Using second fundamental theorem of calculus

I = F(3)-F(2)

[Tex]=\frac{1}{2}[ \,log|\frac{3-1}{3+1}|-log|\frac{2-1}{2+1}|] \,[/Tex]

[Tex]=\frac{1}{2}[ \,log|\frac{2}{4}|-log|\frac{1}{3}|] \,[/Tex]

[Tex]=\frac{1}{2}[ \,log|\frac{1}{2}|-log|\frac{1}{3}|] \,[/Tex]

[Tex]=\frac{1}{2}[ \,log|\frac{3}{2}|] \,[/Tex]

Question 12: [Tex]\int_{0}^{\frac{\pi}{2}} cos^2x \,dx [/Tex]

Answer:

Let [Tex]I = \int_{0}^{\frac{\pi}{2}} cos^2x \,dx [/Tex]

[Tex]\int cos^2x \,dx =\int( \,\frac{1+cos2x}{2}) \,dx[/Tex]

[Tex]\frac{x}{2}+\frac{sin2x}{4} = \frac{1}{2}( \,x+\frac{sin2x}{2}) \,=F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I=[ \,F(\frac{\pi}{2})-F(0)] \,[/Tex]

[Tex]=\frac{1}{2}[ \,( \,\frac{\pi}{2}-\frac {sin\pi}{2}) \,-( \,0+\frac{sin0}{2}) \,] \,[/Tex]

[Tex]=\frac{1}{2}[ \,\frac{\pi}{2}+0-0-0] \,[/Tex]

[Tex]=\frac{\pi}{4}[/Tex]

Question 13: [Tex]\int_{2}^{3} \frac{x\,dx}{x^2+1}[/Tex]

Answer:

Let [Tex]I=\int_{2}^{3} \frac{x}{x^2+1}dx[/Tex]

[Tex]\int\frac{x}{x^2+1}dx=\frac{1}{2}\int\frac{2x}{x^2+1}dx[/Tex]

[Tex]\frac{1}{2}log(1+x^2)=F(x)[/Tex]

Using second fundamental theorem of calculus
I = F(3)-F(2)

[Tex]=\frac{1}{2}[ \,log(1+(3)^2)-log(1+(2)^2)] \,[/Tex]

[Tex]=\frac{1}{2}[ \,log(10)-log(5)] \,[/Tex]

[Tex]=\frac{1}{2}log( \,\frac{10}{5}) \,=\frac{1}{2}log2[/Tex]

Question 14: [Tex]\int_{0}^{1} \frac{2x+3}{5x^2+1}dx[/Tex]

Answer:

Let [Tex]I=\int_{0}^{1} \frac{2x+3}{5x^2+1}dx[/Tex]

[Tex]\int \frac{2x+3}{5x^2+1}dx[/Tex][Tex]=\frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx[/Tex]

[Tex]=\frac{1}{5}\int \frac{(10x+15)}{5x^2+1}dx[/Tex]

[Tex]=\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5x^2+1}dx[/Tex]

[Tex]=\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5( \,x^2+\frac{1}{5}) \,}dx[/Tex]

[Tex]=\frac{1}{5}log(5x^2+1)+\frac{3}{5}.\frac{1}{\frac{1}{\sqrt{5}}}tan^{-1}\frac{x}{\frac{1}{\sqrt{5}}}[/Tex]

[Tex]=\frac{1}{5}log(5x^2+1)+\frac{3}{\sqrt{5}}tan^{-1}(\sqrt5x) = F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I = F(1)-F(0)[/Tex]

[Tex]=\{\frac{1}{5}log(5+1)+\frac{3}{\sqrt{5}}tan^{-1}(\sqrt5) \}-\{\frac{1}{5}log(1)+\frac{3}{\sqrt{5}}tan^{-1}(0) \}[/Tex]

[Tex]=\frac{1}{5}log6+\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}[/Tex]

Question 15: [Tex]\int_{0}^{1} xe^{x^2}dx[/Tex]

Answer:

Let [Tex]I=\int_{0}^{1} xe^{x^2}dx[/Tex]

Put x² = t=>2x dx=dt

As x->0,t->0 and as x->1,t->1,

[Tex]I=\frac{1}{2}\int_{0}^{1} e^tdt[/Tex]

[Tex]I=\frac{1}{2}\int e^tdt=\frac{1}{2}e^t = F(t)[/Tex]

Using second fundamental theorem of calculus

I = F(1)-F(0)

[Tex]=\frac{1}{2}e-\frac{1}{2}e^0[/Tex]

[Tex]=\frac{1}{2}(e-1)[/Tex]

Question 16: [Tex]\int_{1}^{2} \frac{5x^2}{x^2+4x+3}[/Tex]

Answer:

Let [Tex]I=\int_{1}^{2} \frac{5x^2}{x^2+4x+3}dx[/Tex]

[Tex]Dividing\,5x^2 \,byx^2+4x+3,we \,obtain[/Tex]

[Tex]I=\int_{1}^{2} \{5-\frac{20x+15}{x^2+4x+3}\}dx[/Tex]

[Tex]I=\int_{1}^{2} 5dx-\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx[/Tex]

[Tex]I={[ \,5x]}_{1}^{2}-\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx[/Tex]

[Tex]I={5-I}_1,where \,I =\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx \,\,………(i)[/Tex]

Consider [Tex]{I}_1=\int_{1}^{2} \frac{20x+15}{x^2+4x+3}dx[/Tex]

[Tex]Let\,20x+15=A\frac{d}{dx}(x^2+4x+3)+B[/Tex]

[Tex]=2Ax+(4A+B)[/Tex]

Equating the coefficient of x and constant term

A = 10 and B = -25

[Tex]=> I_1=10\int_{1}^{2} \frac{2x+4}{x^2+4x+3}dx-25\int_{1}^{2} \frac{1}{x^2+4x+3}dx[/Tex]

[Tex]Let\, x^2+4x+3=t\\=>(2x+4)dx=dt\\=> I_1=10\int\frac{dt}{t}-25\int\frac{dx}{(x^2+2)^2-1^2}[/Tex]

[Tex]=10logt-25[ \,\frac{1}{2}log( \,\frac{x+2-1}{x+2+1})] \,[/Tex]

[Tex]=[ \,10log(x^2+4x+3)]_{1}^{2} \,-25[ \,\frac{1}{2}log( \,\frac{x+1}{x+3})]_{1}^{2} \,[/Tex]

[Tex]=[ \,log15-10log8] \,-25[ \,\frac{1}{2}log\frac{3}{5}-\frac{1}{2}log\frac{2}{4}] \,[/Tex]

[Tex]=[ \,log(3*5)-10log(4*2)] \,-\frac{25}{2}[ \,log3-log5-log2+log4] \,[/Tex]

[Tex][ \,10+\frac{25}{2}] \,log5+[ \,-10-\frac{25}{2}] \,log4+[ \,10-\frac{25}{2}] \,log3+[ \,-10+\frac{25}{2}] \,log2[/Tex]

[Tex]=\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2}[/Tex]

Substituting the value of [Tex]I_1[/Tex] in (1),we obtain
[Tex]I={5-[ \,\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2}}] \, \\ =5-\frac{5}{2}[ \,9log\frac{5}{4}-log\frac{3}{2}] \,[/Tex]

Question 17: [Tex]\int_{0}^{\frac{\pi}{4}} (2sec^2x+x^3+2)dx[/Tex]

Answer:

Let [Tex]I=\int_{0}^{\frac{\pi}{4}} (2sec^2x+x^3+2)dx[/Tex]

[Tex]\int(2sec^2x+x^3+2)dx[/Tex] [Tex]=2tanx+\frac{x^4}{4}+2x=F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I = F({\frac{\pi}{4}})-F(0)[/Tex]

[Tex]=[ \,\{2tan\frac{\pi}{4}+\frac{1}{4}(\frac{\pi}{4})^4+2(\frac{\pi}{4})\}-( \,2tan0+0+0) \,] \,[/Tex]

[Tex]=2tan\frac{\pi}{4}+\frac{\pi ^4}{4^5}+\frac{\pi}{2}[/Tex]

[Tex]=2+\frac{\pi}{2}+\frac{\pi ^4}{1024}[/Tex]

Question 18:[Tex]\int_{0}^{\pi} ( \,sin^2\frac{x}{2}-cos^2\frac{x}{2}) \, \,dx [/Tex]

Answer:

Let [Tex]I=\int_{0}^{\pi} ( \,sin^2\frac{x}{2}-cos^2\frac{x}{2}) \, \,dx [/Tex]

[Tex]=-\int_{0}^{\pi} ( \,cos^2\frac{x}{2}-sin^2\frac{x}{2}) \, \,dx [/Tex]

[Tex]\int_{0}^{\pi} cosx \, \,dx [/Tex]

[Tex]\int cosx \,dx = sinx=F(x)[/Tex]

Using second fundamental theorem of calculus

[Tex]I=F(\pi)-F(0)[/Tex]

[Tex]=sin\pi-sin0=>0[/Tex]

Question 19: [Tex]\int_{0}^{2} \frac{6x+3}{x^2+4}dx[/Tex]

Answer:

Let [Tex]I=\int_{0}^{2} \frac{6x+3}{x^2+4}dx[/Tex]

[Tex]\int \frac{6x+3}{x^2+4}dx=3\int \frac{2x+1}{x^2+4}dx[/Tex]

[Tex]=3\int \frac{2x}{x^2+4}dx+3\int \frac{1}{x^2+4}dx[/Tex]

[Tex]=3log(x^2+4)+\frac{3}{2}tan^{-1}\frac{x}{2}=F(x)[/Tex]

Using second fundamental theorem of calculus

I = F(2)-F(0)

[Tex]=\{3log(2^2+4)+\frac{3}{2}tan^{-1}\frac{2}{2}\}-\{3log(0^2+4)+\frac{3}{2}tan^{-1}\frac{0}{2}\}[/Tex]

[Tex]=3log(8)+\frac{3}{2}tan^{-1}1-3log(4)-\frac{3}{2}tan^{-1}0[/Tex]

[Tex]=3log(8)+\frac{3}{2}\frac{\pi}{4}-3log(4)-0[/Tex]

[Tex]=3log( \,\frac{8}{4}) \,+\frac{3\pi}{4}[/Tex]

[Tex]=3log2+\frac{3\pi}{4}[/Tex]

Question 20: [Tex]\int_{0}^{1} ( \,xe^x+sin \frac{\pi x}{4}) \,dx[/Tex]

Answer:

Let [Tex]I=\int_{0}^{1} ( \,xe^x+sin \frac{\pi x}{4}) \,dx[/Tex]

[Tex]\int ( \,xe^x+sin \frac{\pi x}{4}) \,dx[/Tex]

[Tex]=\int ( \,xe^x)dx +\int (sin \frac{\pi x}{4}) \,dx[/Tex]

[Tex]=x\int e^x dx-\int \{( \,\frac{d}{dx}x) \,\int e^xdx\}dx+\{\frac{-cos\frac{\pi x}{4}}{\frac{\pi}{4}}\}[/Tex]

[Tex]=xe^x-\int e^xdx-\frac{4}{\pi}cos\frac{\pi x}{4}[/Tex]

[Tex]=xe^x-e^x-\frac{4}{\pi}cos\frac{\pi x}{4}[/Tex] = F(x)

Using second fundamental theorem of calculus

I = F(1)-F(0)

[Tex]=( \,1e^1-e^1-\frac{4}{\pi}cos\frac{\pi }{4}) \,-( \,0e^0-e^0-\frac{4}{\pi}cos\frac{\pi *0}{4}) \,[/Tex]

[Tex]=e-e-\frac{4}{\pi}( \,\frac{1}{\sqrt{2}}) \,+1+\frac{4}{\pi}[/Tex]

[Tex]=1+\frac{4}{\pi}-\frac{2\sqrt2}{\pi}[/Tex]

Question 21: [Tex]\int_{1}^{\sqrt{3}}\frac{dx}{1+x^2}\,equals\\\\(A).\frac{\pi}{3}\\\\(B).\frac{2\pi}{3}\\\\(C).\frac{\pi}{6}\\\\(D).\frac{\pi}{12}[/Tex]

Correct Answer is [Tex](D).\frac{\pi}{12}[/Tex]

Question 22: [Tex]\int_{0}^{\frac{2}{3}} \frac{dx}{4+9x^2}\,equals[/Tex]

[Tex]\\\\(A).\frac{\pi}{6}\\\\(B).\frac{\pi}{12}\\\\(C).\frac{\pi}{24}\\\\(D).\frac{\pi}{4}[/Tex]

Correct Answer is [Tex](C).\frac{\pi}{24}[/Tex]

Solutions for NCERT Class 12 Maths Chapter 7 Exercise 7.9 (Integrals) are given in the article today and the main formula used while solving this exercise is shown below:

[Tex]If\, \int f(x)=F(x)\,then\\ \int_{a}^{b} f(x) \,dx=[ \,F(x)] \,_{a}^{b}=F(b)-F(a)[/Tex]