NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.11

NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.11

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.11 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination. 

Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.11

Exercise 7.11 

Q1. By using the properties of definite integrals, evaluate the integral. 0π2cos2xdx

Answer. I=0π2cos2xdx ……..(1) I=0π2cos2(π2x)dx(0af(x)dx=0af(ax)dx) I=0π2sin2xdx ………..(2) 

Q2. By using the properties of definite integrals, evaluate the integral. 

Answer. 0π2sinxsinx+cosxdx Let I=0π2sinxsinx+cosxdx.......(1) I=0π2sin(π2x)sin(π2x)+cos(π2x)dx(0af(x)dx=0af(ax)dx) I=0π2coscos+sinxdx......(2) 

Q3. By using the properties of definite integrals, evaluate the integral. 0π2sin32xdxsin32x+cos32x

Answer. Let I=0π2sin32xsin32x+cos32xdx..(1) I=0π2sin32(π2x)sin32(π2x)+cos32(π2x)dx(0af(x)dx=0af(ax)dx) I=0π2cos32xsin32x+cos32xdx.....(2) Adding (1) and (2), we obtain 

Q4. By using the properties of definite integrals, evaluate the integral. 

Answer. LetI=0π2cos5xsin5x+cos5xdx......(1) I=0π2cos5(π2x)sin5(π2x)+cos5(π2x)dx(0af(x)dx=0af(ax)dx) I=0π2sin5xsin5x+cos5xdx.....(2) 

Q5. By using the properties of definite integrals, evaluate the integral. 

Answer. I=55|x+2|dx It can be seen that (x+2) 0 on [5,2] and (x+2)0 on [2,5] I=52(x+2)dx+25(x+2)dx(abf(x)=acf(x)+cbf(x)) I=[x22+2x]52+[x22+2x]25 =[(2)22+2(2)(5)222(5)]+[(5)22+2(5)(2)222(2)] 

Q6. By using the properties of definite integrals, evaluate the integral. 28|x5|dx

Answer.  Let I=26|x5|dx It can be seen that (x5)0 on [2,5] and (x5)0 on [5,8] I=2s(x5)dx+2s(x5)dx(abf(x)=atf(x)+cbf(x)) 

Q7. By using the properties of definite integrals, evaluate the integral. 01x(1x)ndx

Answer.  Let I=01x(1x)ndxI=01(1x)(1(1x))ndx =01(1x)(x)ndx=01(xnxn+1)dx =[xn+1n+1xn+2n+2]01(10f(x)dx=10f(ax)dx) 

Q8. By using the properties of definite integrals, evaluate the integral. 0π4log(1+tanx)dx

Answer. I=0π4log(1+tanx)dx.....(1) I=0π4log[1+tan(π4x)]dx(0af(x)dx=0af(ax)dx) I=0π4log{1+tanπ4tanx1+tanπ4tanx}dxI=0π4log{1+1tanx1+tan}dx}dx I=0π4log2(1+tanx)dxI=0π4log2dx0π4log(1+tanx)dx I=0π4log2dxI 

Q9. By using the properties of definite integrals, evaluate the integral. 02x2xdx

Answer.  Let I=02x2xdxI=02(2x)xdx(0af(x)dx=0af(ax)dx) =02{2x12x32}dx=[2(x322)x522]02 =[43x3225x52]02=43(2)3225(2)52 

Q10. By using the properties of definite integrals, evaluate the integral. 

Answer.  Let I=0π2(2logsinxlogsin2x)dxI=0x2{2logsinxlog(2sinxcosx)}dx I=0π2{2logsinxlogsinxlogcosxlog2}dxI=0π2{logsinxlogcosxlog2}dx.....(1) It is known that, (0af(x)dx=0af(ax)dx) I=0π2{logcosxlogsinxlog2}dx(2) Adding (1) and (2), we obtain  2I=0π2(log2log2)dx2I=2log20π1dxI=log2[π2] 

Q11. By using the properties of definite integrals, evaluate the integral. π2π2sin2xdx

Answer.  Let I=π2π2sin2xdx As sin2(x)=(sin(x))2=(sinx)2=sin2x, therefore, sin2x is an even function.  It is known that if f(x) is an even function, then aaf(x)dx=20af(x)dx 

Q12. By using the properties of definite integrals, evaluate the integral. 

Answer. I=0πxdx1+sinx.....(1) I=0π(πx)1+sin(πx)dx(0af(x)dx=0af(ax)dx) I=0π(πx)1+sinxdx...(2) Adding (1) and (2), we obtain  2I=0=π1+sinxdx2I=π0π(1sinx)(1+sinx)(1sinx)dx2I=π0=1sinxcos2xdx 

Q13. By using the properties of definite integrals, evaluate the integral. 

Answer. Let I=π2π2sin7xdx....(1) As sin7(x)=(sin(x))7=(sinx)7=sin7x, therefore, sin2x is an odd function. 

Q14. By using the properties of definite integrals, evaluate the integral. 

Answer.  Let I=02πcos5xdx(1)cos5(2πx)=cos5x It is known that, 02af(x)dx=21af(x)dx, if f(2ax)=f(x)=0 if f(2ax)=f(x) I=20πcos5xdx 

Q15. By using the properties of definite integrals, evaluate the integral. 0π2sinxcosx1+sinxcosxdx

Answer. Let I=0π2sinxcosx1+sinxcosxdx....(1) I=0πsin(π2x)cos(π2x)1+sin(π2x)cos(π2x)dx(0af(x)dx=0af(ax)dx) I=0π2cosxsinx1+sinxcosxdx....(2) 

Q16. By using the properties of definite integrals, evaluate the integral. 0πlog(1+cosx)dx

Answer. I=0πlog(1+cosx)dx.....(1) I=0πlog(1+cosx)(10f(x)dx=10f(ax)dx) I=1πlog(1cosx)dx....(2)  Adding (1) and (2), we obtain 2I=0π{log(1+cosx)+log(1cosx)}dx2I=0πlog(1cos2x)dx2I=0πlogsin2xdx2I=20πlogsinxdx I=0πlogsinxdxsin(πx)=sinx.....(3) I=20π2logsinxdx...(4)I=20xlogsin(π2x)dx=20π2logcosxdx....(5)  Adding (4) and (5), we obtain 2I=20π2(logsinx+logcosx)dx I=0x2(logsinx+logcosx+log2log2)dx I=0π2(log2sinxcosxlog2)dxI=0π2logsin2xdx0π2log2dx  Let 2x=t2dx=dt When x=0,t=0 and when x=π2,π=I=1π20πlogsintdtlog22log2 

Q17. By using the properties of definite integrals, evaluate the integral. 

Answer. Let I=00xx+axdx....(1) It is known that, (0af(x)dx=0af(ax)dx) I=0aaxax+xdx.....(2)  Adding (1) and (2), we obtain 2I=0ax+axx+axdx2I=0a1dx 

Q18. By using the properties of definite integrals, evaluate the integral. 

Answer. I=04|x1|dx It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4 I=04|x1|dx+4|x1|dx(abf(x)=acf(x)+tbf(x)) =01(x1)dx+01(x1)dx =[xx22]01+[x22x]14 =112+(4)22412+1 

Q19. By using the properties of definite integrals, evaluate the integral. Show that 0af(x)g(x)dx=20af(x)dx, if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4

Answer. I=0af(x)g(x)dx....(1) I=10f(ax)g(ax)dx(10f(x)dx=00f(ax)dx) I=0af(x)g(ax)dx.....(2) Adding (1) and (2), we obtain  2I=0a{f(x)g(x)+f(x)g(ax)}dx2I=0af(x){g(x)+g(ax)}dx 2I=0πf(x)×4dx[g(x)+g(ax)=4] 

Q20. Choose the correct answer The value of π2π2(x3+xcosx+tan5x+1)dxis (A) 0 (B) 2 (C) π (D) 1

Answer.  Let I=π2π2(x3+xcosx+tan5x+1)dxI=π2π2x3dx+π2π2cosx+2π2tan5xdx+2π21dx It is known that if f(x) is an even function, then aaf(x)dx=20af(x)dx I=0+0+0+20π21dx=2[x]0π2=2π2π= Hence, the correct Answer is C.

Q21. Choose the correct answer The value of 0π2log(4+3sinx4+3cosx)dxis  (A) 2 (B) 34 (C) 0 (D) 2

Answer. Let I=02log(4+3sinx4+3cosx)dx.....(1) I=0πlog[4+3sin(π2x)4+3cos(π2x)]dx(0af(x)dx=0af(ax)dx) I=0π2log(4+3cosx4+3sinx)dx.....(2)  Adding (1) and (2), we obtain 2I=0π2{log(4+3sinx4+3cosx)+log(4+3cosx4+3sinx)}dx2I=0π2log(4+3sinx4+3cosx×4+3cosx4+3sinx)dx 2I=0π2log1dx2I=0π20dxI=0 Hence, the correct Answer is C

Chapter-7 (integrals)