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NCERT Solutions Class 12 Maths Chapter-8 (Application Of Integrals)Exercise 8.1

NCERT Solutions Class 12 Maths Chapter-8 (Application Of Integrals)Exercise 8.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-8 (Application Of Integrals)Exercise 8.1 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-8 (Application Of Integrals)Exercise 8.1
NCERT Question-Answer

Class 12 Mathematics

Chapter-8 (Application Of Integrals)

Questions and answers given in practice

Chapter-8 (Application Of Integrals)

 Exercise 8.1

Q1. Find the area of the region bounded by the curve y2=x and the lines x = 1, x = 4 and the x-axis in the first quadrant.

Answer. Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.  Area of ABCD=4ydx=xdx=[3232]1 

Q2. Find the area of the region bounded by y2=9x, x = 2, x = 4 and the x-axis in the first quadrant.

Answer. Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.  Area of ABCD=24ydx=243xdx=3[32]24=2[x32]24 

Q3. Find the area of the region bounded by x2=4y, y = 2, y = 4 and the y-axis in the first quadrant.

Answer. Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis is the area ABCD.  Area of ABCD=24xdy=242ydy=224ydy=2[322]2 

Q4. Find the area of the region bounded by the ellipse x216+y29=1

Answer. The given equation of the ellipse, x216+y29=1, can be represented as Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area of OAB  Area of OAB=04ydx=041x216dx=340416x2dx =34[x216x2+162sin1x4]04=34[21616+8sin1(1)08sin1(0)]=34[8π2]=34[4π]=3π Therefore, area bounded by the ellipse = 4 × 3π = 12π units

Q5. Find the area of the region bounded by the ellipse x24+y29=1

Answer. The given equation of the ellipse can be represented as Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 x24+y29=1y=31x24 It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area OAB  Area of OAB =02ydx=021x243dx[ Using (1)]=32024x2dx 

Q6. Find the area of the region in the first quadrant enclosed by x-axis, line x=3y and the circle x2+y2=4

Answer. The area of the region bounded by the circle,x2+y2=4,x=3y , and the x-axis is the area OAB. Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 The point of intersection of the line and the circle in the first quadrant is (3,1) . Area OAB = Area ΔOCA + Area ACB  Area of OAC=12×OC×AC=12×3×1=32(i) Area of ABC=32ydx =324x2dx=[x24x2+42sin1x2]52=[2×π232432sin1(32)]=[π322(13)]=[π322π3]=[π332]...(ii) 

Q7. Find the area of the smaller part of the circle x2+y2=a2 cut off by the line x=a2

Answer.  The area of the smaller part of the circle, x2+y2=a2, cut off by the line,  area ABCDA.  Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC Area of ABC=aπydx=2πa2x2dx=[x2a2x2+a22sin1xa]a2a=[a22(π2)a22a2a22a22sin1(12)] =a2π4a22a2a22(π4)=a2π4a24a2π8=a24[π1π2]=a24[π21]Area ABCD=2[a22(π21)]=a22(π21) 

Q8. The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Answer.

Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1

Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1 It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC Area OED=Area EFCD Area OED=0tydx=0πxdx=[32]0a=23(a)32   ...(i)  Area of EFCD=04xdx=[3232]0=23[8a32]   (ii) 

Q9. Find the area of the region bounded by the parabola y = x2 and y=|x 

 Find the area of the region bounded by the parabola y=x2 and y=|x| Answer  The area bounded by the parabola, x2=y, and the line, y=|x|, can be represented as  Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1  The given area is symmetrical about y -axis.  Area OACO = Area ODBO  The point of intersection of parabola, x2=y, and line, y=x, is A (1,1) .  Area of OACO= Area ΔOAB Area OBACO   Area of ΔOAB=12×OB×AB=12×1×1=12  Area of OBACO =0ydx=x2dx=[x33]01=13 Area of OACO = Area of ΔOAB Area of OBACO =1213=16Therefore,required area 

Q10. Find the area bounded by the curve x2 = 4y and the line x = 4y – 2

Answer.
 Find the area bounded by the curve x2=4y and the line x=4y2 Answer  The area bounded by the curve, x2=4y, and line, x=4y2, is represented by the  shaded area OBAO.  Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.1