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NCERT Solutions Class 12 Maths Chapter-8 (Application Of Integrals)Exercise 8.2

NCERT Solutions Class 12 Maths Chapter-8 (Application Of Integrals)Exercise 8.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-8 (Application Of Integrals)Exercise 8.2 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 8.2

Exercise 8.2

Q1. Find the area of the circle 4x2+4y2=9 which is interior of the parabola x2=4y

Answer. The required area is represented by the shaded area OBCDO. Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.2  Solving the given equation of circle, 4x2+4y2=9, and parabola, x2=4y, we obtain the  point of intersection as (2,12) and D(2,12) It can be observed that the required area is symmetrical about y-axis.   Area OBCDO = 2× Area OBCO  We draw BM perpendicular to OA.  Therefore, the coordinates of M are (2,0) .  Therefore, Area OBCO = Area OMBCO - Area OMBO  =02(94x2)4dxx24dx=120294x2dx1402x2dx=14[x94x2+92sin12x3]0214[x33]02 =14[298+92sin1223]112(2)3=24+98sin122326=212+98sin1223=12(26+94sin1223) 

Q2. Find the area bounded by curves (x1)2+y2=1 and x2+y2=1

Answer.  The area bounded by the curves, (x1)2+y2=1 and x2+y2=1, is represented by  the shaded area as  Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.2  On solving the equations, (x1)2+y2=1 and x2+y2=1, we obtain the point of  intersection as A(12,32) It can be observed that the required area is symmetrical about x-axis.   Area OBCAO =2× Area OCAO  We join AB, which intersects oC at M, such that AM is perpendicular to oc.  The coordinates of M are (12,0)  Area OCAO= Area OMAO+Area MCAM =[0121(x1)2dx+1211x2dx]=[x121(x1)2+12sin1(x1)]012+[x21x2+12sin1x]121=[141(12)2+12sin1(121)12sin1(1)]+ [12sin1(1)141(12)212sin1(12)] =[38+12(π6)12(π2)]+[12(π2)3812(π6)]=[34π12+π4+π4π12]+[12(π2)3812(π6)]=[34π6+π2]=[2π634] Therefore ,required area OBCAO 

Q3. Find the area of the region bounded by the curves y=x2+2,y=x,x=0 and x=3

Answer.  The area bounded by the curves, y=x2+2,y=x,x=0, and x=3, is represented by  the shaded area OCBAO as  Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.2  Then, Area OCBAO= Area ODBAO - Area ODCO =03(x2+2)dx03xdx=[x33+2x]03[x22]03 

Q4. 

Answer.  BL and CM are drawn perpendicular to x -axis.  It can be observed in the following figure that,  Area (ΔACB)= Area (ALBA)+ Area (BLMCB) Area (AMCA)(1) Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.2  Equation of line segment AB is y0=301+1(x+1)y=32(x+1) Area (ALBA)=1+32(x+1)dx=32[x22+x]11=32[12+112+1]=3 units   Equation of line segment BC is y3=2331(x1)y=12(x+7) Area (BLMCB)=1312(x+7)dx=12[x22+7x]13=12[92+21+127]=5 units  

Q5.  Using integration find the area of the triangular region whose sides have the equations y=2x+1,y=3x+1 and x=4 . 

Answer.  The equations of sides of the triangle are y=2x+1,y=3x+1, and x=4 .  On solving these equations, we obtain the vertices of triangle as A(0,1),B(4,13), and C(4,9). Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.2 

Q6. 

Answer.  The smaller area enclosed by the circle, x2+y2=4, and the line, x+y=2, is  represented by the shaded area ACBA as  Solutions Class 12 maths Chapter-8 (Application Of Integrals)Exercise 8.2  It can be observed that,  Area ACBA= Area OACBO - Area (\DeltaOAB) =024x2dx02(2x)dx 

Q7. Area lying between the curve y2=4x and y=2x is  A. 23 B. 13