## Using integration, find the area of the region

##
{(x,Β y) : x
^{
2
}
+ y
^{
2
}
β€ 1,Β x + y β₯ 1,Β x β₯ 0, y β₯ 0 }

CBSE Class 12 Sample Paper for 2020 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10 Important

Question 11 Important

Question 12

Question 13

Question 14 (OR 1st Question)

Question 14 (OR 2nd Question) Important

Question 15 (OR 1st Question)

Question 15 (OR 2nd Question) Important

Question 16

Question 17 Important

Question 18 (OR 1st Question)

Question 18 (OR 2nd Question)

Question 19

Question 20

Question 21 (OR 1st Question) Important

Question 21 (OR 2nd Question) Important

Question 22

Question 23

Question 24 (OR 1st Question)

Question 24 (OR 2nd Question)

Question 25

Question 26 Important

Question 27

Question 28 (OR 1st Question) Important

Question 28 (OR 2nd Question)

Question 29

Question 30 Important

Question 31 (OR 1st Question) Important

Question 31 (OR 2nd Question)

Question 32 Important

Question 33 (OR 1st Question) Important

Question 33 (OR 2nd Question) Important

Question 34 You are here

Question 35 (OR 1st Question) Important

Question 35 (OR 2nd Question) Important

Question 36 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 24, 2019 by Teachoo

Question 34 Using integration, find the area of the region {(π₯, π¦)": x2 + y2 " β€" 1, x + y " β₯" 1, x " β₯" 0, y " β₯" 0" } Here, we are given A circle and a line And we need to find area enclosed Circle - "x2 + y2 "β€" 1" Circle is π₯2+π¦2 =1 (π₯β0)2+(π¦β0)2 =1^2 So, Center = (0, 0) & Radius = 1 And "x2 + y2 "β€" 1" means area enclosed inside the circle Line -" x + y "β₯" 1" We draw line x + y = 1 And "x + y"β₯" 1" means area on right side of line We see that circle and line intersect at two points (1, 0) and (0, 1) Now, {(π₯, π¦)": x2 + y2 " β€" 1, x + y " β₯" 1, x " β₯" 0, y " β₯" 0" } is the blue shaded region Area required Area required = Area OBCA β Area OAB Area OBCA Area OBCA = β«1_0^1βγπ¦ ππ₯γ y β equation of circle π₯^2 + π¦^2 = 1 π¦^2= 1 β π₯^2 y = β(1βπ₯^2 ) Therefore, Area ACBO = β«1_0^1βγβ(1βπ₯^2 ) " " ππ₯γ = β«1_0^1βγβ(12βπ₯^2 ) ππ₯γ = [π₯/2 β(12βπ₯^2 )+12/2 sin^(β1)β‘γπ₯/1γ ]_0^1 = [π₯/2 β(1βπ₯^2 )+1/2 sin^(β1)β‘π₯ ]_0^1 = [1/2 β(1β1^2 )+1/2 sin^(β1)β‘1 ] β [0/2 β(1β0^2 )+1/2 sin^(β1)β‘0 ] = [0+1/2 sin^(β1)β‘1 β0β0] = 1/2 sin^(β1)β‘1 = 1/2 Γ π/2 = π/4 Area OAB Area OAB = β«1_0^1βγπ¦ ππ₯γ y β equation of line π₯ + y = 1 y = 1 β x Therefore, Area OAB = β«1_0^1β(1βπ₯)ππ₯ = [π₯βπ₯^2/2]_0^1 = ["1 β " 1^2/2] β [0β0/2] = 1 β 1/2 β 0 β 0 = 1/2 Thus, Area required = Area OACB β AREA OAB = (π /πβπ/π) square units