Class 12

Math

Calculus

Application of Integrals

Smaller area enclosed by the circle $x_{2}+y_{2}=4$ and the line $x+y=2$ is

- $2(π−2)$
- $π−2$
- $2π−1$
- $2(π+2)$

It can be observed that,

$AreaACBA=AreaOACBO−Area(ΔOAB)$

$=∫_{0}4−x_{2} dx−∫_{0}(2−x)dx$

$=[2x 4−x_{2} +24 sin_{−1}2x ]_{0}−[2x−2x_{2} ]_{0}$

$=[2⋅2π ]−[4−2]$

$=(π−2)$sq. units

Thus, the correct answer is $B$.