Intergration Question Help (1 Viewer)

[chingyloke]

Find the exact area enlosed between the curve y=sqroot(4-x^2) and the line x-y+2=0


can i get a worked solution off someone? thankyou.

 

[jet]

To do this we have to first solve for the points of intersection to get the bounds of the integral.
√(4-x^2)= x + 2 (since x-y+2=0 can be written as y=x+2)
4-x^2 = (x + 2)^2
4 - x^2 = x^2 + 4x + 4
2x^2 + 4x=0
x(2x + 4) = 0
x = 0, x = -2
Next, we find the integral:
Area = ∫(-2 to 0) √(4 - x^2) - (x + 2)dx
=∫(-2 to 0) √(4 - x^2)dx - ∫x + 2 dx
Now, if you can see it, the left integral is actually a quarter of the area of a circle, radius 2
= (1/4)(πr^2) - [x + 2](-2 to 0)
=(π(4))/(4) - (0 + 2 - (-2 + 2))
=π - 2 units squared.

 

[chingyloke]

ahaaaaa. i see now. thanks for your help.