Sums and Differences of Areas (Integration) (1 Viewer)

[jkerr138]

So I have the question;
Find the exact area enclosed between the curve y= sqrt (4-x^2) and the line x-y+2=0
At the moment, I know I can integrate it using integration by substitution and trig identities, along with inverse sin.

Is there a faster or simpler way to integrate it?

Thanks

 

[FrankXie]

So I have the question;
Find the exact area enclosed between the curve y= sqrt (4-x^2) and the line x-y+2=0
At the moment, I know I can integrate it using integration by substitution and trig identities, along with inverse sin.

Is there a faster or simpler way to integrate it?

Thanks

If you don't have to use integration, here is one way: the enclosed area = area of the quarter circle - area of the triangle.