Let the integral be I

`I = int_-2^2sqrt(4-x^2)dx`

this can be done using a trignometric substiution,

let `x = 2sin(u)` then `dx = 2cos(u)` and the limits,

`x =-2, sin(u) =-1 and u=(-pi)/2`

`x =2, sin(u) =1 and u=pi/2`

Therefore the integral changes,

`I = int_((-pi)/2)^(pi/2)(sqrt(4-4sin^2(u))*2*cos(u)du`

`I =4int_((-pi)/2)^(pi/2)cos^2(u)du`

but you know from trignometry, `cos(2u) = 2cos^2(u)-1`

therefore,

`I =4int_((-pi)/2)^(pi/2)((1+cos(2u))/2) du`

`I =2int_((-pi)/2)^(pi/2)(1+cos(2u))du`

`2int(1+cos(2u))du = 2(u-sin(2u)/2)`

therefore

`I = 2(pi/2 - sin(pi)/2 - ((-pi)/2 - sin(-pi)/2))`

`I =2(pi/2-0+pi/2+0)`

`I =2pi`

therefore, `I = int_-2^2sqrt(4-x^2)dx = 2pi`