
Yefim S. answered 05/11/21
Math Tutor with Experience
a) ∫1/√(4-x^2)dx = sin-1(x/2) + C
b) Area A = ∫011/(4 - x2)dx = sin-1(x/2)01 = sin-1(1/2) - sin-1(0) = π/6
c) ∫011/√(4 - x2)dx=∫1a1/√(4 - x2)dx; sin-1(x/2)01 = sin-1(x/2)1a; sin-1(1/2) = sin-1(a/2) - sin-1(1/2)
sin-1(a/2) = π/3; a/2 = √3/2; a = √3