Question

Evaluate `int_0^1 sqrt(3 - 2x - x^2) dx`

Answer 1

`int_0^1 sqrt(3 - 2x - x^2) dx = int_0^1 sqrt(-(x^2 + 2x - 3)) dx`

= `int_0^1 sqrt(-(x^2 + 2x + 1 - 4)) dx`

= `int_0^1 sqrt((2)^2 - (x + 1)^2) dx`

Let x + 1 = dt

dx = dt

When x = 0, t = 1

x = 1, t = 2

= `int_1^2 sqrt((2)^2 - t^2) dt`

`["Using" int sqrt(a^2 - x^2) dx = x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1 (x/a)]`

= `[t/2 sqrt((2)^2 - t^2) + (2)^2/2 sin^-1  t/2]^2`

= `[2/2 sqrt(4 - 4) + 2 sin^-1  2/2] - [1/2 sqrt(4 - 1) + 2 sin^-1  1/2]`

= `0 + 2(π/2) - [sqrt(3)/2 + 2(π/6)]`

= `π - sqrt(3)/2 - π/3`

= `(2π)/3 - sqrt(3)/2`

= `(4π - 3sqrt(3))/6`