Question

`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.

Options

• `pi/2`

• `pi/3`

• π

• 2π

Answer 1

`int_0^3 1/sqrt(3x - x^2)"d"x` = π.

Explanation:

`int_0^3 1/sqrt(3x - x^2)"d"x`

= `int_0^3 1/sqrt(9/4 - 9/4 + 3x - x^2) "d"x`

= `int_0^3 1/sqrt(9/4 - (x^2 - 3x + 9/4))"d"x`

= `int_0^3 1/sqrt((3/2)^2 - (x - 3/2)^2) "d"x`

= `[sin^-1 ((x - 3/2)/(3/2))]_0^3`

= `sin^-1 (1) - sin^-1 (-1)`

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

∴ I = π