Question

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

पर्याय

• `pi/6`

• `pi/3`

• `pi/4`

• `pi/2`

Answer 1

`int_0^1 1/(sqrt(3 + 2x - x^2))dx` = `underline(pi/6)`

Explanation:

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

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

= `int_0^1 1/(sqrt((2)^2 - (x - 1)^2))dx = [sin^-1((x - 1)/2)]_0^1`

= `sin^-1(0) - sin^-1(-1/2) = 0 - (-pi/6)`

= `pi/6`