Question

Evaluate `int_0^1sqrt(sqrtx-x)dx`

Answer 1

Let  I = `int_0^1sqrt(sqrtx-x)dx`

I = `int_0^1sqrt(sqrtx-sqrtx.sqrtx)dx`

Take `sqrtx` common

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

Put `x6(1/2)=t`

Squaring both sides,

`therefore x=t^2`

Differentiate w.r.t x,

∴ 𝒅𝒙 = 𝟐𝒕.𝒅𝒕

Limits after substitution : Lim ⟶[ 0,1 ]

`therefore I=int_0^1t^(1/2)sqrt(1-t).2.tdt`

`=2int_0^1t^(3/2)sqrt(1-t)  dt`

`=2beta(5/2,3/2)` ...........`{int_0^1t^m.(1-t)^n=beta(m+1,n+1)}`

`therefore I = pi/8`