Evaluate the definite integral between the x limits 0, 1 if the function is given by y=max(1/4,x^2).

We notice that if x = 1/2, then there is no maximum, the 2
given functions having the same value.

Therefore, we'll
evaluate the definite integral over 2 ranges.

If x belongs
to the interval [0,1/2], then 1/4 > x^2.

If x
belongs to the interval [1/2,1], then x^2 > 1/4.

The
definite integral of the given function is calculated over the identified
ranges.

I = Int dx/4 (0->1/2) + Int x^2dx
(1/2->1)

We'll apply Leibniz Newton formula to
determine the definite integrals:

Int dx/4 = F(1/2) -
F(0)

Int dx/4 = x/4

F(1/2) =
1/8 and  F(0) = 0

F(1/2) - F(0) = 1/8
(*)

Int x^2dx = F(1) -
F(1/2)

Int x^2dx = x^3/3

F(1)
= 1/3

F(1/2) = 1/24

F(1) -
F(1/2) = 1/3 - 1/24

F(1) - F(1/2) =
(8-1)/24

F(1) - F(1/2) = 7/24
(**)

We'll add (*) and (**) to find out
I:

I = 1/8 + 7/24

I =
(3+7)/24

I = 10/24

I =
5/12

The requested definite integral of the
given function is I = 5/12.