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

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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 andÂ 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.**