Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 25

You need to determine first the points of intersection between curves y = sqrt x and y = (1/2)x , by solving the equation, such that:
sqrt x = (1/2)x => x = x^2/4 => x^2/4 - x = 0
Factoring out x yields:
x(x/4 - 1) = 0 => x = 0 orx/4 - 1 = 0
Hence, the endpoints of integral are x = 0 and x = 4.
You need to decide what curve is greater than the other on the interval [0,4]. You need to notice that(1/2)x < sqrt x on the interval [0,4], hence, you may evaluate the area of the region enclosed by the given curves, such that:
int_a^b (f(x) - g(x))dx , where f(x) > g(x) for x in [a,b]
int_0^4 (sqrt x - x/2)dx = int_0^4 sqrt x dx - (1/2)int_0^4 x dx
int_0^4 (sqrt x - x/2)dx = ((2/3)x*sqrt x - x^2/4)|_0^4
int_0^4 (sqrt x - x/2)dx = (2/3)(4sqrt 4 - 0*sqrt 0) - (4^2/4 - 0^2/4)
int_0^4 (sqrt x - x/2)dx = 16/3 - 4
int_0^4 (sqrt x - x/2)dx = (16-12)/3
int_0^4 (sqrt x - x/2)dx = 4/3
Hence, evaluating the area of the region enclosed by the given curves, yields int_0^4 (sqrt x - x/2)dx = 4/3.

The area of the region enclosed by the given curves is found between the red and black curves, for x in [0,4].