Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 14

You need to determine first the points of intersection between curves y = x^2 and y = 4x - x^2 , by solving the equation, such that:
x^2 = 4x - x^2 => 2x^2 - 4x = 0
Factoring out 2x yields:
2x(x - 2) = 0 => 2x = 0 or x - 2 = 0
Hence, the endpoints of integral are x = 0 and x = 2.
You need to decide what curve is greater than the other on the interval [0,2]. You need to notice that x^2 < 4x - x^2 on the interval [0,2], 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_0^2 (4x - x^2 - x^2)dx = int_0^2 (4x - 2x^2)dx
int_0^2 (4x - 2x^2)dx = int_0^2 (4x)dx - int_0^2 (2x^2)dx
int_0^2 (4x - 2x^2)dx = 4x^2/2|_0^2 - 2x^3/3|_0^2
int_0^2 (4x - 2x^2)dx = 2x^2|_0^2 - 2x^3/3|_0^2
int_0^2 (4x - 2x^2)dx = 2(2^2 - 0^2) - (2/3)(2^3 - 0^3)
int_0^2 (4x - 2x^2)dx = 8 - 16/3
int_0^2 (4x - 2x^2)dx = (24-16)/3
int_0^2 (4x - 2x^2)dx = 8/3
Hence, evaluating the area of the region enclosed by the given curves, yields int_0^2 (4x - 2x^2)dx = 8/3.

The area of the region enclosed by the given curves is found between the red and orange curves, for x in [0,2] .