Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 42

$$\frac{4}{3} $$

Given $$y = 2x - x ^2, \ \ \ \ y = 0$$ First, we find the intersection points \begin{aligned} 2x-x^2&=0\\ x(2-x)&=0 \end{aligned} Then $$x=0,\ \ \ \ x=2 $$ and $$ 2x-x^2\geq0\ \ \ \ \text{for}\ \ \ 0\leq x\leq 2 $$ Then \begin{aligned} \text{Area}&= \int_0^2(2x-x^2)dx\\ &=x^2-\frac{1}{3}x^3\bigg|_0^22 &= 4-\frac{8}{3}\\ &=\frac{4}{3} \end{aligned}