# Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 257: 2

$A=\frac{4}{3}$

#### Work Step by Step

$f(x)=2x-x^2$ $Area=\int_0^2(2x-x^2)dx$ $A=[x^2-\frac{x^3}{3}]_0^2$ $A=4-\frac{8}{3}$ $A=\frac{4}{3}$

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