Chapter 13 - Section 13.4 - The Definite Integral: Algebraic Viewpoint and the Fundamental Theorem of Calculus - Exercises - Page 999: 46

$$1$$

Here, the area $(A)$ can be expressed as: $A=\int_0^{1} 2x \ dx$ In order to solve the above integral, we will use the following formula such as: $\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$ Now, we have $A=\int_0^{1} 2x \ dx \\ =2 \int_0^{1} x \ dx\\=2 [\dfrac{x^2}{2}]_0^1\\ =[x^2]_0^1 \\=1-0 \\=1$ Therefore, the required area is: $Area=1$