Which of the following integrals correctly represents the area of the region enclosed by the curves $y=x^2$ and $y=2x$ for $x\in [0,2]$?

• A. ${\int }_0^2{x}^{2}dx$
• B. ${\int }_0^2(x^2-2x)dx$
• C. ${\int }_0^2(2x-x^2)dx$
• D. ${\int }_0^2(2x+x^2)dx$

Answer: (C)

Verified step by step guidance

1. Step 1: Understand the problem. The goal is to find the area of the region enclosed by the curves y = x^2 and y = 2x over the interval [0, 2]. To do this, we need to determine which curve is above the other within the given interval.
2. Step 2: Analyze the curves. Compare y = x^2 and y = 2x by finding their intersection points. Set x^2 = 2x and solve for x. This gives x(x - 2) = 0, so the intersection points are x = 0 and x = 2.
3. Step 3: Determine which curve is above the other. For values of x in [0, 2], substitute a test point (e.g., x = 1) into both equations. At x = 1, y = x^2 = 1 and y = 2x = 2. Since 2x > x^2 in this interval, the curve y = 2x is above y = x^2.
4. Step 4: Set up the integral for the area. The area between two curves is given by the integral of the difference between the top curve and the bottom curve. In this case, the top curve is y = 2x and the bottom curve is y = x^2. Therefore, the integral is ∫₀² (2x - x²) dx.
5. Step 5: Verify the options provided. Among the given choices, the correct integral representing the area is ∫₀² (2x - x²) dx, as it correctly accounts for the difference between the top curve (2x) and the bottom curve (x²) over the interval [0, 2].