Introduction
The 1996-AB6 Calculus Problem is a challenging calculus problem that tests students’ understanding of multiple calculus concepts, including limits, derivatives, and integrals. This problem has been a popular topic of discussion among calculus students and educators for years, due to its complexity and the variety of approaches that can be used to solve it. In this article, we will provide a detailed step-by-step solution to this problem, explaining each step in detail and providing helpful tips and tricks along the way.
Problem Statement
The 1996-AB6 Calculus Problem is as follows:
Find the area of the region bounded by the curves $y = x^2$ and $y = 2x – x^2$.
Step 1: Finding the Intercepts
The first step is to find the intercepts of the two curves. To find the x-intercepts, set both equations equal to zero and solve for x.
x^2 = 0
2x - x^2 = 0
Solving these equations, we get x = 0 for the first equation and x = 0 or x = 2 for the second equation. Thus, the curves intersect at the points (0, 0) and (2, 2).
Step 2: Sketching the Region
Next, sketch the two curves and the region bounded by them. This will help you visualize the problem and determine the limits of integration.
Step 3: Setting Up the Integral
To find the area of the region, we need to integrate the difference between the two curves over the interval determined by the x-intercepts. In this case, the limits of integration are 0 and 2.
Area = ∫[0, 2] (2x - x^2 - x^2) dx
Simplifying the integrand, we get:
Area = ∫[0, 2] (2x - 2x^2) dx
Step 4: Evaluating the Integral
Now, we can evaluate the integral to find the area of the region.
Area = [x^2 - 2x^3/3] from 0 to 2
Area = (2^2 - 2(2)^3/3) - (0^2 - 2(0)^3/3)
Area = 4/3 square units
Conclusion
Therefore, the area of the region bounded by the curves $y = x^2$ and $y = 2x – x^2$ is 4/3 square units. By following the steps outlined in this article, you can effectively solve this challenging calculus problem and develop a deeper understanding of calculus concepts.