Given the parametric equations and , for , find the area enclosed by the curve and the y-axis.
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- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
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- 1. Limits and Continuity2h 2m
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- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
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- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
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- 16. Parametric Equations & Polar Coordinates7h 58m
9. Graphical Applications of Integrals
Area Between Curves
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Which of the following integrals correctly represents the area of the region enclosed by the curves and for ?
A
B
C
D

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.
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.
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. For 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.
Step 4: Set up the integral for the area. The area between two curves is given by the integral of the difference between the upper curve and the lower curve. Here, the upper curve is y = 2x and the lower curve is y = x^2. Therefore, the integral is ∫[0,2] (2x - x^2) dx.
Step 5: Verify the options provided. The correct integral representation for the area is ∫[0,2] (2x - x^2) dx, as it correctly accounts for the difference between the upper and lower curves over the interval [0, 2].
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