Find the area enclosed by one loop of the polar curve .
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- 0. Functions7h 55m
- Introduction to Functions18m
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- Properties of Functions9m
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- Combining Functions27m
- Exponent rules32m
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- Introduction to Trigonometric Functions38m
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- 1. Limits and Continuity2h 2m
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- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
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- 10. Physics Applications of Integrals 3h 16m
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- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
9. Graphical Applications of Integrals
Area Between Curves
Multiple Choice
Which of the following integrals correctly represents the area of the region enclosed by the curves , , and for ?
A
B
+
C
D
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Verified step by step guidance1
Step 1: Understand the problem. The goal is to find the area of the region enclosed by the curves y = 2x, y = 8x, and y = 18x for x > 0. This involves determining the correct integral representation for the area between these curves.
Step 2: Analyze the curves. For x > 0, the curves y = 2x, y = 8x, and y = 18x are linear functions with increasing slopes. The curve y = 2x is the lowest, y = 8x is in the middle, and y = 18x is the highest.
Step 3: Break the region into subregions. The area enclosed by these curves can be split into two parts: (1) the area between y = 8x and y = 2x, and (2) the area between y = 18x and y = 8x.
Step 4: Write the integral expressions for each subregion. For the area between y = 8x and y = 2x, the integral is \( \int_{0}^{1} (8x - 2x) \, dx \). For the area between y = 18x and y = 8x, the integral is \( \int_{0}^{1} (18x - 8x) \, dx \).
Step 5: Combine the integrals. The total area is the sum of the two integrals: \( \int_{0}^{1} (8x - 2x) \, dx + \int_{0}^{1} (18x - 8x) \, dx \). This represents the correct integral expression for the area of the region enclosed by the curves.
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