Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

9. Graphical Applications of Integrals

Area Between Curves

Calculus

9. Graphical Applications of Integrals

Area Between Curves

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Multiple Choice

Which of the following integrals correctly represents the area of the region enclosed by the curves $y = 4 x$ , $y = 16 x$ , and $y = \frac{1}{16} x$ for $x > 0$ ?

\int_{x = 1}^{x = 4} (16 x - 4 x) d x

\int_{x = 0}^{x = 1} (16 x - 4 x) d x

\int_{x = 0}^{x = 1} (4 x - (\frac{1}{16}) x) d x

\int_{x = 0}^{x = 1} (16 x - (\frac{1}{16}) x) d x

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Verified step by step guidance

Step 1: Understand the problem. We are tasked with finding the integral that represents the area of the region enclosed by the curves y = 4x, y = 16x, and y = 1/16x for x > 0. The integral should represent the difference between the upper curve and the lower curve over the specified interval.

Step 2: Analyze the curves and their relationships. For x > 0, y = 16x is the upper curve, and y = 1/16x is the lower curve in the region of interest. The area is calculated by subtracting the lower curve from the upper curve.

Step 3: Determine the interval of integration. The problem specifies that the correct integral is over the interval x = 0 to x = 1. This interval corresponds to the region where the curves y = 16x and y = 1/16x enclose the area.

Step 4: Set up the integral. The area is given by the integral of the difference between the upper curve and the lower curve over the interval x = 0 to x = 1. This can be expressed as:

\int x_{0} x^{1} (16 x - \frac{1}{16 x}) d x

Step 5: Verify the setup. The integral correctly represents the area of the region enclosed by the curves y = 16x and y = 1/16x over the interval x = 0 to x = 1. The subtraction ensures that the area is calculated as the difference between the upper and lower curves.