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

8. Definite Integrals

Introduction to Definite Integrals

Calculus

8. Definite Integrals

Introduction to Definite Integrals

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Multiple Choice

Find the area of the region enclosed by one loop of the curve $r = \sin (8 θ)$ .

\frac{π}{32}

\frac{π}{8}

\frac{1}{8}

\frac{π}{16}

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Verified step by step guidance

Step 1: Recognize that the problem involves finding the area enclosed by one loop of a polar curve. The formula for the area enclosed by a polar curve is A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta, where r is the polar function and \theta_1, \theta_2 are the bounds of integration.

Step 2: Identify the polar function r = \sin(8\theta). To find the bounds of integration, determine the values of \theta where one loop of the curve is completed. For r = \sin(8\theta), a loop is completed when \sin(8\theta) returns to zero after one full oscillation. Solve \sin(8\theta) = 0 to find \theta_1 and \theta_2.

Step 3: Substitute r = \sin(8\theta) into the area formula. The integral becomes A = \frac{1}{2} \int_{\theta_1}^{\theta_2} (\sin(8\theta))^2 \, d\theta. Use the trigonometric identity (\sin(x))^2 = \frac{1 - \cos(2x)}{2} to simplify the integrand.

Step 4: Simplify the integral further. The integral now becomes A = \frac{1}{2} \int_{\theta_1}^{\theta_2} \frac{1 - \cos(16\theta)}{2} \, d\theta. Break this into two separate integrals: A = \frac{1}{4} \int_{\theta_1}^{\theta_2} 1 \, d\theta - \frac{1}{4} \int_{\theta_1}^{\theta_2} \cos(16\theta) \, d\theta.

Step 5: Evaluate each integral. The first integral \int_{\theta_1}^{\theta_2} 1 \, d\theta gives the length of the interval \theta_2 - \theta_1. The second integral \int_{\theta_1}^{\theta_2} \cos(16\theta) \, d\theta can be solved using the formula \int \cos(ax) \, dx = \frac{\sin(ax)}{a}. Combine the results to find the area A.