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 part of the plane $5 x + 4 y + z = 20$ that lies in the first octant.

40

30

24

20

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Verified step by step guidance

Step 1: Understand the problem. The plane 5x + 4y + z = 20 intersects the coordinate axes in the first octant. The first octant is where x, y, and z are all positive. To find the area of the triangular region formed by the plane in the first octant, we need to determine the points where the plane intersects the x-axis, y-axis, and z-axis.

Step 2: Find the intercepts of the plane. Set two variables to zero at a time to find the intercepts: - For the x-intercept, set y = 0 and z = 0, solve 5x = 20 to get x = 4. - For the y-intercept, set x = 0 and z = 0, solve 4y = 20 to get y = 5. - For the z-intercept, set x = 0 and y = 0, solve z = 20 to get z = 20.

Step 3: Visualize the triangular region. The intercepts (4, 0, 0), (0, 5, 0), and (0, 0, 20) form a triangle in the first octant. The area of this triangle can be calculated using the formula for the area of a triangle in 3D space: $ \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| $, where $ \vec{AB} $ and $ \vec{AC} $ are vectors formed by the vertices of the triangle.

Step 4: Compute the vectors $ \vec{AB} $ and $ \vec{AC} $. Using the points (4, 0, 0), (0, 5, 0), and (0, 0, 20): - $ \vec{AB} = (0 - 4, 5 - 0, 0 - 0) = (-4, 5, 0) $. - $ \vec{AC} = (0 - 4, 0 - 0, 20 - 0) = (-4, 0, 20) $.

Step 5: Calculate the cross product $ \vec{AB} \times \vec{AC} $. Use the determinant formula for the cross product: $ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 5 & 0 \\ -4 & 0 & 20 \end{vmatrix} $. Then, find the magnitude of the resulting vector and divide by 2 to get the area of the triangle.