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

Estimating Area with Finite Sums

Calculus

8. Definite Integrals

Estimating Area with Finite Sums

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2:23 minutes

Problem 5.1.71b

Textbook Question

Displacement from a velocity graph Consider the velocity function for an object moving along a line (see figure).
(b) Use geometry to find the displacement of the object between t = 0 and t = 2.

Verified step by step guidance

Step 1: Understand the problem. The displacement of an object can be found by calculating the area under the velocity-time graph between t = 0 and t = 2 seconds. This is because displacement is the integral of velocity over time.

Step 2: Analyze the graph. Between t = 0 and t = 2 seconds, the velocity graph forms two geometric shapes: a triangle from t = 0 to t = 1 and a rectangle from t = 1 to t = 2.

Step 3: Calculate the area of the triangle. The triangle has a base of 1 second (from t = 0 to t = 1) and a height of 20 m/s (velocity at t = 1). Use the formula for the area of a triangle: A = (1/2) × base × height.

Step 4: Calculate the area of the rectangle. The rectangle spans from t = 1 to t = 2 seconds, with a width of 1 second and a constant height of 20 m/s. Use the formula for the area of a rectangle: A = width × height.

Step 5: Add the areas of the triangle and rectangle. The total displacement is the sum of these areas, which represents the total area under the velocity graph from t = 0 to t = 2 seconds.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity and Displacement

Velocity is the rate of change of displacement with respect to time, indicating how fast an object is moving in a specific direction. Displacement, on the other hand, is the overall change in position of the object, which can be calculated by integrating the velocity function over a given time interval. In this context, understanding the relationship between velocity and displacement is crucial for solving the problem.

Area Under the Curve

In a velocity-time graph, the displacement of an object can be determined by calculating the area under the velocity curve between two time points. Each segment of the graph represents a different velocity, and the area can be computed using geometric shapes such as rectangles and triangles. This geometric approach simplifies the process of finding displacement without needing to perform calculus directly.

Geometric Shapes in Graphs

When analyzing a velocity graph, different segments can form various geometric shapes, such as rectangles and triangles. The area of these shapes corresponds to the displacement during specific time intervals. For example, a rectangle's area is calculated as base times height, while a triangle's area is one-half base times height, allowing for straightforward calculations of displacement based on the graph's features.

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