Definite Integrals and Area✅

Topic 5: The Definite Integral

Scenario: A car travels on a straight line at a constant speed V meters per second.
Distance Formula: Distance = Speed × Time = V \times (T2 - T1), where T1 and T2 are the initial and final times, respectively.
Geometric Interpretation:
- The formula V \times (T2 - T1) represents the area of a rectangle.
- The height of the rectangle is the constant speed V, and the width is the time interval (T2 - T1).
- Thus, distance traveled can be visualized as the area of a rectangle on a speed-time graph.

Scenario: The speed of a car varies over time.
Data Collection: Record the car's speed every 10 seconds from time 0 to 60 seconds.
Approximation Method:
- Calculate the average speed for each 10-second interval.
- Determine the distance traveled in each interval using the average speed.
- Sum the distances from all intervals to estimate the total distance traveled.
Mathematical Representation:
- If vi and v{i+1} are the speeds at the beginning and end of an interval, the average speed for that interval is \frac{vi + v{i+1}}{2}.
- The distance traveled during that interval is approximately \frac{vi + v{i+1}}{2} \times 10.
- The total approximate distance is the sum of these distances over all intervals.

Lower Estimate:
- Draw a line connecting consecutive speed observations on a graph.
- Form rectangles below these lines.
- The sum of the areas of these rectangles provides a lower estimate of the distance traveled.
Upper Estimate:
- Form rectangles above the lines connecting the speed observations.
- The sum of these areas gives an upper estimate of the distance.
Mid Estimate:
- Use the average speed in each interval to determine the height of the rectangles.
- This approach typically yields an estimate between the lower and upper estimates.
Accuracy: Recording speed more frequently (e.g., every 5 seconds or every second) improves the accuracy of the distance approximation.

Concept: The definite integral is used to find the exact area under a curve by using the concept of a limit.
Setup:
- Consider a function y = f(x) defined on the interval [a, b].
- Divide the interval [a, b] into n subintervals at points x0, x1, x2, …, x{n-1}, xn, where x0 = a and x_n = b.
- Assume the subintervals are equally spaced, with a length of \Delta x for each.
Area Estimation:
- In each subinterval, approximate the area under the curve with a rectangle.
- The base of the rectangle is \Delta x, and the height is f(xi) for some point xi in the interval.
- The area of the rectangle is f(x_i) \Delta x.
- The sum of these rectangular areas is an approximation of the total area under the curve.
Definite Integral as a Limit:
- To find the exact area, take the limit as the width of the subintervals approaches zero (\Delta x \to 0).
- This limit is the definite integral of f(x) from a to b.
Riemann Sum: The sum \sum{i=1}^{n} f(xi) \Delta x is called the Riemann sum.

Formula: The definite integral of a function f(x) from a to b is defined as:
\int{a}^{b} f(x) dx = \lim{\Delta x \to 0} \sum{i=1}^{n} f(xi) \Delta x
Interpretation: The definite integral represents the area under the curve of f(x) between x = a and x = b.

Notation: The area A under the curve y = f(x) between x = a and x = b is written as:
A = \int_{a}^{b} f(x) dx
Calculation: The area under the curve can be calculated using Riemann sums or other integration techniques.

Example 1: Find the definite integral from 0 to 1 of the function f(x) = 2.
- Graph the function y = 2. It's a horizontal line intersecting the y-axis at 2.
- The area under the curve between x = 0 and x = 1 is a rectangle with base 1 and height 2.
- The definite integral is the area of this rectangle, which is 1 \times 2 = 2.
- \int_{0}^{1} 2 dx = 2
Example 2: Find the definite integral from -1 to 1 of the function f(x) = x.
- Graph the function y = x. It's a straight line passing through the origin with a slope of 1.
- The area between the curve and the x-axis from -1 to 0 is below the x-axis, while the area from 0 to 1 is above the x-axis.
- The areas are symmetric and equal in magnitude but have opposite signs.
- The definite integral is 0 because the positive and negative areas cancel each other out.