4.3
Definite Integral
Consider a real-valued function defined on a closed interval. One of the fundamental objectives in calculus is to determine the area under the graph o…
Consider a function over a closed interval. To approximate the area under the curve, divide the interval into equal subintervals, forming rectangles of equal width.
Select an arbitrary point within the subinterval and use the function value there as the rectangle’s height. Multiplying the height by the width gives the area of that rectangle. The total area under the curve is approximated by summing the areas of all rectangles and is called a Riemann sum.
The definite integral of a function over a given interval is defined as the limit of its Riemann sum as the number of subintervals approaches infinity and the width of each rectangle becomes infinitesimally small, provided that this limit exists.
For example, consider a car traveling with a varying velocity. Divide the total travel time into equal time intervals.
In each interval, select an arbitrary time point and take the corresponding velocity. Multiply the velocity by the interval width to estimate the distance traveled.
The sum of these forms a Riemann sum, and in this example, the area under the curve is physically interpreted as the distance traveled by the car. When the velocity remains positive, the total distance is found by integrating the velocity function within the limits.
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Q1: How does dividing an interval into subintervals help approximate area under a curve?
Dividing a closed interval into equal subintervals creates rectangles with uniform width. By selecting an arbitrary point within each subinterval and using the function value there as the rectangle's height, you can calculate each rectangle's area by multiplying height by width. Summing all rectangle areas produces a Riemann sum, which approximates the total area under the curve.
Q2: What is the relationship between Riemann sums and definite integrals?
A definite integral is defined as the limit of Riemann sums as the number of subintervals approaches infinity and each rectangle's width becomes infinitesimally small. This limiting process refines the approximation into an exact value representing the area under the curve over a specified interval, provided the limit exists.
Q3: How can you use integration to find the distance traveled by a moving object?
When a car travels with varying velocity, divide the total travel time into equal intervals and estimate distance in each interval by multiplying velocity at a chosen moment by the interval width. Summing these distances creates a Riemann sum. When velocity remains positive, integrating the velocity function over the time interval yields the exact total distance traveled.
Q4: Why does accuracy improve when you increase the number of subintervals?
As the number of subintervals increases, their widths decrease, making each rectangle's height more representative of the function's actual behavior over that interval. This finer partition reflects the curve's shape more closely, reducing the gap between the approximation and the true area, thereby improving accuracy.
Q5: What does it mean for a rectangle's height to be determined by an arbitrary point within a subinterval?
Within each subinterval, you can choose any point and evaluate the function at that location. The resulting function value becomes the rectangle's height for that subinterval. This flexibility in point selection is fundamental to constructing Riemann sums and demonstrates how different choices can yield different approximations of the area.
Q6: How does the definite integral represent a limit process?
The definite integral emerges when the partition of an interval becomes infinitely fine. As the number of rectangles grows without bound and each rectangle's width approaches zero, the Riemann sum converges to a limiting value. This limiting value represents the exact area under the curve and defines the definite integral.
Q7: What physical interpretation does the area under a velocity curve represent?
The area under a velocity-time curve represents the total distance traveled during that time interval. When velocity remains positive throughout, integrating the velocity function over the specified time limits yields the exact distance covered. This connection between integration and motion demonstrates how calculus models real-world phenomena.