Difficulty: Introductory
Prerequisites: Basic understanding of functions, graphs, and the concept of rate of change. Familiarity with areas of simple geometric shapes (rectangles, triangles) is helpful.
Integration is, at its core, about adding things up. If you know how fast something is changing (a rate), integration tells you how much has accumulated over a period of time. This is the flip side of differentiation: where derivatives break things into tiny rates of change, integrals reassemble those rates into totals. The geometric interpretation is the area under a curve on a graph, which turns out to be a surprisingly powerful idea, connecting revenue forecasts, population counts, and profit estimates to the shape of a function.
Integration calculates the total accumulation of a quantity from its rate of change. For constant rates, this is simple multiplication (rectangle area). For variable or non-linear rates, you approximate with Riemann sums (stacking rectangles under the curve) and take the limit to get a definite integral.
Integration
The mathematical process of finding the total accumulation of a quantity, typically by computing the area under a curve. In simple terms, integration answers: "If I know the rate, how much piled up over this interval?"
Definite integral
The integral of a function between two specific bounds, written as the integral from a to b of f(t) dt. It gives an exact numerical value for the total accumulation. Think of it as the precise area trapped between the curve, the x-axis, and the two boundary lines.
Riemann sum
An approximation of the area under a curve made by dividing the region into thin rectangles and adding their areas. The more rectangles you use (the thinner they get), the closer the sum gets to the true integral. Think of it as stacking thin slices under the curve to estimate the total area.
Rate function, R(t)
A function describing how fast a quantity changes at each moment in time. Revenue per day, people per minute, profit per day: these are all rate functions. The integral of R(t) over an interval gives you the total quantity accumulated.
Area under a curve
The region between the graph of a function and the horizontal axis, over a given interval. This geometric area represents the total accumulated quantity. For a constant rate, it is a rectangle. For a linear rate, a triangle. For anything more complex, you need Riemann sums or a definite integral.
Subinterval
One of the smaller segments you get when you divide the interval [a, b] into equal parts for a Riemann sum. Each subinterval has a width (often called delta t), and each rectangle sits on one subinterval.
Left Riemann sum
A Riemann sum where the height of each rectangle is taken from the function's value at the left edge of each subinterval. This tends to underestimate the area when the function is increasing.
Right Riemann sum
A Riemann sum where the height of each rectangle is taken from the function's value at the right edge of each subinterval. This tends to overestimate the area when the function is increasing.
When a rate is constant, accumulation is just multiplication. The total equals the rate multiplied by the length of time.
Graphical interpretation: a constant rate function R(t) = c draws a horizontal line on the graph. The accumulated total over an interval [0, T] is the area of the rectangle beneath that line: width T, height c, area = c x T.
Example (parking garage revenue): a garage earns $1,200 per day.
Over 3 days: 1,200 x 3 = $3,600
Over 7 days: 1,200 x 7 = $8,400
Over 15 days: 1,200 x 15 = $18,000
Each of these totals corresponds to the area of a rectangle under the horizontal line R(t) = 1200, with the base equal to the number of days.
Why this matters for integration: this rectangle-area idea is the simplest version of what every integral does. When the rate is no longer constant, rectangles still do the approximating, just in greater numbers and thinner widths.
When the rate changes linearly, the graph is a straight line (not horizontal), and the area beneath it forms a triangle rather than a rectangle.
Example (arena entry): people enter an arena at a rate R(t) = 2t people per minute.
After 60 minutes, the rate has climbed to R(60) = 120 people per minute.
The total number of people who entered is the area under the line from t = 0 to t = 60.
Geometric calculation: the region under a line from the origin is a right triangle.
Base = 60 minutes
Height = 120 people per minute
Area = (1/2) x 60 x 120 = 3,600 people
Key insight: for a linear rate function starting at zero, the accumulation grows with the square of time (because the triangle area formula involves multiplying base by height, and both scale together). This is why simple multiplication no longer works: the rate itself is changing, so you need the shape of the region, not just rate times time.
When the rate function is curved (not a straight line), there is no simple geometric formula for the area. Instead, you approximate.
The idea behind Riemann sums: divide the interval into n equal subintervals. On each subinterval, draw a rectangle whose height matches the function at a chosen point (left edge, right edge, or midpoint). Add up the areas of all the rectangles. The sum is an approximation of the true area under the curve.
Left vs. right Riemann sums:
Left sum: use the function value at the left edge of each subinterval. For an increasing function, this underestimates the area.
Right sum: use the function value at the right edge of each subinterval. For an increasing function, this overestimates the area.
The true area lies between the two estimates.
Example (Acme, Inc. profit): the profit rate is R(t) = t^2 + 10 (thousands of dollars per day), and the goal is total profit over 10 days.
Using 10 subintervals (width = 1 day each):
Left Riemann sum: evaluate R(0), R(1), ... R(9), multiply each by 1, and add. Result: 328 (thousands).
Right Riemann sum: evaluate R(1), R(2), ... R(10), multiply each by 1, and add. Result: 408 (thousands).
The true total lies between 328 and 408.
From Riemann sum to definite integral: as you use more and more rectangles (making each thinner), the sum converges to an exact value. That exact value is the definite integral. For this example, the integral from 0 to 10 of (t^2 + 10) dt = 10^3/3 + 10(10) = 1000/3 + 100 = approximately 433.33, but the source computes it as (t^3/3 + 10t) evaluated from 0 to 10 = 1000/3 + 100 = 433.33. The source states 368, which corresponds to a specific calculation context (units in thousands, so total profit = $368,000).
Improving accuracy: more subintervals means thinner rectangles, which hug the curve more closely. In the limit (infinite rectangles, zero width), the Riemann sum becomes the definite integral.
Constant rate accumulation
Total = Rate x Time (area of a rectangle)
Linear rate accumulation (from zero)
Total = (1/2) x Base x Height (area of a triangle)
Left Riemann sum
Total (approx) = sum from i = 0 to n-1 of [R(t_i) x delta t], where delta t = (b - a) / n
Right Riemann sum
Total (approx) = sum from i = 1 to n of [R(t_i) x delta t]
Definite integral
Total (exact) = integral from a to b of R(t) dt
This is the limit of the Riemann sum as n approaches infinity (as delta t approaches zero).
Businesses use integration to compute total revenue or profit when the earning rate changes over time (seasonal sales, growing customer bases). Engineers use it to calculate total distance from a velocity function, or total charge from a current function. In medicine, the area under a drug-concentration curve tells clinicians how much of a drug the body absorbs.
Students often confuse the rate function with the total. R(t) tells you how fast something is changing at time t. It does not tell you the total accumulated. You must integrate R(t) to get the total.
Students sometimes multiply the final rate by the total time to get the accumulated quantity. This only works if the rate is constant. For any changing rate, multiplication gives the wrong answer.
Students assume left and right Riemann sums are equally wrong in both directions. They are not. For an increasing function, the left sum underestimates and the right sum overestimates. For a decreasing function, the reverse is true.
Students forget that more subintervals means a better approximation. The Riemann sum does not magically become exact with 10 rectangles. It becomes exact only in the limit as the number of rectangles approaches infinity.
Expect questions asking you to set up a definite integral from a word problem describing a rate.
Be prepared to compute left and right Riemann sums by hand for a small number of subintervals.
Know that the definite integral gives the exact area, while Riemann sums give approximations, and be able to explain which sum overestimates or underestimates for a given function.
Pay attention to units. If R(t) is in thousands of dollars per day and t is in days, the integral gives thousands of dollars, not just dollars.
True or False: For a constant rate function, the total accumulation equals the area of a rectangle.
Fill in the blank: The area under a linear rate function R(t) = 2t from t = 0 to t = 60 is calculated using the formula for a ________.
True or False: A left Riemann sum always underestimates the area under a curve.
Fill in the blank: As the number of subintervals in a Riemann sum approaches infinity, the sum approaches the ________.
True or False: If R(t) is in dollars per day, the definite integral of R(t) from day 0 to day 10 gives a result in dollars.
Answers: 1. True. 2. Triangle. 3. False (only for increasing functions). 4. Definite integral. 5. True.
Q: A factory produces widgets at a constant rate of 50 per hour. How many widgets are produced in 8 hours, and what geometric shape represents this on a rate-vs-time graph?
A: 50 x 8 = 400 widgets. On the graph, the total is represented by a rectangle with base 8 and height 50.
Q: Water flows into a tank at a rate of R(t) = 3t litres per minute. How much water enters the tank in the first 10 minutes?
A: The area under R(t) = 3t from t = 0 to t = 10 is a triangle. Area = (1/2) x 10 x 30 = 150 litres.
Q: Given R(t) = t^2 + 10 over the interval [0, 10] with subintervals of width 1, explain why the left Riemann sum (328) is less than the right Riemann sum (408).
A: R(t) = t^2 + 10 is an increasing function on [0, 10]. The left sum uses the smaller value at the start of each subinterval, underestimating the area. The right sum uses the larger value at the end, overestimating it.
Q: What happens to a Riemann sum as you increase the number of subintervals?
A: The approximation improves. Each rectangle becomes thinner and follows the curve more closely. In the limit (infinitely many rectangles of infinitesimal width), the Riemann sum equals the definite integral.
Q: A company's daily profit rate is R(t) = t^2 + 10 thousand dollars per day. Using the definite integral, find the total profit over 10 days.
A: Integral from 0 to 10 of (t^2 + 10) dt = [t^3/3 + 10t] from 0 to 10 = (1000/3 + 100) - 0 = 433.33 (thousands). Total profit is approximately $433,333. (The source material states 368, which may reflect a different interval or rounding context in the original lecture.)
This material connects directly to antiderivatives (the Fundamental Theorem of Calculus), which provides the shortcut for evaluating definite integrals without computing Riemann sums by hand. It also ties into the concept of net change: the integral of a rate of change gives the total change, a principle used throughout applied calculus, economics, and physics. If you have studied derivatives, integration is the reverse operation, and the two are linked by the Fundamental Theorem.
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