Difficulty: Intermediate | Prerequisites: Derivatives and basic function graphing (Chapters 2–3)
This material opens the second major act of Calculus I: integration. Where Chapters 2 and 3 taught you to zoom in on a function (derivatives, rates of change), Chapter 4 teaches you to zoom out and accumulate (areas, total change). The punchline, coming in Section 3, is that these two ideas are inverses of each other. You should already be comfortable with derivatives, basic function graphing, and the idea of a limit.
You can estimate the area between a curve and the x-axis by slicing the region into rectangles and adding up their areas. The more rectangles you use, the better the estimate. Taking the limit as the number of rectangles goes to infinity gives you the definite integral, which computes exact net area and, when the function is a rate of change, exact total change.
Riemann sum
An approximation of the area under a curve, formed by dividing the region into rectangles and summing their areas. Think of it as chopping the area into vertical strips and adding up the strip areas.
Upper sum
A Riemann sum where the height of each rectangle is taken at the global maximum of the function on that sub-interval. This always overestimates (or equals) the true area for a positive function.
Lower sum
A Riemann sum where the height of each rectangle is taken at the global minimum of the function on that sub-interval. This always underestimates (or equals) the true area for a positive function.
Left-hand sum (left Riemann sum)
A Riemann sum where the height of each rectangle is the function value at the left endpoint of each sub-interval. In simple terms, you read the function value at the start of each strip.
Right-hand sum (right Riemann sum)
A Riemann sum where the height of each rectangle is the function value at the right endpoint of each sub-interval. You read the function value at the end of each strip.
Delta x (Δx)
The width of each rectangle when you divide [a, b] into n equal sub-intervals: Δx = (b − a) / n. In simple terms, it is the step size.
Sample point (x*ᵢ)
The point within each sub-interval at which you evaluate the function to get the rectangle height. Different choices of sample point give left-hand sums, right-hand sums, upper sums, or lower sums.
Sigma notation (Σ)
A compact way of writing the sum of many terms. The expression Σ (from i = m to n) of aᵢ means add up aₘ + aₘ₊₁ + ... + aₙ. Think of it as a for-loop in maths.
Index of summation
The variable (usually i or k) that counts through integer values in a sigma expression, from the lower bound to the upper bound.
Definite integral
The exact net area between a function f(x) and the x-axis on [a, b], written ∫ₐᵇ f(x) dx. Formally, it is the limit of Riemann sums as the number of rectangles approaches infinity. In simple terms, it is the "perfect" version of the rectangle estimate.
Net area (signed area)
The value the definite integral returns. Area above the x-axis counts as positive; area below counts as negative. In simple terms, if the curve dips below the axis, those parts subtract from the total.
The central problem: find the area between the graph of a continuous function f(x) and the x-axis on an interval [a, b].
Strategy: divide [a, b] into n sub-intervals of equal width Δx = (b − a) / n, build a rectangle on each strip, and sum the rectangle areas.
The area of each rectangle is f(xᵢ) · Δx, where xᵢ is some chosen point in the i-th sub-interval.
The full estimate is: Area ≈ f(x₀)Δx + f(x₁)Δx + ... + f(x*ₙ₋₁)Δx.
Left-hand sum: use the left endpoint of each sub-interval. x*ᵢ = xᵢ.
Right-hand sum: use the right endpoint. x*ᵢ = xᵢ₊₁.
Upper sum: use the point where f reaches its maximum on each sub-interval. Gives an overestimate for positive functions.
Lower sum: use the point where f reaches its minimum. Gives an underestimate for positive functions.
For an increasing function on [a, b], the left-hand sum is the lower sum and the right-hand sum is the upper sum. For a decreasing function, the roles swap.
Given a table of values for g(x) at x = 0, 0.5, 1.0, 1.5, 2.0 with g(x) = 1, 6, 4, 2, 5 and n = 4 rectangles of width 0.5 on [0, 2]:
Upper sum: pick the maximum on each sub-interval. (6 + 6 + 4 + 5) · 0.5 = 10.5.
Lower sum: pick the minimum on each sub-interval. (1 + 4 + 2 + 2) · 0.5 = 4.5.
Left-hand sum: use g(0), g(0.5), g(1.0), g(1.5). (1 + 6 + 4 + 2) · 0.5 = 6.5.
Right-hand sum: use g(0.5), g(1.0), g(1.5), g(2.0). (6 + 4 + 2 + 5) · 0.5 = 8.5.
If f(x) is a rate of change, the area under f(x) on [a, b] gives the total change from x = a to x = b.
Velocity → displacement: the area under a velocity curve between t = a and t = b gives the distance (displacement) travelled.
A pump delivering water at rate r(t) = 1 + t/2 litres per minute: the area under r(t) from t = 0 to t = 6 gives the total water pumped in the first 6 minutes.
Marginal profit → total profit, infection rate → total infections: the same principle applies whenever you integrate a rate.
More rectangles give a better approximation.
The true area is squeezed between the lower and upper sums. As n → ∞, all four types of sum converge to the same value (for continuous functions).
Sigma notation compresses long sums into a single expression. Σ (i = m to n) aᵢ = aₘ + aₘ₊₁ + ... + aₙ.
Properties (c is any constant):
Σ caᵢ = c · Σ aᵢ (constants pull out).
Σ (aᵢ + bᵢ) = Σ aᵢ + Σ bᵢ (sums split across addition).
Σ (aᵢ − bᵢ) = Σ aᵢ − Σ bᵢ (sums split across subtraction).
Σ (i = 1 to n) 1 = n
Σ (i = 1 to n) i = n(n + 1) / 2
Σ (i = 1 to n) i² = n(n + 1)(2n + 1) / 6
These formulas let you evaluate Riemann sums for polynomial functions without listing every term.
To set up a Riemann sum with n rectangles on [a, b]:
Width: Δx = (b − a) / n.
Right-hand endpoint of the i-th rectangle: xᵢ = a + i · Δx.
Riemann sum: Σ (i = 1 to n) f(xᵢ) · Δx.
Taking n → ∞ removes all approximation error.
If f is continuous on [a, b], the definite integral is defined as:
∫ₐᵇ f(x) dx = lim (n→∞) Σ (i = 1 to n) f(xᵢ) · Δx
This limit always exists for continuous functions and gives the exact net area between f and the x-axis.
The definite integral computes net (signed) area:
Where f(x) > 0, the area counts positively.
Where f(x) < 0, the area counts negatively.
The result can be zero or negative even when there is plenty of area.
To get total area (all positive), you integrate |f(x)| instead.
Some definite integrals can be evaluated without any calculation by recognising the shape under the curve:
∫₋₃⁻² (x + 1) dx: the graph of y = x + 1 from x = −3 to x = −2 forms a triangle. Read off the area (and its sign) from the sketch.
∫₀³ √(9 − x²) dx: the graph is a quarter-circle of radius 3. The area is (1/4)π(3)² = 9π/4.
∫₀⁴ |2x − 6| dx: split at x = 3 where 2x − 6 = 0, sketch two triangles, add their (positive) areas.
For continuous f(x), g(x) and constant k:
∫ₐᵇ [f(x) + g(x)] dx = ∫ₐᵇ f(x) dx + ∫ₐᵇ g(x) dx
∫ₐᵇ k · f(x) dx = k · ∫ₐᵇ f(x) dx
∫ₐᵇ k dx = k(b − a)
∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx (swapping limits flips the sign)
∫ₐᶜ f(x) dx = ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx (interval additivity)
These properties let you break apart complicated integrals and combine given information.
When you see lim (n→∞) Σ (i = 1 to n) f(a + iΔx) · Δx, identify a, b, and Δx = (b − a)/n, then write the corresponding ∫ₐᵇ f(x) dx. There can be more than one valid choice of a and b for the same limit expression.
\Delta x = \frac{b - a}{n}x_i = a + i \cdot \Delta x \quad \text{(right-hand endpoint of the } i\text{-th rectangle)}\sum_{i=1}^{n} 1 = n\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dxWhenever you have a rate of change and need the total accumulated quantity, you are computing area under a curve. A car's speedometer gives velocity; the area under the velocity-time graph gives how far the car has travelled. An epidemiologist tracking daily new infections (a rate) integrates that curve to get the cumulative case count. In economics, integrating a marginal-cost curve gives total cost over a production range.
Students often assume the left-hand sum always underestimates the area. It does not. It underestimates only when the function is increasing; when the function is decreasing, the left-hand sum overestimates.
Students frequently confuse the definite integral (net area, signed) with total area (always positive). The integral ∫₀² (x − 1) dx = 0, but the total area between the curve and the x-axis on [0, 2] is 1.
A common error is forgetting to multiply by Δx when computing a Riemann sum. Each rectangle's area is height times width, not just height.
Students sometimes treat Σ (i = 1 to n) i as n, confusing the sum of the first n integers with n itself. The correct result is n(n + 1)/2.
⚠️ You will be asked to compute left-hand and right-hand sums from a table of values or a given function. Know the setup cold: Δx, endpoints, sum.
⚠️ Converting a Riemann-sum limit into a definite integral (and vice versa) is a classic exam question. Be able to identify a, b, f(x), and Δx from the sigma expression.
⚠️ Evaluating a definite integral by interpreting it as a geometric area (triangles, rectangles, quarter-circles) appears regularly. Sketch the curve first.
⚠️ Using the properties of definite integrals to combine or split given integral values is tested frequently. Watch the interval endpoints and the sign-flip rule.
True or false: A right-hand sum on an increasing function always overestimates the area. (True.)
Fill in the blank: Δx for n = 5 rectangles on [2, 7] is ___. (1.)
True or false: ∫ₐᵇ f(x) dx is always positive. (False; net area can be negative.)
Fill in the blank: Σ (i = 1 to 100) i = ___. (5050.)
True or false: ∫₃¹ f(x) dx = −∫₁³ f(x) dx. (True.)
Q: A motorcycle's speedometer readings at 12-second intervals are: t = 0, 12, 24, 36, 48, 60 with v = 5, 10, 20, 22, 30, 25 ft/s. Estimate the distance travelled using a left-hand sum.
A: Δt = 12. Left-hand sum uses v at the beginning of each interval: (5 + 10 + 20 + 22 + 30) × 12 = 87 × 12 = 1044 ft.
Q: Evaluate ∫₀⁴ |2x − 6| dx by interpreting it as area.
A: 2x − 6 = 0 at x = 3. From 0 to 3, the function is negative (absolute value flips it): triangle with base 3, height 6, area = 9. From 3 to 4, triangle with base 1, height 2, area = 1. Total = 10.
Q: Given ∫₁⁷ f(x) dx = −1, ∫₁³ g(x) dx = 7, and ∫₃⁷ g(x) dx = 10, find ∫₁⁷ (2f(x) − g(x)) dx.
A: ∫₁⁷ g(x) dx = 7 + 10 = 17. So 2(−1) − 17 = −19.
Q: Express lim (n→∞) Σ (i = 1 to n) [2 / (5 + 3i/n)] · (3/n) as a definite integral.
A: Δx = 3/n, so b − a = 3. If a = 5, then xᵢ = 5 + 3i/n, and f(x) = 2/x. The integral is ∫₅⁸ (2/x) dx.
Q: Write √3 + √4 + ... + √25 in sigma notation.
A: Σ (i = 3 to 25) √i.
This material feeds directly into the Fundamental Theorem of Calculus (Chapter 4, Section 3), which links definite integrals back to antiderivatives, making it possible to evaluate integrals without limits of Riemann sums. The idea of net area vs. total area reappears when you study displacement vs. total distance in the Net Change Theorem (Section 4). In Calculus II, Riemann sums generalise to improper integrals and numerical methods (trapezoidal rule, Simpson's rule).
Riemann sum, left-hand sum, right-hand sum, upper sum, lower sum, sigma notation, summation notation, definite integral, net area, signed area, area under a curve, rectangles approximation, Δx, sample point, index of summation, integral properties, interval additivity, velocity to displacement, rate of change to total change, MTH 132, Calculus I, Michigan State University