Difficulty: Intermediate | Prerequisites: Derivatives (Ch. 2–3), basic function graphing, algebra of polynomials and trig functions.
This material introduces the second pillar of calculus: integration. Where derivatives answered "how fast is it changing?", integrals answer "how much has accumulated?" You will learn to approximate the area under a curve using rectangles, compress those sums with sigma notation, and then take the limit to arrive at the definite integral. These ideas connect directly to real problems: if you know velocity, integration gives displacement; if you know a rate of infection, integration gives total infected. Everything here sets the stage for the Fundamental Theorem of Calculus, which ties integration back to derivatives.
Area under a rate-of-change curve equals total change. We approximate that area with rectangles (Riemann sums), write those sums compactly with sigma notation, then take the limit as the number of rectangles approaches infinity to define the definite integral. The definite integral gives the net (signed) area between a function and the x-axis.
Area under a curve
The region between the graph of a non-negative function f(x) and the x-axis over an interval [a, b]. In simple terms, it is the "total amount" that the function accumulates across that interval.
Riemann sum
An approximation of the area under a curve formed by dividing the interval into sub-intervals, building a rectangle on each one, and adding up their areas. Think of it as slicing the region into thin vertical strips and pretending each strip is a rectangle.
Delta x (Δx)
The width of each sub-interval when you divide [a, b] into n equal pieces: Δx = (b – a) / n. In simple terms, it is how wide each rectangle is.
Left-hand sum
A Riemann sum where the height of each rectangle is determined by the function value at the left endpoint of its sub-interval.
Right-hand sum
A Riemann sum where the height of each rectangle is determined by the function value at the right endpoint of its sub-interval.
Upper sum
A Riemann sum where the height of each rectangle is set to 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 set to the global minimum of the function on that sub-interval. This always underestimates (or equals) the true area for a positive function.
Sigma notation (Σ)
A compact way of writing a sum of many terms. The expression Σ (i = m to n) a_i means "add up a_m + a_{m+1} + ... + a_n". Think of it as a shorthand instruction that says "start at i = m, plug into the formula, and keep adding until i = n."
Index of summation
The variable (commonly i, j, or k) that steps through consecutive integer values in a sigma notation expression.
Definite integral
The limit of Riemann sums as the number of rectangles approaches infinity: ∫ from a to b of f(x) dx = lim (n→∞) Σ f(x_i) Δx. In simple terms, it is the exact net area between the curve and the x-axis on [a, b].
Net area (signed area)
The value the definite integral returns. Regions above the x-axis count as positive area; regions below the x-axis count as negative area. The integral adds them together with their signs.
If f(x) is a rate of change, then the area under f(x) on the interval [a, b] equals the total change from x = a to x = b.
The classic physics version: distance = rate × time. For a constant velocity, the distance travelled is the area of a rectangle under the horizontal velocity line. When velocity varies, that rectangle becomes a curved region, and you need calculus to find its area.
Common rate-to-total pairings:
Velocity → displacement
Rate of infection → total number of people infected
Marginal profit → total profit
Water flow rate → total volume delivered
Suppose f(x) is continuous and positive on [a, b]. Divide [a, b] into n sub-intervals of equal width Δx = (b – a)/n. Pick a sample point x*_i in each sub-interval. Then:
Area ≈ f(x*_0)Δx + f(x*1)Δx + ... + f(x*{n−1})Δx
The choice of sample point gives different named sums:
Left-hand sum: x*_i = left endpoint of each sub-interval
Right-hand sum: x*_i = right endpoint of each sub-interval
Upper sum: x*_i = whichever point in the sub-interval gives the global maximum of f
Lower sum: x*_i = whichever point in the sub-interval gives the global minimum of f
For an increasing function on [a, b]:
Left-hand sum = lower sum (under-estimate)
Right-hand sum = upper sum (over-estimate)
For a decreasing function on [a, b], the roles reverse.
For a function that is neither purely increasing nor purely decreasing, left-hand and upper sums are not the same thing. You must check which point in each sub-interval actually gives the max or min.
As n increases, each rectangle gets narrower and fits the curve more closely. The error between the Riemann sum and the true area shrinks. In the limit as n → ∞, the approximation becomes exact.
Σ (i = m to n) a_i = a_m + a_{m+1} + ... + a_{n−1} + a_n
The letter i is the index of summation. It takes on every integer value from m up to n, inclusive.
For any constant c:
Σ (i = m to n) c·a_i = c · Σ (i = m to n) a_i (constants pull out)
Σ (i = m to n) (a_i + b_i) = Σ a_i + Σ b_i (sums split)
Σ (i = m to n) (a_i – b_i) = Σ a_i – Σ b_i (differences split)
These three formulas let you evaluate large sums without writing out every term:
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
A right-hand Riemann sum for f on [a, b] with n rectangles is:
Σ (i = 1 to n) f(x_i) · Δx
where Δx = (b – a)/n and x_i = a + iΔx.
Sigma notation lets you write 100 or 1,000,000 terms in a single line, then use the closed-form formulas to collapse them into an expression in n. That expression is what you take the limit of to get the definite integral.
If f is continuous on [a, b], the definite integral of f from a to b is:
∫ (from a to b) f(x) dx = lim (n→∞) Σ (i = 1 to n) f(x_i) Δx
where Δx = (b – a)/n and x_i = a + iΔx (right-hand endpoints).
The definite integral gives the net (signed) area between f and the x-axis:
Where f(x) > 0, the region contributes positive area.
Where f(x) < 0, the region contributes negative area.
The integral is the algebraic sum of these signed areas.
For instance, if a function encloses area 5 above the axis and area 3 below it, the definite integral over that interval is 5 + (−3) = 2.
Some definite integrals can be computed without antiderivatives by recognising the shape under the curve:
∫ (from a to b) k dx = k(b – a) (rectangle)
A linear function on [a, b] forms a trapezoid (or triangle).
∫ (from 0 to r) √(r² – x²) dx = πr²/4 (quarter-circle of radius r)
For continuous f, g and constant k:
Linearity (sum): ∫ (f + g) dx = ∫ f dx + ∫ g dx
Linearity (scalar): ∫ k·f dx = k · ∫ f dx
Constant function: ∫ (from a to b) k dx = k(b – a)
Reversing limits: ∫ (from a to b) f dx = –∫ (from b to a) f dx
Splitting intervals: ∫ (from a to c) f dx = ∫ (from a to b) f dx + ∫ (from b to c) f dx
These properties let you break apart complicated integrals and combine known values to evaluate new ones.
To express a limit of Riemann sums as a definite integral, identify:
Δx = the factor that looks like (constant)/n
a = the starting value (often the number added to i · Δx)
b = a + nΔx
f(x) = the function being evaluated at x_i
To go the other direction (integral to Riemann sum), reverse the process: choose n, compute Δx and x_i, then write the sigma expression.
Δx = (b – a) / n
x_i = a + iΔx
Area ≈ Σ (i = 1 to n) f(x*_i) Δx
∫ (from a to b) f(x) dx = lim (n→∞) Σ (i = 1 to n) f(x_i) Δx
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
∫ (a to b) [f(x) + g(x)] dx = ∫ f dx + ∫ g dx
∫ (a to b) k f(x) dx = k ∫ f dx
∫ (a to b) k dx = k(b – a)
∫ (a to b) f dx = –∫ (b to a) f dx
∫ (a to c) f dx = ∫ (a to b) f dx + ∫ (b to c) f dx
A pump delivers water at a rate r(t) litres per minute. The total water delivered over the first 6 minutes is the area under the r(t) curve from t = 0 to t = 6. The same logic applies any time you have a rate and want a total: speedometer readings give total distance, marginal revenue gives total revenue, infection rates give total cases.
Students often assume left-hand sums always under-estimate and right-hand sums always over-estimate. This is only true for increasing functions. For decreasing functions, the roles flip. For functions that are neither, neither sum type consistently over- or under-estimates.
Students confuse upper/lower sums with right/left sums. Upper and lower sums use the max and min on each sub-interval, which may not fall at an endpoint if the function changes direction within that sub-interval.
Students forget that the definite integral gives net (signed) area, not total area. If the curve dips below the x-axis, that portion counts as negative. To get total area, integrate the absolute value.
When converting a Riemann-sum limit to a definite integral, students often misidentify a and b. Carefully extract Δx and x_i from the expression before naming the bounds.
⚠️ Computing left-hand, right-hand, upper, and lower sums from a table of values is a standard exam question. Know the definitions cold.
⚠️ Evaluating a definite integral by geometry (triangles, trapezoids, quarter-circles) appears regularly. Recognise ∫ (0 to r) √(r² – x²) dx as a quarter-circle area.
⚠️ Using integral properties to combine or split known integral values into a new one is heavily tested.
⚠️ Writing a Riemann sum as a definite integral (and vice versa) is a frequent exam problem. Practise both directions.
True or False: For a decreasing function, a left-hand sum gives an over-estimate. (True)
Fill in the blank: Σ (i = 1 to n) i = ____. (n(n+1)/2)
True or False: ∫ (from 3 to 3) f(x) dx = 0. (True)
Fill in the blank: If ∫ (1 to 5) f dx = 7, then ∫ (5 to 1) f dx = ____. (−7)
True or False: The definite integral always gives a positive number. (False, it gives net/signed area.)
Q: Given the table x = {0, 0.5, 1.0, 1.5, 2.0} and g(x) = {1, 6, 4, 2, 5}, find the left-hand sum with 4 rectangles on [0, 2].
A: Δx = 0.5. Use left endpoints: g(0) + g(0.5) + g(1.0) + g(1.5) = 1 + 6 + 4 + 2 = 13. Multiply each by Δx: 0.5(1 + 6 + 4 + 2) = 6.5.
Q: Evaluate ∫ (from −3 to −2) (x + 1) dx by interpreting it as area.
A: f(x) = x + 1 is a line. At x = −3, f = −2; at x = −2, f = −1. Both values are negative, so the region is below the x-axis. The shape is a trapezoid with parallel sides 2 and 1, width 1. Area of the trapezoid = 0.5(2 + 1)(1) = 1.5, but it sits below the axis, so the integral = −1.5.
Q: If ∫ (1 to 7) f(x) dx = −1, ∫ (1 to 3) g(x) dx = 7, and ∫ (3 to 7) g(x) dx = 10, evaluate ∫ (1 to 7) [2f(x) – g(x)] dx.
A: Split using linearity: 2∫ (1 to 7) f dx – ∫ (1 to 7) g dx. First part: 2(−1) = −2. For g, split: ∫ (1 to 7) g dx = ∫ (1 to 3) g dx + ∫ (3 to 7) g dx = 7 + 10 = 17. Result: −2 – 17 = −19.
Q: Express lim (n→∞) Σ (i=1 to n) (7 + 2i/n)³ · (2/n) as a definite integral with a = 7.
A: Here Δx = 2/n and x_i = 7 + 2i/n = 7 + iΔx. So a = 7, b = 7 + 2 = 9, and f(x) = x³. The integral is ∫ (from 7 to 9) x³ dx.
This material connects directly to the Fundamental Theorem of Calculus (Ch. 4, Sec. 3), which provides a shortcut: instead of taking limits of Riemann sums, you evaluate a definite integral using antiderivatives. The concept of net area reappears in the Net Change Theorem (Ch. 4, Sec. 4) and in computing area between curves (Ch. 5, Sec. 1).
Riemann sum, left-hand sum, right-hand sum, upper sum, lower sum, sigma notation, summation notation, index of summation, definite integral, net area, signed area, area under a curve, rectangle approximation, Δx, sub-interval, partition, limit of sums, MTH 132, Calculus I, MSU, Michigan State, integrals and applications, Chapter 4