Sigma notation and the four summation formulas, approximating the area under a curve with rectangles, the left, right and midpoint rules, upper and lower sums bracketing the answer, and the limit of Riemann sums that defines area exactly.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 5 — Integration
Approximating Areas
Objectives
Five outcomes. The first is notation, the middle three are approximations, and the last makes them exact.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 438-455 — the section these objectives are drawn from
Warm-up
You can find the area of a rectangle, a triangle and a circle. You cannot yet find the area under a parabola.
Discussion prompt
The region under y equals x squared from 0 to 2 is bounded by a curve. How could rectangles help?
Hint: You can compute a rectangle's area exactly, and you can compute a great many of them.
Answer:
Cover the region with rectangles whose areas you can compute exactly, and add them up. Four rectangles give a crude answer, sixteen a better one, a thousand better still — but no finite number is ever exact, because rectangles have flat tops and the curve does not.
\[ \text{4 rects: } 3.75, \quad \text{16: } 2.917, \quad \text{1000: } 2.671, \quad \text{true: } \tfrac83 = 2.6\overline{6} \]
The values improve without arriving, which is exactly the situation Chapter 2 built limits for. Archimedes did this by hand two thousand years ago and called it the method of exhaustion; what he lacked was the limit that turns the sequence of approximations into an exact answer.
Concept
The area under a curve is approximated by rectangles whose heights come from the function. As the rectangles grow narrower and more numerous, the approximations approach a single number, and that limit is the area.
Riemann sum — A sum of function values times subinterval widths, approximating the area under a curve. Its limit as the number of subintervals grows without bound is the exact area.
\[ A = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_{i}^{*})\,\Delta x \]
Nothing about this looks connected to derivatives, and that is the point. Section 5.3 will prove the connection, and the surprise is a large part of why the Fundamental Theorem deserves its name.
Figure (svg): Archimedes' method of exhaustion: rectangles filling a curved region ever more finely
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 439-445
Section
Section 1
Concept
Sigma notation writes a patterned sum compactly. Four standard formulas evaluate the commonest such sums in closed form, replacing n terms by a single expression in n.
sigma notation — A sum written with the Greek capital sigma, a starting index below, a stopping index above, and the general term beside it. The index is a placeholder and its name does not matter.
\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6} \]
The formulas are not decoration. A limit cannot be taken of a sum with n terms, because the number of terms is what is changing; the closed form has a fixed shape and a limit can be taken of that.
Figure (svg): Sigma notation, with each part named, and the four summation formulas
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 439-446 — sigma notation and summation formulas
Picture it
Each part named, and the sums that will be needed.
Figure (svg): Sigma notation, with each part named, and the four summation formulas
The last three formulas are proved by induction and the first is immediate. Together they cover every sum this section produces, because rectangle heights for polynomials are powers of the index.
Worked example
Example 5.2. The formulas applied and combined.
\[ \text{Evaluate } \sum_{i=1}^{n}\left(3i^{2}-2i+5\right). \]
Split across the sum
Why: Sums distribute over addition.
\[ 3 \sum i ^{2} - 2 \sum i + \sum 5 \]
Apply the square formula
Why: The third standard formula.
\[ 3 n(n + 1) (2 n + 1) / 6 \]
Apply the linear formula
Why: The second.
\[ -2 n(n + 1) / 2 \]
Apply the constant formula
Why: Adding 5 exactly n times.
\[ 5 n \]
Simplify
Why: Expand and collect.
\[ n ^{3} + (\frac{1}{2}) n ^{2} +...\text{ in closed form} \]
Figure (svg): Sigma notation, with each part named, and the four summation formulas
\[ \frac{n(n+1)(2n+1)}{2}-n(n+1)+5n \]
Verify: test the closed form at a small value of n
Why: At n equal to 2 the original sum is the term at i equal to 1, which is 6, plus the term at i equal to 2, which is 13, giving 19. The closed form gives 2 times 3 times 5 over 2, minus 6, plus 10, which is 15 minus 6 plus 10, also 19. Checking a closed form at one or two small values catches almost every algebra slip and takes seconds.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 442-443
Fill the middle
The sum of the first n integers.
Fill in the blanks
\sum_1^___ i = \frac___})}___
Why: Pairing the first with the last, the second with the second-last, gives n over 2 pairs each summing to n plus 1. The formula is the oldest in the section and the one most often needed.
Worked example
Checkpoint 5.2. A limit that cannot otherwise be taken.
\[ \text{Evaluate } \lim_{n\to\infty}\frac{1}{n^{3}}\sum_{i=1}^{n}i^{2}. \]
Note the difficulty
Why: The number of terms grows with n.
Replace the sum by its closed form
Why: The square formula.
\[ n(n + 1) (2 n + 1) / 6 \]
Substitute
Why: Into the expression.
\[ n(n + 1) (2 n + 1) / (6 n ^{3}) \]
Divide through by n cubed
Why: Section 4.6's technique.
\[ (1) (1 + \frac{1}{n}) (2 + \frac{1}{n}) / 6 \]
Take the limit
Why: The reciprocals vanish.
\[ \frac{2}{6} = \frac{1}{3} \]
Figure (svg): The solution to Worked example why closed form is necessary shown as a ladder of expressions, one row per legal move
\[ \lim_{n\to\infty}\frac{1}{n^{3}}\sum_{i=1}^{n}i^{2} = \frac13 \]
Verify: confirm numerically and note what made it possible
Why: At n equal to 100 the expression is about 0.33835, and at n equal to 1000 about 0.33383 — closing on one third. The essential move was the closed form: a limit cannot be taken term by term when the number of terms is itself the variable, so collapsing the sum into a fixed expression in n was not a convenience but the only route. Every exact area computation in this section depends on it.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 443-444
Trap
\[ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}1 \]
Each term goes to zero, so the sum goes to zero
Why: The student takes the limit inside.
\[ \text{but the sum is } \frac{n}{n} = 1 \text{ for every } n \]
The number of terms grows at exactly the rate the terms shrink, so the two effects cancel and the limit is 1.
\[ \text{evaluate the sum first, then take the limit} \]
Collapse to closed form before the limit
Why: Term-by-term reasoning fails when the term count is the variable.
This is the same competition that made Section 4.8's forms indeterminate: shrinking terms against a growing count. Neither effect wins by default, and only computing the sum settles it.
Matching
The four standard formulas.
Match the pairs
Why: The fourth is the square of the second, which is a pleasant coincidence worth remembering. The third is the one this section uses most, since rectangle heights for a parabola are squares of the index.
Two truths and a lie
All three are about sigma notation.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. When n is both the number of terms and the limiting variable, taking the limit inside is invalid — the sum of n copies of one over n is 1 for every n, though each term tends to zero. The closed form must be found first, and that is what the four formulas are for.
Prediction
Commit before reasoning.
Predict first
Why must a sum be put in closed form before its limit is taken?
Correct: Because the number of terms is the limiting variable.
\[ \frac1n\sum_{i=1}^{n}1 = 1 \text{ for every } n, \text{ not } 0 \]
Why: A limit describes behaviour as one quantity varies, and here that quantity controls both the size of each term and how many there are. Those two effects compete — exactly the situation Section 4.8 called indeterminate — and only evaluating the sum reveals which wins. The closed form has a fixed algebraic shape with n appearing as an ordinary variable, and a limit of that is straightforward.
Section
Section 2
Concept
Divide the interval into n equal subintervals of width the interval length over n. On each, build a rectangle whose height is the function's value at a chosen point, and add the areas.
the partition and the subinterval width — Dividing an interval into n equal pieces gives each a width of the total length divided by n, and the division points are the left endpoint plus multiples of that width.
\[ \Delta x = \frac{b-a}{n}, \qquad x_{i} = a + i\,\Delta x \]
The choice of which point in each subinterval supplies the height is genuinely free. The left endpoint, the right endpoint and the midpoint are the standard choices, and they give different approximations of the same area.
Figure (svg): Left, right and midpoint rectangles on the same curve and partition
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 444-452 — approximating area with rectangles
Picture it
The same curve and partition, three rules.
Figure (svg): Left, right and midpoint rectangles on the same curve and partition
The left rule underestimates and the right overestimates for this increasing function, while the midpoint rule lands remarkably close. All three converge to the same limit.
Worked example
Example 5.4. Four rectangles on a parabola.
\[ \text{Approximate the area under } f(x)=x^{2} \text{ on } [0,2] \text{ with 4 left rectangles.} \]
Compute the width
Why: Interval length over the count.
\[ \frac{2}{4} = 0.5 \]
List the left endpoints
Why: Starting at 0.
\[ 0, 0.5, 1, 1.5 \]
Compute the heights
Why: Square each.
\[ 0, 0.25, 1, 2.25 \]
Sum the heights
Why: Add.
\[ 3.5 \]
Multiply by the width
Why: Common factor.
\[ 3.5 \times 0.5 = 1.75 \]
Figure (svg): Left, right and midpoint rectangles on the same curve and partition
\[ L_{4} = 1.75 \]
Verify: compare with the true area and explain the direction of the error
Why: The true area is 8 over 3, about 2.667, so this underestimates by about 0.92. The reason is visible in the picture: the function increases, so its value at each subinterval's left end is the smallest it takes there, and every rectangle sits entirely below the curve. That is not luck but a guarantee for any increasing function, and it is what makes the left sum a lower bound rather than merely a low answer.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 447-448
Fill the middle
Dividing an interval into equal pieces.
Fill in the blanks
\Delta x = \frac0.5___ = \frac______ = ___
Why: Each of the four subintervals is half a unit wide. Every rectangle's area is its height times this width, and forgetting the multiplication is the section's commonest error.
Worked example
Checkpoint 5.4. Three rules on the same partition.
\[ \text{Compute } R_{4} \text{ and } M_{4} \text{ for the same function and interval.} \]
List the right endpoints
Why: Shifted by one width.
\[ 0.5, 1, 1.5, 2 \]
Compute and sum the heights
Why: Square each and add.
\[ 0.25 + 1 + 2.25 + 4 = 7.5 \]
Multiply by the width
Why: The right sum.
\[ 3.75 \]
List the midpoints
Why: Halfway across each subinterval.
\[ 0.25, 0.75, 1.25, 1.75 \]
Compute the midpoint sum
Why: Square, add, multiply.
\[ 5.25 \times 0.5 = 2.625 \]
Figure (svg): The solution to Worked example the right and midpoint rules compared shown as a ladder of expressions, one row per legal move
\[ L_{4}=1.75, \; M_{4}=2.625, \; R_{4}=3.75 \]
Verify: compare the three errors and explain why the midpoint wins
Why: The errors are 0.92, 0.04 and 1.08 respectively — the midpoint rule is more than twenty times better than either endpoint rule on the same partition. The reason is that a midpoint rectangle cuts the curve, overshooting on one side of the midpoint and undershooting on the other, so the two errors largely cancel within each rectangle. The endpoint rules have no such cancellation, since every rectangle errs in the same direction.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 449-451
Error analysis
A student computes a left-endpoint approximation.
Annotate
On: \( L_{4} = 0 + 0.25 + 1 + 2.25 = 3.5 \)
Adding heights rather than areas is the commonest slip here, and it survives because the number looks reasonable. Checking the answer against a bounding rectangle catches it immediately.
Ranking
A rectangle approximation.
Put in order
Why: Step e is the one skipped. Adding heights gives a number that looks like an answer, and only comparing it with a bounding rectangle reveals it is too large by a factor of one over the width.
Sorting
For an increasing function.
Sort into buckets
Sort each rule by what it does to the area of an increasing function.
For a DECREASING function the first two swap over, which is why the reliable descriptions are the fourth and fifth — smallest and largest value — rather than left and right. Those give bounds whatever the function does.
Prediction
Commit before reasoning.
Predict first
On the same partition the midpoint rule was twenty times more accurate. Why?
Correct: Because the errors within each rectangle largely cancel.
\[ \text{errors: } 0.92, \; 0.04, \; 1.08 \quad (L_{4}, M_{4}, R_{4}) \]
Why: A midpoint rectangle is too low on the side where the curve is higher and too high where the curve is lower, so most of the error is compensated within the rectangle itself. Endpoint rules have no such cancellation — every rectangle errs the same way and the errors accumulate. The advantage is general rather than special to parabolas, though the exact factor varies with the function.
Section
Section 3
Concept
Taking the smallest function value on each subinterval gives a sum guaranteed below the true area; taking the largest gives one guaranteed above. The true area is trapped between them.
lower and upper sums — Sums built from the minimum and maximum values of the function on each subinterval. Every Riemann sum for the same partition lies between them, and so does the true area.
\[ L \le A \le U \]
A bracket is a much stronger statement than an estimate. One number says the area is about 2.6; a bracket says it is certainly between 2.31 and 3.02, which can be acted on.
Figure (svg): Lower and upper sums bracketing the true area
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 448-455 — upper and lower sums
Picture it
Six rectangles below and six above.
Figure (svg): Lower and upper sums bracketing the true area
Every Riemann sum for this partition lies somewhere between the two pictures, and so does the true area. Increasing the count squeezes the bracket without either bound ever crossing the answer.
Worked example
Example 5.5. Lower and upper sums with six rectangles.
\[ \text{Bracket the area under } f(x)=x^{2} \text{ on } [0,2] \text{ using } n=6. \]
Compute the width
Why: Two divided by six.
\[ \frac{1}{3} \]
Identify the smallest value on each subinterval
Why: The function increases, so the left end.
Compute the lower sum
Why: Heights times width.
\[ \text{about } 2.31 \]
Identify the largest value on each
Why: The right end.
Compute the upper sum
Why: Heights times width.
\[ \text{about } 3.02 \]
Figure (svg): Lower and upper sums bracketing the true area
\[ 2.31 \le A \le 3.02 \]
Verify: check the true value lies inside and measure the bracket's width
Why: The true area is 8 over 3, about 2.667, comfortably inside. The bracket is 0.71 wide, which is not very precise but is a guarantee rather than a guess. Note that the two sums differ by exactly the width times the total rise of the function, since the rectangles' differences stack into a single column — so doubling n halves the bracket, predictably.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 450-452
Fill the middle
Lower and upper sums with six rectangles.
Fill in the blanks
2.31 \le A \le 3.02
Why: The true area of 8 over 3 lies inside. Unlike a single estimate, a bracket is a guarantee: the answer cannot be outside it, whatever else is unknown.
Worked example
Checkpoint 5.5. Left and right are no longer the answer.
\[ \text{How are upper and lower sums found for a function that rises then falls?} \]
Note the difficulty
Why: Left is not always smallest.
Work subinterval by subinterval
Why: On each one separately.
Where the function increases
Why: Left is smallest.
Where it decreases
Why: The reverse.
Where a turning point lies inside
Why: Use the extreme value there.
\[ \text{Section } 4.3' s\text{ method} \]
Figure (svg): The solution to Worked example bounds for a non-monotonic function shown as a ladder of expressions, one row per legal move
\[ L=\sum\min_{[x_{i-1},x_{i}]}f\cdot\Delta x, \; U=\sum\max f\cdot\Delta x \]
Verify: connect this to the Extreme Value Theorem
Why: That the minimum and maximum exist on each closed subinterval is not obvious — it is Section 4.3's Extreme Value Theorem, which needs continuity. For a continuous function the bounds always exist and the bracket is always available; for a discontinuous one the whole construction can fail. That is one reason continuity is assumed throughout this chapter.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 452-453
Trap
\[ f(x) = 4-x^{2} \text{ on } [0,2], \; \text{left endpoints} \]
Call the left sum the lower sum
Why: The student generalises from the increasing case.
\[ \text{but } f \text{ decreases here, so left endpoints give the LARGEST values} \]
The left sum is the upper sum for a decreasing function, and calling it a lower bound inverts the guarantee.
\[ \text{lower sum uses the MINIMUM on each subinterval} \]
Define the bounds by minimum and maximum, not by position
Why: Which endpoint achieves them depends on the direction of the function.
The definitions in terms of minimum and maximum work for any continuous function, monotonic or not. Left and right are convenient shortcuts for the monotonic case and nothing more.
Two truths and a lie
All three are about bounds.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. For a decreasing function the left endpoints give the largest values, making the left sum the UPPER one. The reliable definitions use the minimum and maximum on each subinterval, which work whatever the function does.
Sorting
It depends on the direction.
Sort into buckets
Sort each situation.
The last case is why the definitions are stated with minimum and maximum rather than left and right, and why finding them can require Section 4.3's method for extreme values on a closed interval.
Prediction
Commit before reasoning.
Predict first
The midpoint rule gave 2.625, much closer than either bound. Why bother with the bracket?
Correct: Because the bracket is a guarantee.
\[ \text{estimate: } \approx 2.625; \quad \text{bracket: } 2.31 \le A \le 3.02 \text{ certainly} \]
Why: The midpoint value was indeed excellent, but nothing in the computation revealed that — its closeness was only visible by comparison with an answer already known. A bracket needs no such comparison: it states that the area is certainly between two numbers, and squeezing it by increasing n produces certainty rather than confidence. That distinction is the whole reason mathematical bounds exist.
Section
Section 4
Concept
A Riemann sum takes any point in each subinterval as the sample. For a continuous function every choice gives the same limit, so the area is a property of the function rather than of the approximation scheme.
Riemann sum — The sum of the function's value at a chosen sample point in each subinterval, times the subinterval's width. The sample points may be chosen freely.
\[ \sum_{i=1}^{n} f(x_{i}^{*})\,\Delta x, \qquad x_{i}^{*} \in [x_{i-1},x_{i}] \]
That all choices agree is the substantial content. It is what allows the limit to be called the area rather than merely the left-endpoint area or the midpoint area, and it is a theorem about continuous functions.
Figure (svg): The limit of Riemann sums, with the width shrinking and the count growing
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 452-458 — Riemann sums
Picture it
From partition to limit.
Figure (svg): The limit of Riemann sums, with the width shrinking and the count growing
The last line carries the weight. Without it there would be several different notions of area depending on how one sampled, and the theory would be far less useful.
Worked example
Example 5.6. General n, right endpoints.
\[ \text{Write the right-endpoint Riemann sum for } f(x)=x^{2} \text{ on } [0,2] \text{ with } n \text{ subintervals.} \]
Write the width
Why: In terms of n.
\[ \frac{2}{n} \]
Write the right endpoints
Why: Left end plus i widths.
\[ x _{i} = 2 i / n \]
Write the heights
Why: Square them.
\[ (2 i / n) ^{2} = 4 i ^{2} / n ^{2} \]
Assemble the sum
Why: Height times width, summed.
\[ \sum\text{ of } (4 i ^{2} / n ^{2}) (\frac{2}{n}) \]
Pull the constants out
Why: Only i varies.
\[ (8 / n ^{3}) \sum\text{ of } i ^{2} \]
Figure (svg): Computing the limit exactly with a summation formula
\[ R_{n} = \frac{8}{n^{3}}\sum_{i=1}^{n}i^{2} \]
Verify: check the general form against the earlier numerical answer
Why: At n equal to 4 the sum of squares is 30, so the expression gives 8 times 30 over 64, which is 3.75 — exactly the right-endpoint sum computed earlier by listing rectangles. The general form reproduces the specific one, which is the check worth making before taking a limit. Note that pulling the constants out was essential: it left a sum the standard formula can evaluate.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 454-455
Fill the middle
Right endpoints on an interval starting at zero.
Fill in the blanks
x_2i/n = a + i\,\Delta x = 0 + i\cdot\frac______ = ___
Why: The i-th right endpoint is i widths from the start. Substituting this into the function gives the height, and the whole Riemann sum follows mechanically from there.
Worked example
Checkpoint 5.6. Three rules, one limit.
\[ \text{Compare } L_{n}, M_{n} \text{ and } R_{n} \text{ as } n \text{ grows.} \]
Compare at n = 4
Why: The three values.
\[ 1.75, 2.625, 3.75 \]
Compare at n = 100
Why: Much closer.
\[ 2.627, 2.6667, 2.707 \]
Compare at n = 1000
Why: Closer still.
\[ 2.663, 2.66667, 2.671 \]
Note the pattern
Why: All three approach the same value.
\[ \frac{8}{3} \]
State the theorem
Why: For a continuous function.
Figure (svg): The solution to Worked example the choice of sample point does not matter shown as a ladder of expressions, one row per legal move
\[ \lim L_{n} = \lim M_{n} = \lim R_{n} = \tfrac83 \]
Verify: see why the differences must vanish
Why: The left and right sums differ by exactly the width times the total rise of the function, which for this case is 2 over n times 4 — and that tends to zero. Since every Riemann sum lies between them, all are squeezed to the same limit by Section 2.3's squeeze theorem. So the agreement is not a numerical coincidence but a consequence of the bracket closing, and it holds for every continuous function.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 455-457
Error analysis
A student sets up a Riemann sum and tries to evaluate it.
Annotate
On: \( R_{n} = \sum_{i=1}^{n}\frac{8i^{2}}{n^{3}} = \frac{8}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}\cdot n \)
Pulling a factor out of a sum removes it entirely; it is not also multiplied by the number of terms. Checking the general expression at n equal to 4 against the known value of 3.75 catches this immediately.
Ranking
Writing a Riemann sum in general n.
Put in order
Why: Step e is what makes the closed-form formulas applicable: after it, what remains inside is a pure power of i. Without it the sum matches no standard formula and the limit cannot be taken.
Sorting
For a continuous function.
Sort into buckets
Sort each statement.
The fourth is subtle: the midpoint rule does converge fastest for most functions, but not for all, so as a blanket claim it is false. What is guaranteed is that they all reach the same destination.
Prediction
Commit before reasoning.
Predict first
Why do all Riemann sums for a continuous function have the same limit?
Correct: Because every sum is trapped between bounds whose difference tends to zero.
\[ L_{n} \le \sum f(x_{i}^{*})\Delta x \le U_{n}, \quad U_{n}-L_{n} \to 0 \]
Why: Any sample value lies between the minimum and maximum on its subinterval, so every Riemann sum lies between the lower and upper sums. For a continuous function on a closed interval the gap between those bounds shrinks to zero as the subintervals narrow, and Section 2.3's squeeze theorem forces everything between them to the same limit. It is a theorem with real content rather than a definition.
Section
Section 5
Concept
Write the Riemann sum in general n, collapse it with a summation formula, simplify, and take the limit. The result is the exact area, with no approximation anywhere.
area as a limit — The exact area under a continuous non-negative function is the limit of its Riemann sums as the number of subintervals grows without bound.
\[ A = \lim_{n\to\infty}\frac{8}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6} = \frac83 \]
The method works but is laborious, and it needs a summation formula for whatever powers appear. Section 5.3 will replace the whole computation with an antiderivative evaluated at two points.
Figure (svg): Computing the limit exactly with a summation formula
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 455-462 — area as a limit of Riemann sums
Picture it
The parabola's area, computed.
Figure (svg): Computing the limit exactly with a summation formula
No rectangle count was ever chosen and no approximation remains. The summation formula in the third line is what makes the limit in the fifth possible at all.
Worked example
Example 5.8. The full computation.
\[ \text{Find the exact area under } f(x)=x^{2} \text{ on } [0,2]. \]
Write the Riemann sum
Why: Right endpoints, general n.
\[ (8 / n ^{3}) \sum\text{ of } i ^{2} \]
Apply the summation formula
Why: The third standard formula.
\[ (8 / n ^{3}) n(n + 1) (2 n + 1) / 6 \]
Simplify
Why: Cancel one n and expand.
\[ (\frac{4}{3}) (1 + \frac{1}{n}) (2 + \frac{1}{n}) \]
Take the limit
Why: The reciprocals vanish.
\[ (\frac{4}{3}) (1) (2) \]
State
Why: The exact area.
\[ \frac{8}{3} \]
Figure (svg): Computing the limit exactly with a summation formula
\[ A = \frac{8}{3} \]
Verify: check against the numerical approximations
Why: The rectangle counts gave 3.75, 2.917 and 2.671 for n equal to 4, 16 and 1000 — all converging on 2.6667, which is 8 over 3. So the exact answer matches the numerical evidence. Note what was required: a summation formula for squares. For a cubic the formula for cubes would be needed, and for a sine no such formula exists at all — which is exactly why a better method is wanted.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 456-458
Fill the middle
The simplified Riemann sum, with n growing.
Fill in the blanks
\frac43\left(1+\frac1n\right)\left(2+\frac1n\right) \;\longrightarrow\; \frac43\cdot 1\cdot 2 = 8/3
Why: Both reciprocals vanish, leaving four thirds times one times two. That is the exact area, and it matches the numerical approximations converging on 2.6667.
Worked example
Checkpoint 5.8. The limits of this approach.
\[ \text{Why is this method impractical for } \int\sin x \text{ or } \int e^{x}? \]
Set up the Riemann sum for the sine
Why: Heights are sines of multiples of the width.
Look for a summation formula
Why: The four standard ones cover powers.
Note a formula does exist
Why: A trigonometric identity gives one.
Consider a general function
Why: Most have no closed-form sum at all.
Draw the conclusion
Why: A different approach is needed.
Figure (svg): The solution to Worked example where the method runs out shown as a ladder of expressions, one row per legal move
\[ \text{no closed form for } \sum \Longrightarrow \text{ no limit} \]
Verify: state precisely what the obstacle is
Why: The obstacle is not the limit itself but the closed form that must precede it. Every step of the method is mechanical except collapsing the sum, and that step depends on having a formula for the particular pattern of terms. Powers have such formulas; almost nothing else does. Section 5.3's theorem removes the sum entirely, replacing it with an antiderivative evaluated twice — which is why it deserves the name Fundamental.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 458-460
Trap
\[ \lim_{n\to\infty}\frac{8}{n^{3}}\sum_{i=1}^{n}i^{2} \]
The factor in front goes to zero, so the limit is zero
Why: The student takes the limit of one factor only.
The sum grows like n cubed, so the two factors compete and neither wins by default — this is an infinity-times-zero form.
\[ \frac{8}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6} \to \frac83 \]
Evaluate the sum first; only then is a limit meaningful
Why: The competition is resolved by computing, not by inspection.
This is Section 4.8's indeterminate form appearing in a new guise. The sum's growth rate exactly matches the prefactor's decay, and the finite non-zero limit is what remains — which is precisely why the area is finite.
Ranking
Finding an area as a limit.
Put in order
Why: Step c is the pivot and the bottleneck. Everything before it is mechanical and everything after it is routine, but it needs a summation formula for the particular pattern — which exists for powers and almost nothing else.
Sorting
Ask whether a summation formula exists.
Sort into buckets
Sort each integrand.
The fourth does have an obscure closed form from a trigonometric identity, and the fifth has none at all — nor does it even have an elementary antiderivative. The method's reach is genuinely narrow, which is the case for the next two sections.
Prediction
Commit before reasoning.
Predict first
The limit method is exact but laborious and narrow. What does Chapter 5 replace it with?
Correct: The Fundamental Theorem, using an antiderivative.
\[ \text{Section 5.3: } A = F(b)-F(a) \text{ where } F'=f \]
Why: Section 5.3 proves that the area under a curve equals an antiderivative evaluated at the two endpoints and subtracted — turning a page of summation into two substitutions. That the answer to an area question is Chapter 4's antidifferentiation is genuinely surprising, since nothing in this section's rectangles suggested derivatives at all. Better approximations and faster computation miss the point: the theorem gives exact answers with no sum whatever.
Comparison
Fill the blanks. All three converge to the same limit.
Comparison matrix
| Rule | Sample point | For an increasing function |
|---|---|---|
| Left | the left end of each subinterval | underestimates |
| Right | the right end of each subinterval | overestimates |
| Midpoint | halfway across each subinterval | errors largely cancel |
| Any point | anywhere in each subinterval | lies between the bounds; same limit |
The final row is the theorem that makes area well defined. If different sampling gave different limits there would be no single answer to call the area.
Pattern
Given an area under a curve to compute exactly.
Step four is the bottleneck. Steps one to three and step five are mechanical for any function; step four needs a formula that exists for powers and for very little else.
Stewart, Calculus: Early Transcendentals 8e, §5.1 Areas and Distances §5.1, pp. 366-377
Check
Sigma notation.
Check your understanding
What is the closed form for the sum of the first n squares?
Answer: A
Why: This is the third standard formula, and the one used most in this section.
Check
Rectangles.
Check your understanding
For an increasing function, which sum is guaranteed to underestimate the area?
Answer: A
Why: On an increasing function the left end gives the smallest value on each subinterval.
Check
The limit.
Check your understanding
Why do left, right and midpoint sums all have the same limit?
Answer: A
Why: The squeeze theorem forces everything between the closing bounds to one limit.
Real world
A hospital infusion pump logs the delivery rate in millilitres per minute every thirty seconds during a two-hour infusion. The pharmacist needs the total volume delivered, and the pump records only rates.
Discussion prompt
Explain how the total is recovered from the rate log, which rule is appropriate, and what a bracket would add.
Hint: Rate times time is volume, and the log gives a rate on each interval.
Answer:
Each half-minute contributes approximately the logged rate times half a minute — a rectangle. Summing them over the two hours is exactly a Riemann sum, with the log entries as sample points and 0.5 minutes as the width.
\[ V \approx \sum_{i=1}^{240} r(t_{i})\,\Delta t, \qquad \Delta t = 0.5 \]
The midpoint rule is the appropriate one if the pump logs the rate at the middle of each interval, and it will be far more accurate than using the value at either end — the same twenty-fold improvement seen for the parabola, and for the same reason.
The bracket adds something an estimate cannot. Taking the smallest and largest rate on each interval gives a range the true volume certainly lies within. For a drug where over-delivery is dangerous, a guaranteed upper bound is worth more than a good estimate: it supports a statement about what could not have happened, rather than what probably did.
Note what makes this problem unavoidable rather than merely convenient. There is no formula for the rate — it is measured data, one number per half minute — so no antiderivative exists to apply. Section 5.3's theorem will not help here, and the Riemann sum is not an approximation to a better method but the only method there is.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why must a Riemann sum be put in closed form before its limit is taken?
Correct: Because the number of terms is the limiting variable.
\[ \frac{8}{n^{3}} \to 0 \text{ but } \sum i^{2} \sim \frac{n^{3}}{3}: \; \text{the product is } \tfrac83 \]
Why: As n grows, each term shrinks and there are more of them — two effects competing, exactly the indeterminate situation of Section 4.8. The sum of n copies of one over n is 1 for every n even though each term tends to zero, which shows term-by-term reasoning fails outright. Only evaluating the sum in closed form, so that n appears as an ordinary variable in a fixed expression, makes the limit meaningful.
Explain it
They added the four rectangle heights and reported that as the area.
Discussion prompt
In four sentences or fewer, show them what is missing.
Hint: Ask what a rectangle's area is.
Answer:
Ask them for the area of a single rectangle: it is height times width, not height alone. Each of their rectangles is half a unit wide, so the total of the heights must be multiplied by 0.5.
A quick check makes it obvious: the whole region fits inside a box 2 wide and 4 tall, area 8, and it is clearly well under half of that. Their answer of 3.5 was nearly half the box, which should have looked wrong immediately.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the formulas, write all four out and check each at n equal to 3. For rectangles, always compute the width first and multiply at the end. For bounds, remember they are defined by minimum and maximum, not left and right. For the limit, pull constants out first so the sum matches a standard formula. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, sketch the same curve three times with four rectangles each, using left, right and midpoint heights, and write the three sums beneath with the true area beside them. To the right, write sigma notation with its three parts labelled and list the four summation formulas. In the middle of the page, draw one curve twice with lower and upper rectangles, and write the bracket as an inequality with the true area inside it — then note in a few words why left is not always the lower one. In the lower half, work the exact computation for the parabola in five lines: the Riemann sum in general n, constants pulled out, the summation formula applied, simplification, and the limit. Beside it write what step would fail for a sine, and why. At the bottom, write one sentence saying what Section 5.3 will replace all of this with.
If your three rectangle pictures look identical, look again — the left rectangles should sit entirely below the curve and the right ones entirely above, and that visible difference is what the bracket is built on.
Recap
Five things, and the last is exact rather than approximate.
| If you see | Then |
|---|---|
| A sum with a pattern | Write it in sigma notation |
| A sum whose limit is wanted | Find its closed form first |
| An increasing function | Left underestimates, right overestimates |
| A guarantee needed | Bracket with lower and upper sums |
| A general n required | Width is (b-a)/n and x_i = a + i times that |
| Constants inside a sigma | Pull them out before applying a formula |
| An integrand that is not a power | The limit method will probably stall |
Section 5.2 gives this limit a name and a symbol — the definite integral — and extends it to functions that go below the axis, where the answer becomes signed area rather than area.
OpenStax Calculus Volume 1, §5.1 Approximating Areas §5.1, pp. 438-455 — everything on these slides traces back here
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