The definite integral as the limit of Riemann sums, its notation and the meaning of each part, net signed area against total area, which functions are integrable, the properties inherited from sums, and the average value of a function.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 5 — Integration
The Definite Integral
Objectives
Five outcomes. The first is a definition, and the rest are what can be done with it before any antiderivative appears.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 456-471 — the section these objectives are drawn from
Warm-up
Section 5.1 built rectangle approximations and showed their limit exists and does not depend on the sample points chosen.
Discussion prompt
That limit deserves a name and a symbol. What should the symbol look like?
Hint: It is replacing a sum of heights times widths.
Answer:
The limit replaces a sigma with an elongated S — the same letter, stretched — and the width, which shrank to nothing, becomes the dx. The heights stay as the function.
\[ \lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\,\Delta x \;\longrightarrow\; \int_{a}^{b}f(x)\,dx \]
Every symbol in the new notation traces back to one in the old. Beyond notation, this section establishes what can be done with these objects — splitting them, comparing them, bounding them — all before Section 5.3 shows how to compute one quickly.
Concept
The definite integral of a function over an interval is the limit of its Riemann sums. It represents net signed area, counting regions below the axis as negative, and it obeys a set of properties inherited from sums.
definite integral — The limit of Riemann sums for a function on a closed interval, when that limit exists independently of the sample points. It is a number, not a family of functions.
\[ \int_{a}^{b}f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\,\Delta x \]
The word definite distinguishes this from Section 4.10's indefinite integral. That one produced a family of functions; this produces a single number, and the two are connected by the theorem of Section 5.3.
Figure (svg): Definite integral notation, with each part named and the Riemann sum it came from
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 462-470
Section
Section 1
Concept
The integral sign is a stretched sigma, the integrand is the height, the differential is the vanished width, and the numbers on the sign are the interval's ends.
limits of integration — The numbers at the bottom and top of the integral sign, naming the interval's left and right ends. They are not limits in the Chapter 2 sense, despite the name.
\[ \int_{a}^{b}f(x)\,dx: \; a \text{ lower}, \; b \text{ upper}, \; f \text{ integrand}, \; x \text{ variable} \]
The variable of integration is a placeholder like a sum's index: the integral of f of x with respect to x and the integral of f of t with respect to t over the same interval are the same number.
Figure (svg): Definite integral notation, with each part named and the Riemann sum it came from
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 462-470 — definition and notation
Picture it
Each part named.
Figure (svg): Definite integral notation, with each part named and the Riemann sum it came from
Notice the differential's role has shifted. In Section 4.10 it named the variable; here it does that and also records where the width went, which is why it must not be omitted.
Worked example
Example 5.10. Reading a Riemann sum backwards.
\[ \text{Express } \lim_{n\to\infty}\sum_{i=1}^{n}\left(2+\frac{3i}{n}\right)^{2}\frac{3}{n} \text{ as an integral.} \]
Identify the width
Why: The factor not depending on the function.
\[ \frac{3}{n},\text{ so } b - a = 3 \]
Identify the sample points
Why: The expression inside.
\[ x _{i} = 2 + 3 i / n \]
Read off the left end
Why: At i equal to 0.
\[ a = 2 \]
Read off the right end
Why: Left end plus the total width.
\[ b = 5 \]
Identify the function
Why: What is applied to the sample point.
\[ f(x) = x ^{2} \]
Figure (svg): Definite integral notation, with each part named and the Riemann sum it came from
\[ \int_{2}^{5}x^{2}\,dx \]
Verify: check the width against the interval
Why: The interval from 2 to 5 has length 3, and dividing it into n pieces gives width 3 over n — which is exactly the factor in the sum. The sample points run from 2 plus 3 over n up to 5, confirming right endpoints on that interval. Reading a Riemann sum backwards like this is the skill Section 5.3 will need in reverse, and the width factor is always the giveaway.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 466-467
Fill the middle
A Riemann sum whose width factor is three over n.
Fill in the blanks
\Delta x = \frac3___ \;\Longrightarrow\; b-a = ___
Why: The width is the interval length over n, so the length is 3. Combined with the sample points starting at 2, the interval is from 2 to 5.
Worked example
Checkpoint 5.10. Renaming changes nothing.
\[ \text{Are } \int_{0}^{2}x^{2}dx \text{ and } \int_{0}^{2}t^{2}dt \text{ the same?} \]
Write the Riemann sum for the first
Why: Sample points and width.
Write it for the second
Why: The same construction.
Compare
Why: Identical numbers throughout.
Take limits
Why: Identical limits.
Conclude
Why: The letter is a placeholder.
Figure (svg): The solution to Worked example the variable is a placeholder shown as a ladder of expressions, one row per legal move
\[ \int_{0}^{2}x^{2}dx = \int_{0}^{2}t^{2}dt = \tfrac83 \]
Verify: contrast this with the indefinite integral
Why: For an INDEFINITE integral the letter does matter in the answer: Section 4.10 showed the antiderivative of t cubed with respect to t is t to the fourth over 4, in the variable t. But a definite integral is a number, and a number carries no variable — so the letter vanishes with the limit. This is why the variable of integration is called a dummy variable, exactly like a sum's index.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 467-468
Trap
\[ \int_{0}^{2}x^{2}\,dx = \frac{x^{3}}{3}\bigg|_{0}^{2} = \frac{x^{3}}{3} \]
Leave the variable in the final answer
Why: The student stops before substituting.
A definite integral is a number. An answer containing x is not a number and cannot be one.
\[ = \frac{8}{3}-0 = \frac{8}{3} \]
Substitute both limits and subtract, leaving a number
Why: The variable disappears with the evaluation.
A quick check on any definite integral: if the answer contains the variable of integration, something has been left undone. That single test catches this error every time.
Matching
Each part came from the Riemann sum.
Match the pairs
Why: The notation was designed so that every piece of the sum survives visibly in the limit. Leibniz chose it deliberately, and it is why the symbol is so easy to read once the sum behind it is understood.
Two truths and a lie
All three are about 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. The word limit is doing two different jobs: the limits of integration are just the interval's endpoints, while the limit in the definition is the Chapter 2 kind, taken as n grows. The clash of terminology is unfortunate but standard.
Prediction
Commit before reasoning.
Predict first
Why can the variable of integration be renamed without changing a definite integral?
Correct: Because the result is a number.
\[ \int_{0}^{2}x^{2}dx = \tfrac83, \quad \int x^{2}dx = \tfrac{x^{3}}{3}+C \]
Why: The variable exists only inside the construction — it labels the sample points that get summed — and both the sum and its limit produce a single number with no variable left in it. Contrast the indefinite integral, which produces a function and where the letter genuinely survives into the answer. That difference is the clearest way to keep the two objects apart.
Section
Section 2
Concept
Where the function is negative, its rectangles have negative height, so those contributions subtract. The integral therefore gives net signed area, which can be zero or negative even for a region of substantial size.
net signed area — The area above the axis minus the area below it. It is what the integral computes, and it differs from total area whenever the function changes sign.
\[ \int_{0}^{2\pi}\sin x\,dx = 0 \]
To get total area instead, integrate the absolute value — which in practice means splitting at the zeros and negating the pieces below the axis before adding.
Figure (svg): Net signed area: regions below the axis counted negative
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 468-474 — net signed area
Picture it
A full period of the sine.
Figure (svg): Net signed area: regions below the axis counted negative
The integral is exactly zero although the region has area 4. Nothing has gone wrong: the integral was never a measure of total area, and asking it for one is asking the wrong question.
Worked example
Example 5.12. The same region, two questions.
\[ \text{For } f(x)=\sin x \text{ on } [0,2\pi], \text{ find the integral and the total area.} \]
Find where the function changes sign
Why: The zeros inside the interval.
Note the first half is above the axis
Why: Positive contribution.
\[ \text{area } 2 \]
Note the second half is below
Why: Negative contribution.
\[ -2 \]
Add for the integral
Why: The signed total.
\[ 0 \]
Add magnitudes for total area
Why: Negating the lower piece.
\[ 2 + 2 = 4 \]
Figure (svg): Net signed area: regions below the axis counted negative
\[ \int_{0}^{2\pi}\sin x\,dx = 0, \qquad \text{total area} = 4 \]
Verify: confirm the cancellation is exact and say why
Why: The sine's second half is the exact reflection of its first through the point at pi, so the two regions are congruent and the cancellation is exact rather than approximate. Note that the integral being zero says something true and useful — over a full period the sine's net contribution vanishes, which is why alternating currents deliver no net charge — but it is a different fact from the region's size.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 470-471
Sorting
What is the question actually asking?
Sort into buckets
Sort each question by which quantity answers it.
The displacement-against-distance pair is the clearest case. A particle returning to its start has zero displacement and a substantial distance travelled, and the same integral answers only the first.
Worked example
Checkpoint 5.12. Split at the zeros.
\[ \text{Find the total area between } f(x)=x-1 \text{ and the axis on } [0,3]. \]
Find the zero
Why: Set the function to zero.
\[ x = 1 \]
Split the interval there
Why: Two pieces.
\[ [0, 1]\text{ and } [1, 3] \]
Compute the first piece
Why: A triangle below the axis.
\[ \text{signed value } -\frac{1}{2} \]
Compute the second
Why: A triangle above.
\[ \text{signed value } 2 \]
Add magnitudes
Why: Negate the first.
\[ \frac{1}{2} + 2 = \frac{5}{2} \]
Figure (svg): The solution to Worked example computing total area shown as a ladder of expressions, one row per legal move
\[ \text{total} = \tfrac52, \qquad \int_{0}^{3}(x-1)dx = \tfrac32 \]
Verify: check both against the geometry
Why: The two triangles have bases 1 and 2 and heights 1 and 2, giving areas one half and 2 — matching. The integral, at three halves, is the difference rather than the sum, which is exactly two times one half less than the total. Splitting at the zeros is the essential move: attempting the total area in one integral gives the signed value instead, and the difference is invisible in the arithmetic.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 471-473
Error analysis
A student is asked for the total area between a curve and the axis.
Annotate
On: \( \text{total area} = \int_{0}^{2\pi}\sin x\,dx = 0 \)
The answer of zero should itself have been the warning. An area is never negative and is zero only for an empty region, so any signed-looking answer to an area question means the sign changes were not handled.
Fill the middle
A sine over a full period, split at its interior zero.
Fill in the blanks
\text4 = 2 + |___| = ___
Why: The two halves each have area 2, and the total counts both positively. The integral, which subtracts them, gives zero — a different and equally correct answer to a different question.
Ranking
A function that crosses the axis.
Put in order
Why: Step d is what distinguishes this from the signed computation, and step a is what makes it possible. Skipping the split and taking one magnitude at the end gives the magnitude of the signed value, which is not the total area.
Prediction
Commit before reasoning.
Predict first
The integral of the sine over a full period is zero. Has something gone wrong?
Correct: No: the integral gives signed area and the halves cancel.
\[ \int_{0}^{2\pi}\sin x\,dx = 0 \quad \text{but} \quad \int_{0}^{2\pi}|\sin x|\,dx = 4 \]
Why: The region genuinely has area 4, and the integral genuinely is zero — both are correct answers to different questions. The integral counts area below the axis negatively, which is exactly what makes it the right tool for displacement, net charge and net change, where opposite contributions really do cancel. For geometric area the integrand must be split at its zeros first.
Section
Section 3
Concept
Every continuous function on a closed bounded interval is integrable, and so is every bounded function with finitely many discontinuities. Integrability fails when different sample choices give different limits.
integrable — A function is integrable on an interval when the limit of its Riemann sums exists and is the same for every choice of sample points. Continuity on a closed bounded interval guarantees it.
\[ f \text{ continuous on } [a,b] \;\Longrightarrow\; f \text{ integrable on } [a,b] \]
The standard non-integrable example is the function that is 1 at rational inputs and 0 elsewhere. Choosing rational sample points gives every sum the value 1; choosing irrational ones gives 0, so no limit exists.
Figure (svg): What can be integrated, and one function that cannot
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 469-476 — integrable functions
Picture it
Two columns, with the standard counterexample.
Figure (svg): What can be integrated, and one function that cannot
The failure is not that the function is strange to look at but that the sample points alone decide the answer. Without agreement between choices there is no single number to call the integral.
Worked example
Example 5.13. A jump does not prevent integration.
\[ \text{Is the step function equal to } 1 \text{ on } [0,1) \text{ and } 2 \text{ on } [1,2] \text{ integrable?} \]
Note the discontinuity
Why: A jump at 1.
Check boundedness
Why: Values are 1 and 2.
Count the discontinuities
Why: One point.
Apply the criterion
Why: Bounded with finitely many jumps.
Compute the value
Why: Two rectangles.
\[ 1 + 2 = 3 \]
Figure (svg): What can be integrated, and one function that cannot
\[ \int_{0}^{2}f = 1\cdot 1 + 2\cdot 1 = 3 \]
Verify: see why one point cannot matter
Why: Any subinterval containing the jump has width tending to zero, and the function is bounded there, so its contribution to the sum tends to zero however the sample point is chosen inside it. A single point has no width and therefore no area — which is why changing a function's value at finitely many points does not change its integral at all. The criterion of finitely many jumps is exactly this observation made general.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 472-473
Sorting
Check boundedness and the discontinuities.
Sort into buckets
Sort each function.
The reciprocal fails for a different reason from the last one: it is unbounded near zero rather than badly discontinuous. Both requirements — boundedness and few discontinuities — are genuinely needed.
Worked example
Checkpoint 5.13. The standard counterexample.
\[ \text{Show that the function equal to } 1 \text{ at rationals and } 0 \text{ elsewhere is not integrable on } [0,1]. \]
Choose rational sample points
Why: Every subinterval contains one.
\[ \text{all heights are } 1 \]
Compute that sum
Why: Total width times 1.
\[ 1,\text{ for every } n \]
Choose irrational sample points
Why: Every subinterval contains one too.
\[ \text{all heights are } 0 \]
Compute that sum
Why: Total width times 0.
\[ 0,\text{ for every } n \]
Conclude
Why: Two different limits.
Figure (svg): The solution to Worked example a function that is not integrable shown as a ladder of expressions, one row per legal move
\[ \lim = 1 \text{ or } 0, \text{ depending on the choice} \]
Verify: identify precisely which requirement failed
Why: The function is bounded, between 0 and 1, so boundedness was not the problem. What failed is the requirement that the limit be independent of the sample points — and this function is discontinuous at every point, not finitely many. The example is what forced mathematicians to state that independence requirement explicitly rather than assuming it, and it is why Section 5.1's theorem about agreement was worth proving.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 473-474
Trap
\[ f \text{ has a jump at } 1 \;\Longrightarrow\; \text{not integrable} \]
Reject the function
Why: The student overgeneralises from the counterexample.
A bounded function with finitely many jumps is integrable, because each jump sits in a subinterval whose width tends to zero.
\[ \text{bounded} + \text{finitely many discontinuities} \;\Longrightarrow\; \text{integrable} \]
Check boundedness and count the discontinuities
Why: Continuity is sufficient but not necessary.
The Dirichlet function fails because it is discontinuous EVERYWHERE, not because it is discontinuous somewhere. The distinction matters in applications, where measured data with occasional jumps is common and perfectly integrable.
Fill the middle
The simplest guarantee of integrability.
Fill in the blanks
f \textcontinuous \; ___ \text___ [a,b] \;\Longrightarrow\; f \text___
Why: Continuity on a closed bounded interval is sufficient. It is not necessary — step functions are integrable too — but it covers almost everything met in practice.
Two truths and a lie
All three are about integrability.
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. Step functions are discontinuous and perfectly integrable, and so is any bounded function with finitely many jumps. What matters is whether the discontinuities are few enough that the sample choice cannot change the limit.
Prediction
Commit before reasoning.
Predict first
For the function that is 1 at rationals and 0 elsewhere, which requirement of the definition fails?
Correct: That the limit be independent of the sample points.
\[ \text{rational choice} \to 1, \quad \text{irrational choice} \to 0 \]
Why: The function is bounded between 0 and 1, and the interval is perfectly ordinary. What fails is agreement: rational sample points give every sum the value 1 and irrational ones give 0, so there are two candidate limits and no single number to call the integral. This example is precisely why that independence clause appears in the definition at all.
Section
Section 4
Concept
Equal limits give zero, swapping the limits flips the sign, sums split, constants come outside, and adjacent intervals join. Each is the limit of the corresponding fact about finite sums.
the additive property — The integral over an interval equals the sum of the integrals over any two pieces meeting at a point, and the sign convention makes it hold even when that point lies outside the interval.
\[ \int_{a}^{b} = \int_{a}^{c} + \int_{c}^{b}, \qquad \int_{b}^{a} = -\int_{a}^{b} \]
The reversal rule is a definition rather than a theorem, chosen precisely so the additive property holds for every c. Without it the splitting rule would need a case distinction for every position of c.
Figure (svg): The properties of the definite integral, grouped by what they do
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 474-482 — properties of the definite integral
Picture it
Each with its origin.
Figure (svg): The properties of the definite integral, grouped by what they do
Only the second is a convention; the rest are theorems, and each is proved by writing the corresponding statement about Riemann sums and taking limits.
Worked example
Example 5.15. Combining several rules.
\[ \text{Given } \int_{0}^{3}f = 7 \text{ and } \int_{0}^{5}f = 12, \text{ find } \int_{5}^{3}f. \]
Split the larger interval
Why: At the interior point 3.
\[ \int 0\text{ to } 5 = \int 0\text{ to } 3 + \int 3\text{ to } 5 \]
Substitute the known values
Why: Two of the three are given.
\[ 12 = 7 + \int 3\text{ to } 5 \]
Solve
Why: Subtract.
\[ \int 3\text{ to } 5 = 5 \]
Apply the reversal rule
Why: Swapping the limits flips the sign.
\[ \int 5\text{ to } 3 = -5 \]
State
Why: The answer.
\[ -5 \]
Figure (svg): The properties of the definite integral, grouped by what they do
\[ \int_{5}^{3}f = -5 \]
Verify: sanity-check the reversal and the split
Why: The reversal must flip the sign, so a positive integral from 3 to 5 gives a negative one from 5 to 3 — consistent. The split can be checked by adding back: 7 plus 5 gives 12, matching the given value over the whole interval. Note that neither the function nor its formula was ever needed, which is the point of having properties: they answer questions about integrals without computing any.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 477-478
Fill the middle
A known integral, with the limits swapped.
Fill in the blanks
\int_-^___f = ___\int____^___f
Why: Swapping the limits flips the sign. This is a definition chosen so that the additive property holds for every splitting point, including ones outside the interval.
Worked example
Example 5.17. An estimate without computing.
\[ \text{Bound } \int_{1}^{2}\frac{1}{x}\,dx \text{ without evaluating it.} \]
Find the function's range on the interval
Why: It decreases.
\[ \text{between } \frac{1}{2}\text{ and } 1 \]
Apply the lower bound
Why: Minimum times the width.
\[ \text{at least } \frac{1}{2} \]
Apply the upper bound
Why: Maximum times the width.
\[ \text{at most } 1 \]
State the bracket
Why: The width is 1.
\[ \frac{1}{2} \le\text{ integral } \le 1 \]
Compare with the true value
Why: It is the natural logarithm of 2.
\[ \text{about } 0.693 \]
Figure (svg): Comparison and bounding: one curve above another, and both between two lines
\[ \tfrac12 \le \int_{1}^{2}\frac{dx}{x} \le 1 \]
Verify: check the bound is genuine and note its cost
Why: The true value of 0.693 sits comfortably inside. The bounding property cost two function evaluations and no integration at all, which is why it is used constantly in proofs and error estimates — often a bound is all that is needed, and it is available when the integral itself is not computable. Note the connection to Section 5.1's upper and lower sums: this is that idea with a single rectangle.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 479-481
Error analysis
A student uses a known integral with the limits the other way round.
Annotate
On: \( \int_{0}^{2}f = 5 \;\Longrightarrow\; \int_{2}^{0}f = 5 \)
Watching the order of the limits is a habit worth building early, because Section 5.3's evaluation rule subtracts the value at the lower limit from the value at the upper one — and getting them the wrong way round negates every answer.
Matching
The rules of the section.
Match the pairs
Why: The first is immediate — no width, no area — and the second is a convention. The last is the comparison property, and it is what makes bounding possible without any computation.
Sorting
Match the question to the rule.
Sort into buckets
Sort each question.
The fourth is the one worth remembering when a problem looks impossible: an integral with no elementary antiderivative can still be bracketed in two lines, and often a bracket is all that is wanted.
Prediction
Commit before reasoning.
Predict first
Why is the integral with swapped limits defined to be the negative rather than the same?
Correct: So that the additive property holds for every splitting point.
\[ \int_{a}^{b} = \int_{a}^{c}+\int_{c}^{b} \quad \text{for every } c, \text{ inside or outside} \]
Why: With the convention, the integral from a to b equals the sum of the integrals from a to c and c to b for ANY c — including one beyond b, where the second piece runs backwards and subtracts the excess. Without it the rule would need a case distinction for every position of c, and Section 5.3's evaluation formula would need one too. The convention is chosen to make later statements simple.
Section
Section 5
Concept
The average value of a function over an interval is its integral divided by the interval's length — the constant height that would enclose the same area. A continuous function actually attains its average somewhere.
average value — The integral of a function over an interval, divided by the interval's length. It is the height of the rectangle with the same base and the same area.
\[ f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b}f(x)\,dx \]
That the value is attained is the Mean Value Theorem for Integrals, and it is the integral counterpart of Section 4.4's theorem — with the same hypothesis of continuity and the same kind of existence conclusion.
Figure (svg): The average value of a function as the height of an equal-area rectangle
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 482-488 — average value of a function
Picture it
A parabola and its average height.
Figure (svg): The average value of a function as the height of an equal-area rectangle
The rectangle at height four thirds encloses exactly the area under the curve, and the curve crosses that height once — which the theorem guarantees for any continuous function.
Worked example
Example 5.18. Integral over width.
\[ \text{Find the average value of } f(x)=x^{2} \text{ on } [0,2]. \]
Recall the integral
Why: From Section 5.1.
\[ \frac{8}{3} \]
Compute the interval's length
Why: Two minus zero.
\[ 2 \]
Divide
Why: Integral over width.
\[ \frac{4}{3} \]
Interpret
Why: The equal-area rectangle's height.
\[ \text{about } 1.333 \]
Locate where it is attained
Why: Solve x squared equals four thirds.
\[ x\text{ about } 1.155 \]
Figure (svg): The average value of a function as the height of an equal-area rectangle
\[ f_{\text{avg}} = \frac{4}{3} \]
Verify: sanity-check the value against the function's range
Why: The function runs from 0 to 4 on this interval, so an average of 1.333 is plausible — and it is well below the midpoint of that range, which is right because the parabola spends most of the interval near its low values. Note the average is not the average of the endpoints, which would be 2; averaging a function means integrating it, not averaging a few of its values.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 484-485
Fill the middle
An integral of eight thirds over an interval of length two.
Fill in the blanks
f_4/3} = \frac______\cdot\frac______ = ___
Why: The average value is the integral divided by the interval's length. It is the height of the rectangle enclosing the same area over the same base.
Worked example
Checkpoint 5.18. The average is attained.
\[ \text{State the theorem and say why continuity is needed.} \]
State the conclusion
Why: For a continuous function.
Note the average lies between the extremes
Why: By the bounding property.
\[ \min \le\text{ average } \le \max \]
Apply the Intermediate Value Theorem
Why: Section 2.4.
Conclude
Why: The average is such a value.
Test the necessity of continuity
Why: A step function jumping over its average.
Figure (svg): The solution to Worked example the Mean Value Theorem for Integrals shown as a ladder of expressions, one row per legal move
\[ \exists c\in[a,b]: \; f(c) = \frac{1}{b-a}\int_{a}^{b}f \]
Verify: build the counterexample that shows continuity is essential
Why: Take the function equal to 0 on the first half of an interval and 2 on the second. Its average is 1, and the function never takes the value 1 anywhere — it jumps straight past it. So the conclusion genuinely fails without continuity, and the proof shows exactly why: it used the Intermediate Value Theorem, which needs continuity. This is the same dependency pattern as Section 4.4's theorem.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 485-487
Trap
\[ f_{\text{avg}} = \frac{f(0)+f(2)}{2} = \frac{0+4}{2} = 2 \]
Average the two endpoint values
Why: The student treats it like averaging two numbers.
The true average is four thirds. Averaging endpoints ignores everything the function does in between.
\[ f_{\text{avg}} = \frac{1}{2}\int_{0}^{2}x^{2}dx = \frac{1}{2}\cdot\frac83 = \frac43 \]
Integrate and divide by the width
Why: The average of infinitely many values is an integral, not a finite mean.
The endpoint average is exactly what a single trapezoid would give, and the gap between 2 and four thirds measures how far the parabola sits below its chord. Averaging a function requires accounting for all of it.
Ranking
The Mean Value Theorem for Integrals.
Put in order
Why: Step d is where continuity is actually used, and it is why the theorem fails for step functions. The bounding in step b is the property from the previous idea, doing real work in a proof.
Sorting
Continuity is the deciding factor.
Sort into buckets
Sort each function on a closed interval.
The sine case is worth noticing: its average over a full period is zero, and it does attain zero — twice. A zero average does not mean the theorem has nothing to say.
Prediction
Commit before reasoning.
Predict first
Why is the average of a function an integral rather than a mean of some values?
Correct: Because the integral is the limit of averaging n sample values.
\[ \frac{1}{n}\sum f(x_{i}) = \frac{1}{b-a}\sum f(x_{i})\Delta x \;\longrightarrow\; \frac{1}{b-a}\int_{a}^{b}f \]
Why: Averaging n equally spaced values means summing them and dividing by n, and multiplying and dividing by the width turns that into a Riemann sum divided by the interval's length. Taking n to infinity gives exactly the integral over the width. So the definition is not an analogy with finite averages but the actual limit of them, which is why it deserves the name.
Comparison
Fill the blanks. The two objects are different in kind.
Comparison matrix
| Indefinite integral | Definite integral | |
|---|---|---|
| Result | a family of functions | a single number |
| Carries a constant | yes, the plus C | no |
| The variable | survives in the answer | vanishes: it is a placeholder |
| Answers | what has this derivative? | what is the net signed area? |
The two look alike and answer unrelated questions. Section 5.3's theorem is startling precisely because it connects them, and until then nothing suggests they should be related at all.
Pattern
Given a question about a definite integral.
Step one is the one that decides whether an answer is right or answers a different question. An area question whose answer comes out negative or zero has almost certainly skipped it.
Stewart, Calculus: Early Transcendentals 8e, §5.2 The Definite Integral §5.2, pp. 378-391
Check
Signed area.
Check your understanding
The integral of sin x from 0 to 2pi is zero. What is the total area between the curve and the axis?
Answer: A
Why: Each half has area 2, and total area counts both positively.
Check
Properties.
Check your understanding
If the integral of f from 2 to 5 is 5, what is the integral from 5 to 2?
Answer: A
Why: Swapping the limits flips the sign, by definition.
Check
Average value.
Check your understanding
The integral of x^2 from 0 to 2 is 8/3. What is the function's average value there?
Answer: A
Why: Divide the integral by the interval's length of 2.
Real world
A tidal turbine's power output is positive when the tide flows in and negative when it draws power to reposition during slack water. Over a twelve-hour cycle the operator has a continuous power curve.
Discussion prompt
Explain what the integral of power over the cycle gives, how it differs from the total energy handled, and what the average value means.
Hint: Power integrated over time is energy, and the sign records direction.
Answer:
The integral of power over the cycle gives the net energy delivered — generation minus consumption — because the negative stretches genuinely subtract. That is the number the operator gets paid on, and the signed convention is exactly right for it.
\[ E_{\text{net}} = \int_{0}^{12}P(t)\,dt \]
The total energy handled is a different quantity, found by splitting at the zeros of the power curve and adding magnitudes. It matters for a different reason: it determines wear on the mechanism, which does not care about direction. A turbine with zero net output can still be working hard.
The average value is the net energy divided by twelve hours — the constant power output that would deliver the same net energy over the cycle. It is the honest figure to quote for a capacity comparison, and it is generally far below the peak, which is why turbine ratings quoted as peak power are misleading.
Note that all three numbers come from the same curve and answer different questions, and that confusing them has real consequences: quoting the peak sells the turbine, quoting the average describes it, and quoting the net signed value is what the electricity meter reads.
Commit first
Answer, then rate your confidence honestly.
Predict first
An area question produces an integral equal to zero. What is the most likely explanation?
Correct: The integrand changes sign and the parts cancelled.
\[ \int_{0}^{2\pi}\sin x\,dx = 0, \quad \text{total area} = 4 \]
Why: The integral gives net signed area, so any region with as much below the axis as above returns zero. That is the correct value of the integral and the wrong answer to an area question — total area needs the integrand split at its zeros and the magnitudes added. An answer of zero to an area question is a reliable warning sign, since a genuinely empty region is rarely what a problem is about.
Explain it
They computed the integral of the sine over a full period, got zero, and concluded the area under a sine wave is nothing.
Discussion prompt
In four sentences or fewer, sort them out.
Hint: Ask them to sketch it.
Answer:
Have them sketch the curve: there is clearly a hump above the axis and an equal one below. The integral counts the lower hump negatively, so the two cancel — that is what a signed area does, and the answer of zero is correct for what the integral measures.
For the region's actual size, split at pi, integrate each half, and add the magnitudes to get 4. The rule of thumb is that an area question answered with zero or a negative number has skipped the split.
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 reading sums, the width factor tells you the interval length. For area, always ask which question is being posed before computing. For properties, remember reversal is a convention chosen to make additivity universal. For averages, divide by the width and remember the theorem needs continuity. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the definition as a limit of Riemann sums with an arrow to the integral notation, and label every symbol with where it came from. Below, sketch a full sine period with the upper region shaded one way and the lower another, and write both the integral and the total area beside it with one line saying why they differ. In the middle, list the seven properties, marking which one is a convention rather than a theorem and writing in a few words why that convention was chosen. Beside them, work one problem using additivity and reversal together. In the lower half, sketch a parabola with its average-value rectangle drawn to the same area, mark where the curve crosses that height, and write the Mean Value Theorem for Integrals with its hypothesis. At the bottom, write the counterexample showing why continuity is needed there, and one line on what makes a function fail to be integrable.
If your average-value rectangle looks taller than the area under the curve, redraw it — the two regions must be equal by construction, and getting that visibly right is what makes the definition memorable.
Recap
Five things, and none of them yet needs an antiderivative.
| If you see | Then |
|---|---|
| A limit of Riemann sums | Read off the interval from the width factor |
| An area question | Split at the zeros before integrating |
| A zero answer to an area question | The signs cancelled; split and retry |
| Limits the wrong way round | Flip them and negate |
| Adjacent intervals | Their integrals add |
| An integral you cannot compute | Bound it by the extremes times the width |
| The word average | Integrate and divide by the width |
Section 5.3 is the chapter's centre. It proves that this limit of sums is computed by an antiderivative evaluated at the two endpoints — connecting Chapter 4 to Chapter 5 and making almost every integral in this deck a two-line calculation.
OpenStax Calculus Volume 1, §5.2 The Definite Integral §5.2, pp. 456-471 — everything on these slides traces back here
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