The basic integration formulas applied to definite integrals, rewriting integrands the table does not cover, the Net Change Theorem, displacement against distance travelled, and the symmetry shortcuts for even and odd functions.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 5 — Integration
Integration Formulas and the Net Change Theorem
Objectives
Five outcomes. The theorem is one line; the rest is knowing when it answers the question asked.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 488-505 — the section these objectives are drawn from
Warm-up
Section 5.3 proved that a definite integral is an antiderivative evaluated at the two endpoints. Section 4.10 supplied the antiderivative formulas.
Discussion prompt
Put them together: what is the integral of the reciprocal from 1 to e?
Hint: The antiderivative was the one exception to the power rule.
Answer:
The antiderivative of the reciprocal is the natural logarithm of the absolute value, so the integral is the logarithm at e minus at 1.
\[ \int_{1}^{e}\frac{dx}{x} = \ln|x|\Big|_{1}^{e} = 1 - 0 = 1 \]
That is the whole method, and it is startlingly quick. This section's work is not the evaluation but two other things: rewriting integrands the short table does not cover, and reading the theorem in the language of rates so it answers questions about accumulated quantities rather than areas.
Concept
The evaluation formula, read with the integrand as a derivative, says that integrating a rate of change over an interval recovers the net change in the quantity. Rates are what instruments measure; totals are what is wanted.
Net Change Theorem — The definite integral of a rate of change over an interval equals the net change in the quantity over that interval. It is the Fundamental Theorem's Part 2 with the integrand read as a derivative.
\[ \int_{a}^{b}Q'(t)\,dt = Q(b)-Q(a) \]
Nothing new is being proved. The value of the reading is that it turns a geometric theorem into a tool for mechanics, hydraulics, economics and any other field where a quantity and its rate both appear.
Figure (svg): The Net Change Theorem: a rate integrated gives the quantity's total change
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 514-522
Section
Section 1
Concept
Every formula of Section 4.10 becomes a definite-integral method when combined with Part 2. The constant is dropped, and the answer's sign should be checked against the integrand.
evaluation notation — A vertical bar with the limits written beside it, meaning the expression evaluated at the upper limit minus its value at the lower.
\[ \int_{a}^{b}f = F(x)\Big|_{a}^{b} = F(b)-F(a) \]
The table covers powers, the reciprocal, the exponential and the standard trigonometric functions. It covers no products, no quotients and no compositions, which is what makes rewriting the central skill.
Figure (svg): The integration formulas available so far, and their reach
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 514-521 — basic integration formulas
Picture it
Seven entries, and what they do not cover.
Figure (svg): The integration formulas available so far, and their reach
The final line is the honest limitation. Until Section 5.5 introduces substitution, any integrand outside this table must be rewritten into it or left alone.
Worked example
Example 5.30. Split, then evaluate each piece.
\[ \text{Evaluate } \int_{1}^{4}\left(3\sqrt{x}-\frac{2}{x}\right)dx. \]
Rewrite as powers
Why: The radical becomes an exponent.
\[ 3 x ^{\frac{1}{2}} - \frac{2}{x} \]
Antidifferentiate the first term
Why: Raise and divide.
\[ 2 x ^{\frac{3}{2}} \]
Antidifferentiate the second
Why: The logarithm exception.
\[ -2 \ln | x | \]
Evaluate at 4
Why: Substitute.
\[ 16 - 2 \ln 4 \]
Evaluate at 1 and subtract
Why: Substitute and subtract.
\[ 16 - 2 \ln 4 - 2 \]
Figure (svg): The integration formulas available so far, and their reach
\[ 14-2\ln 4 \approx 11.23 \]
Verify: sanity-check the size against a bounding rectangle
Why: The integrand at x equal to 1 is 1 and at x equal to 4 is about 5.5, so over a width of 3 the answer should be somewhere between 3 and 17 — and 11.23 sits comfortably inside. Note that both formulas were needed and the second was the power rule's exception; writing the reciprocal as x to the negative one and applying the power rule would have divided by zero.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 517-518
Fill the middle
An antiderivative, evaluated and subtracted.
Fill in the blanks
\ln|x|\Big|_0^___ = 1 - ___
Why: The logarithm of 1 is zero, so the integral is exactly 1. That the reciprocal accumulates exactly one unit of area from 1 to e is what defines e in this framing.
Worked example
Checkpoint 5.30. The integrand as given fits nothing.
\[ \text{Evaluate } \int_{1}^{2}\frac{x^{2}+1}{x}\,dx. \]
Note there is no quotient rule to reverse
Why: The table has no entry.
Divide term by term
Why: Split the fraction.
\[ x + \frac{1}{x} \]
Antidifferentiate
Why: Power rule and logarithm.
\[ x ^{2} / 2 + \ln | x | \]
Evaluate at 2
Why: Substitute.
\[ 2 + \ln 2 \]
Evaluate at 1 and subtract
Why: The logarithm of 1 is zero.
\[ 2 + \ln 2 - \frac{1}{2} \]
Figure (svg): Rewriting an integrand into a form the table covers
\[ \frac32+\ln 2 \approx 2.193 \]
Verify: check the rewrite recombines correctly
Why: Adding x and one over x over a common denominator gives x squared plus one over x, the original integrand — so nothing was lost. The rewrite is the whole difficulty here: the integral is trivial once the fraction is split, and impossible with the current table if it is not. Recognising which rewrite to attempt is the skill worth building, and dividing term by term is the commonest one.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 518-520
Trap
\[ \int\frac{x^{2}+1}{x}dx = \frac{\int(x^{2}+1)dx}{\int x\,dx} \]
Integrate the top and bottom separately
Why: The student invents a quotient rule.
No such rule exists. Differentiation does not turn quotients into quotients, so there is nothing to reverse.
\[ = \int\left(x+\frac1x\right)dx = \frac{x^{2}}{2}+\ln|x|+C \]
Rewrite the quotient into terms the table covers
Why: Dividing term by term is legitimate algebra.
The same applies to products: there is no product rule for integrals either. Until Section 5.5, the only route past a product or quotient is to rewrite it into a sum of things the table handles.
Sorting
Check whether the integrand appears as written.
Sort into buckets
Sort each integrand.
The fourth could also be handled by substitution once Section 5.5 arrives, but expanding works now and is quicker for a square. Rewriting is not a stopgap — it often remains the fastest route even after better tools exist.
Two truths and a lie
All three are about evaluation.
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. Differentiation turns a quotient into something with a squared denominator, not into a quotient of derivatives, so there is nothing of that shape to reverse. Dividing term by term is the legitimate move, and it works whenever the denominator is a single term.
Prediction
Commit before reasoning.
Predict first
For an integrand that is a fraction with a single term in the denominator, what is the standard rewrite?
Correct: Divide each term of the numerator by the denominator.
\[ \frac{x^{2}+1}{x} = x+\frac1x \]
Why: Splitting the fraction turns one unmanageable quotient into a sum of powers, every one of which the table handles. It works whenever the denominator is a single term, which covers a large fraction of the quotients met at this stage. For a denominator with several terms the technique fails and substitution or partial fractions is needed later.
Section
Section 2
Concept
Reading the integrand as a rate of change makes the evaluation formula a statement about accumulation: integrating the rate over an interval gives the quantity's net change over it.
net change — The difference between a quantity's final and initial values. It equals the integral of the quantity's rate of change over the interval, whatever the quantity measures.
\[ Q(b) = Q(a) + \int_{a}^{b}Q'(t)\,dt \]
The second form is the useful one in practice: a known starting value plus the accumulated change gives the final value. That is how a fuel gauge, a bank balance and a position are all computed from their rates.
Figure (svg): The Net Change Theorem: a rate integrated gives the quantity's total change
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 518-526 — the Net Change Theorem
Picture it
Four pairs, one theorem.
Figure (svg): The Net Change Theorem: a rate integrated gives the quantity's total change
The left column lists things that can be measured continuously; the right lists things that generally cannot. That asymmetry is exactly why the theorem is useful.
Worked example
Example 5.32. Water flowing into a tank.
\[ \text{Water flows in at } r(t)=3t^{2}+2 \text{ litres per minute. How much enters over the first } 4 \text{ minutes?} \]
Recognise the reading
Why: The rate is the volume's derivative.
Set up the integral
Why: Over the given interval.
\[ \int\text{ from } 0\text{ to } 4\text{ of } (3 t ^{2} + 2) \]
Antidifferentiate
Why: Term by term.
\[ t ^{3} + 2 t \]
Evaluate at 4
Why: Substitute.
\[ 64 + 8 = 72 \]
Evaluate at 0 and subtract
Why: Zero.
\[ 72\text{ litres} \]
Figure (svg): The Net Change Theorem: a rate integrated gives the quantity's total change
\[ \int_{0}^{4}(3t^{2}+2)\,dt = 72 \]
Verify: sanity-check against the rate's range
Why: The flow starts at 2 litres per minute and reaches 50 by the fourth minute, so over 4 minutes the total must lie between 8 and 200 — and 72 is comfortably inside, nearer the low end because the rate spends most of the interval well below its final value. Note the answer is a volume although the integrand was a rate: the units multiply, litres per minute times minutes giving litres, which is a useful check in itself.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 521-522
Fill the middle
A tank with a known starting volume and a computed change.
Fill in the blanks
V(5) = 50 + (-25) = 25
Why: The integral gives only the change, so the initial amount must be added. Forgetting it produces a negative volume, which is the signal that something has been left out.
Worked example
Checkpoint 5.32. Accumulation added to an initial amount.
\[ \text{A tank holds } 50 \text{ litres at } t=0 \text{ and drains at } r(t)=-2t \text{ litres per minute. Find the volume at } t=5. \]
Write the accumulation form
Why: Initial plus the integral.
\[ V(5) = 50 + \int\text{ from } 0\text{ to } 5\text{ of } (-2 t) \]
Antidifferentiate
Why: The power rule.
\[ -t ^{2} \]
Evaluate at 5
Why: Substitute.
\[ -25 \]
Evaluate at 0 and subtract
Why: Zero.
\[ -25\text{ litres of change} \]
Add to the initial amount
Why: Fifty minus twenty-five.
\[ 25\text{ litres} \]
Figure (svg): The solution to Worked example a final value from a starting value shown as a ladder of expressions, one row per legal move
\[ V(5) = 50 + (-25) = 25 \]
Verify: check the sign and consider what happens later
Why: The negative integral correctly reflects draining, and 25 litres is a plausible remainder. Worth noting: the rate keeps increasing in magnitude, so by t equal to about 7.07 the accumulated loss reaches 50 and the tank is empty — after which the model stops being physical. Checking when a model breaks down is part of using it, and integrating past that point would give a negative volume.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 522-524
Error analysis
A student computes a tank's volume from its drain rate.
Annotate
On: \( V(5) = \int_{0}^{5}(-2t)\,dt = -25 \text{ litres} \)
The theorem gives a net change, never an absolute amount. Whenever the question asks how much there IS rather than how much it changed, an initial value is needed and must come from the problem's data.
Matching
The theorem across fields.
Match the pairs
Why: In every case the time units cancel and the quantity's own units survive, which is a reliable check on whether an integral has been set up correctly. If the answer's units are wrong, the integrand or the variable is.
Ranking
Finding a quantity's value from its rate.
Put in order
Why: Step d is the one skipped, and it is invisible in the arithmetic — the integral looks like a complete answer. Step e catches it: an amount that comes out negative, or units that do not match the question, means something upstream is wrong.
Prediction
Commit before reasoning.
Predict first
Why does the net change reading matter more in practice than the area reading?
Correct: Because instruments measure rates while totals often cannot be measured directly.
\[ \text{measured rate} \;\xrightarrow{\;\int\;}\; \text{wanted total} \]
Why: A flow meter reads litres per minute, a speedometer reads metres per second, a wattmeter reads power — all instantaneous rates. The corresponding totals, volume delivered and distance travelled and energy used, generally have no direct sensor. The theorem converts what can be measured into what is wanted, and that conversion is why the reading appears in every applied field rather than only in geometry.
Section
Section 3
Concept
Integrating velocity gives displacement, in which backward motion cancels forward motion. Total distance travelled requires integrating the speed, which means splitting where the velocity changes sign.
displacement and distance — Displacement is the net change in position, given by the integral of velocity. Distance travelled is the accumulated path length, given by the integral of the velocity's absolute value.
\[ \text{disp} = \int_{a}^{b}v\,dt, \qquad \text{dist} = \int_{a}^{b}|v|\,dt \]
This is Section 5.2's signed-against-total distinction in its most concrete form. The two answers can differ by any amount, and for a particle returning to its start the displacement is zero however far it went.
Figure (svg): Displacement against distance: the same velocity curve, two questions
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 522-530 — displacement and distance travelled
Picture it
A velocity that changes sign.
Figure (svg): Displacement against distance: the same velocity curve, two questions
The red region subtracts from displacement and adds to distance. Splitting at the crossing is the only way to get the second, and the two answers here differ by more than a factor of two.
Worked example
Example 5.34. Split where the sign changes.
\[ \text{For } v(t)=t^{2}-4 \text{ on } [0,3], \text{ find the displacement and the distance travelled.} \]
Compute the displacement
Why: One integral.
\[ t ^{3} / 3 - 4 t\text{ from } 0\text{ to } 3 = -3 \]
Find where the velocity vanishes
Why: Set it to zero.
\[ t = 2 \]
Integrate over the first piece
Why: Where the velocity is negative.
\[ -\frac{16}{3} \]
Integrate over the second
Why: Where it is positive.
\[ \frac{7}{3} \]
Add magnitudes
Why: For the distance.
\[ \frac{16}{3} + \frac{7}{3} = \frac{23}{3} \]
Figure (svg): Displacement against distance: the same velocity curve, two questions
\[ \text{disp} = -3, \qquad \text{dist} = \tfrac{23}{3} \approx 7.67 \]
Verify: reconcile the two answers with the motion
Why: The particle moved 16 over 3 metres backwards in the first two seconds, then 7 over 3 forwards in the last one — ending 3 metres behind its start having covered 23 over 3. The two answers differ because the return trip only partly undid the outward one. Note the displacement could have been computed without any split, but the distance could not: the split at t equal to 2 is the whole difficulty of the second question.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 525-526
Sorting
Signed, or absolute value inside?
Sort into buckets
Sort each question.
The fifth is worth care: distance FROM the start is the magnitude of the displacement, which is not the distance travelled. Three quantities, easily confused, and only careful reading of the question separates them.
Worked example
Checkpoint 5.34. Zero displacement, substantial distance.
\[ \text{A particle has } v(t)=\sin t \text{ on } [0,2\pi]. \text{ Find both quantities.} \]
Compute the displacement
Why: One integral.
\[ -\cos t\text{ from } 0\text{ to } 2 \pi = 0 \]
Interpret
Why: It returned to its start.
Find where the velocity vanishes
Why: Inside the interval.
\[ t = \pi \]
Integrate each piece
Why: Forward then backward.
\[ 2\text{ and } -2 \]
Add magnitudes
Why: For the distance.
\[ 4 \]
Figure (svg): The solution to Worked example a round trip shown as a ladder of expressions, one row per legal move
\[ \text{disp}=0, \qquad \text{dist}=4 \]
Verify: see why zero displacement is not zero motion
Why: The particle travelled 2 units forward and 2 back, ending exactly where it began — so the displacement is genuinely zero and the distance genuinely 4. This is the clearest case for keeping the two apart: no amount of motion produces displacement if it is undone. A runner completing a lap has zero displacement and a very definite distance, and the same integral answers only the first.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 526-528
Trap
\[ \text{dist} = \left|\int_{0}^{3}(t^{2}-4)\,dt\right| = |-3| = 3 \]
Take the magnitude of the displacement
Why: The student applies the absolute value outside.
The true distance is about 7.67. Taking the magnitude at the end does not undo cancellation that already happened inside the integral.
\[ \text{dist} = \int_{0}^{3}|t^{2}-4|\,dt = \tfrac{16}{3}+\tfrac73 = \tfrac{23}{3} \]
Put the absolute value INSIDE, which means splitting at the zeros
Why: The cancellation must be prevented, not repaired.
The order matters entirely. Inside the integral the absolute value stops opposite contributions cancelling; outside it merely makes an already-cancelled answer positive.
Fill the middle
A velocity that changes sign inside the interval.
Fill in the blanks
v(t)=t^2-4=0 \;\Longrightarrow\; t=___ \text___ [0,3]
Why: The velocity changes sign at t equal to 2, so the distance calculation splits there. Without the split the backward motion cancels the forward motion and the answer is a displacement.
Two truths and a lie
All three are about motion.
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. Cancellation happens inside the integral, and an absolute value applied afterwards cannot undo it — for the worked example it gives 3 rather than the true 7.67. The absolute value must be inside, which in practice means splitting first.
Prediction
Commit before reasoning.
Predict first
Why can the absolute value not simply be applied to the finished integral?
Correct: Because the cancellation happens during the integration.
\[ \left|\int f\right| \le \int|f|, \quad \text{with equality only if } f \text{ keeps one sign} \]
Why: By the time the integral has been evaluated, the backward contribution has already subtracted from the forward one and the information about how much of each there was is gone. Taking a magnitude afterwards only fixes the sign of what remains. Splitting at the zeros keeps the two contributions separate so both can be counted positively, which is what the absolute value inside the integral means.
Section
Section 4
Concept
Over an interval symmetric about zero, an even function's integral is twice the integral over the right half, and an odd function's integral is exactly zero.
even and odd functions — An even function is unchanged by replacing the input with its negative; an odd function's output changes sign. Over a symmetric interval the first doubles and the second cancels.
\[ \int_{-a}^{a}f_{\text{even}} = 2\int_{0}^{a}f, \qquad \int_{-a}^{a}f_{\text{odd}} = 0 \]
Checking symmetry costs one substitution and can eliminate the whole computation. It is worth doing before any integral over an interval symmetric about zero.
Figure (svg): Even and odd functions integrated over a symmetric interval
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 528-535 — integrating even and odd functions
Picture it
An even and an odd function over a symmetric interval.
Figure (svg): Even and odd functions integrated over a symmetric interval
The even function's halves are mirror images with the same sign and add; the odd function's are mirror images with opposite signs and cancel. The pictures make both statements obvious.
Worked example
Example 5.36. No computation required.
\[ \text{Evaluate } \int_{-3}^{3}\left(x^{5}-4x^{3}+x\right)dx. \]
Test for symmetry
Why: Replace x with its negative.
Conclude the function is odd
Why: All exponents are odd.
\[ f(-x) = -f(x) \]
Check the interval is symmetric
Why: From -3 to 3.
Apply the shortcut
Why: Odd over symmetric.
Note what was avoided
Why: A degree-six antiderivative.
Figure (svg): Even and odd functions integrated over a symmetric interval
\[ \int_{-3}^{3}\left(x^{5}-4x^{3}+x\right)dx = 0 \]
Verify: confirm by computing it the long way
Why: The antiderivative is x to the sixth over 6, minus x to the fourth, plus x squared over 2 — every term an even power. Evaluating at 3 and at negative 3 gives identical values, so the subtraction is zero. The shortcut reached the same answer without writing any of that, which is the point: for a symmetric interval, checking the parity first can remove the problem entirely.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 530-531
Sorting
Replace the input with its negative.
Sort into buckets
Sort each function.
The mixed case is not a dead end: split it into its even and odd parts and apply each shortcut separately. Only the even part survives over a symmetric interval.
Worked example
Checkpoint 5.36. Half the work.
\[ \text{Evaluate } \int_{-2}^{2}\left(x^{4}+3x^{2}\right)dx. \]
Test for symmetry
Why: All exponents even.
\[ f(-x) = f(x) \]
Apply the shortcut
Why: Twice the right half.
\[ 2 \int\text{ from } 0\text{ to } 2 \]
Antidifferentiate
Why: Term by term.
\[ x ^{5} / 5 + x ^{3} \]
Evaluate at 2
Why: Substitute.
\[ \frac{32}{5} + 8 \]
Double
Why: The shortcut's factor.
\[ 2(\frac{32}{5} + 8) = \frac{144}{5} \]
Figure (svg): The solution to Worked example an even integrand halved shown as a ladder of expressions, one row per legal move
\[ \int_{-2}^{2}(x^{4}+3x^{2})dx = \frac{144}{5} \]
Verify: check against the direct computation
Why: Evaluating the antiderivative at 2 gives 32 over 5 plus 8, and at negative 2 gives its negative — so subtracting doubles it, matching. The saving here is modest because the antiderivative was easy, but for a harder integrand it halves genuine work, and the substitution to test parity costs almost nothing. Mixed parity, such as x cubed plus x squared, gets no shortcut at all and must be done directly.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 531-533
Error analysis
A student integrates over a symmetric interval.
Annotate
On: \( \int_{-1}^{1}\left(x^{3}+x^{2}\right)dx = 0 \quad \text{(claimed odd)} \)
The right move is to split: apply the odd shortcut to the odd terms and the even shortcut to the even ones. Applying either to a mixed function discards a genuine contribution.
Fill the middle
An even integrand over a symmetric interval.
Fill in the blanks
\int_2^___f = ___\int____^___f \quad (f \text___)
Why: The halves are mirror images with the same sign, so computing one and doubling gives the whole. For an odd function they have opposite signs and the integral is zero.
Ranking
An integral over an interval symmetric about zero.
Put in order
Why: Step a is the trigger: without a symmetric interval no shortcut applies whatever the function's parity. The whole check costs one substitution and occasionally removes the entire computation.
Prediction
Commit before reasoning.
Predict first
Why is an odd function's integral zero over a symmetric interval?
Correct: Because the halves have equal magnitude and opposite sign.
\[ \int_{-a}^{a}x^{3}dx = 0 \quad \text{but} \quad \int_{-a}^{a}|x^{3}|dx = \frac{a^{4}}{2} \]
Why: Reflecting through the origin maps the left half onto the right half upside down, so wherever the function is positive on one side it is equally negative on the other. The signed areas cancel exactly. Note this is a statement about signed area — the TOTAL area is twice one half and is not zero, which is the same distinction as everywhere else in this chapter.
Section
Section 5
Concept
A quantity accumulating from a rate is exactly the area function of Section 5.3: its value at any time is the area so far, and its rate of growth is the current rate.
accumulation function — The quantity's value as a function of time, equal to its initial value plus the integral of the rate from the start to that time.
\[ Q(x) = Q(a) + \int_{a}^{x}Q'(t)\,dt \]
Reading the two pictures together is the skill: the total curve is steepest where the rate curve is highest, flat where the rate is zero, and falling where the rate is negative.
Figure (svg): Accumulation: a rate curve and the running total it produces
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 524-535 — applications of the Net Change Theorem
Picture it
The same information, drawn two ways.
Figure (svg): Accumulation: a rate curve and the running total it produces
The shaded area on the left is the height on the right, and the right curve's steepness at any point is the left curve's height there. That is exactly the two parts of Section 5.3's theorem, drawn.
Worked example
Example 5.38. Qualitative reading before any computation.
\[ \text{A rate is positive, falls to zero at } t=3, \text{ then goes negative. Describe the total.} \]
Where the rate is positive
Why: The total accumulates.
Where the rate is largest
Why: The total grows fastest.
At the rate's zero
Why: No accumulation.
Identify what kind of point that is
Why: Growth stops and reverses.
Where the rate is negative
Why: Accumulation reverses.
Figure (svg): Accumulation: a rate curve and the running total it produces
\[ Q \text{ increases, has a maximum at } t=3, \text{ then decreases} \]
Verify: connect this to Chapter 4's language
Why: The rate IS the total's derivative, so this is exactly Section 4.5's first derivative test: positive derivative means increasing, a sign change from positive to negative at t equal to 3 means a local maximum. Nothing new is being used — Chapter 4's analysis of a function from its derivative applies verbatim to an accumulated quantity and its rate. Reading a rate graph qualitatively is often faster and more informative than computing the integral.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 528-530
Matching
The rate is the total's derivative.
Match the pairs
Why: Every row is Section 4.5 applied to an accumulated quantity. The third is the one most often misread: a peak in the rate is where the total climbs fastest, not where it is highest.
Worked example
Checkpoint 5.38. Net and total from one rate.
\[ \text{Traffic crosses a bridge at } r(t)=100-25t \text{ cars per hour, with negative meaning the other way. Find the net over } [0,6]. \]
Find where the rate changes sign
Why: Set it to zero.
\[ t = 4 \]
Antidifferentiate
Why: Term by term.
\[ 100 t - 12.5 t ^{2} \]
Evaluate at 6
Why: Substitute.
\[ 600 - 450 = 150 \]
Evaluate at 0 and subtract
Why: Zero.
\[ 150\text{ cars net} \]
For the total crossing, split at 4
Why: Add magnitudes.
\[ 200 + 50 = 250 \]
Figure (svg): The solution to Worked example an accumulation with a changing sign shown as a ladder of expressions, one row per legal move
\[ \text{net} = 150, \qquad \text{total} = 250 \]
Verify: interpret both numbers physically
Why: Two hundred cars crossed one way in the first four hours and fifty crossed back in the last two, so the net is 150 and the total number of crossings is 250. The bridge's engineers care about the total, since every crossing loads the structure; a traffic planner counting where cars ended up cares about the net. Both come from the same rate function and the same theorem, and asking which is wanted remains the first step.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 530-532
Trap
\[ \text{the rate is highest at } t=1 \;\Longrightarrow\; \text{the total is highest there} \]
Read the rate curve's peak as the total's peak
Why: The student conflates the two graphs.
The total is still growing at t equal to 1 — it grows fastest there. It peaks where the RATE crosses zero, at t equal to 3.
\[ \text{total peaks where the rate is zero and changing sign} \]
Read the rate as the total's derivative and apply Chapter 4
Why: A peak in the derivative is an inflection in the function, not a maximum.
This is exactly the confusion Section 4.5 warned about between a function and its derivative, appearing again in applied clothing. The rate's maximum marks where the total is steepest, which is an inflection point of the total.
Fill the middle
A rate that crosses zero from positive to negative.
Fill in the blanks
r(t)=100-25t=0 \;\Longrightarrow\; \text4 t=___
Why: The total stops growing when the rate hits zero and starts falling once the rate turns negative, so t equal to 4 is a maximum. This is the first derivative test with the rate playing the derivative's role.
Sorting
Net, or total?
Sort into buckets
Sort each question about the bridge traffic.
The engineer and the planner want different numbers from the same data, and neither is more correct. Identifying which one the situation calls for is the modelling step, and it comes before any integration.
Prediction
Commit before reasoning.
Predict first
A rate rises to a peak at t = 1, falls to zero at t = 3, then goes negative. When is the accumulated total largest?
Correct: At t = 3.
\[ Q' > 0 \text{ until } t=3, \; Q' < 0 \text{ after} \;\Longrightarrow\; \text{max at } 3 \]
Why: The total grows for as long as the rate is positive, which is until t equal to 3, and shrinks afterwards. At t equal to 1 it is growing fastest, which makes that an inflection point of the total rather than a maximum. Confusing a peak in the derivative with a peak in the function is the same error Section 4.5 identified, and it is the commonest misreading of a rate graph.
Comparison
Fill the blanks. The same rate answers both, differently.
Comparison matrix
| Net | Total | |
|---|---|---|
| Integrand | the rate itself | the rate's absolute value |
| Method | one integral | split at the rate's zeros, add magnitudes |
| For motion | displacement | distance travelled |
| A round trip gives | zero | the full path length |
The last row is the clearest test of whether the distinction has landed. A quantity that returns to where it started has zero net change and a total that can be arbitrarily large.
Pattern
Given a rate and a question about the quantity.
Steps one and five are where applied problems go wrong, and neither is visible in the arithmetic. An amount that comes out negative, or units that do not match the question, means one of them was skipped.
Stewart, Calculus: Early Transcendentals 8e, §5.4 Indefinite Integrals and the Net Change Theorem §5.4, pp. 402-411
Check
Net change.
Check your understanding
Water flows in at 3t^2 + 2 litres per minute. How much enters over the first 4 minutes?
Answer: A
Why: The antiderivative is t^3 + 2t, giving 64 + 8 at t = 4.
Check
Distance.
Check your understanding
For v(t) = t^2 - 4 on [0,3], the displacement is -3. What is the distance travelled?
Answer: A
Why: Splitting at t = 2 gives 16/3 backwards and 7/3 forwards, totalling 23/3.
Check
Symmetry.
Check your understanding
What is the integral of x^5 - 4x^3 + x from -3 to 3?
Answer: A
Why: Every exponent is odd, so the halves cancel over the symmetric interval.
Real world
A domestic battery paired with solar panels charges when generation exceeds household demand and discharges when it does not. A meter records the net power to the battery every minute across a day, positive when charging.
Discussion prompt
Explain what the plain integral gives, what the battery's owner also needs, and what determines when the charge is highest.
Hint: Charging and discharging are opposite signs of the same rate.
Answer:
The plain integral over the day gives the net change in stored energy — how much fuller the battery is at midnight than it was at the start. Added to the morning's charge, it gives the evening's.
\[ E(\text{end}) = E(\text{start}) + \int_{0}^{24}P(t)\,dt \]
The total energy cycled is a different number, found by splitting at every sign change and adding magnitudes. It matters because battery lifetime is measured in cycles: a battery that ends the day where it began may have charged and discharged several times over, and that wear is invisible in the net figure.
The charge is highest where the net power crosses zero from positive to negative — typically mid-afternoon, when generation falls below demand. Note that this is not when generation peaks, which is usually around noon: at noon the battery is filling fastest, not fullest. That is the rate-against-total confusion in a form that costs money if a control system is tuned on the wrong one.
Note the practical asymmetry that makes all of this necessary: the meter reads power, an instantaneous rate, because that is what can be measured directly. Stored energy has no equivalent sensor and must be inferred by integrating — which is the Net Change Theorem doing the only job available.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does taking the magnitude of a displacement not give the distance travelled?
Correct: Because the cancellation happened inside the integral.
\[ \left|\int_{0}^{3}(t^{2}-4)dt\right| = 3 \quad \text{but} \quad \int_{0}^{3}|t^{2}-4|\,dt = \tfrac{23}{3} \]
Why: Once the integral is evaluated, backward motion has already subtracted from forward motion and the record of how much of each there was is gone — taking a magnitude afterwards only fixes the sign of what survived. The absolute value must be inside, which in practice means splitting at the velocity's zeros. For the worked example the two answers were 3 and about 7.67, so the difference is not a technicality.
Explain it
They computed a tank's volume from its drain rate and got a negative number.
Discussion prompt
In four sentences or fewer, show them what is missing.
Hint: Ask what the integral actually computed.
Answer:
Ask them what a volume of negative 25 litres would mean — it cannot mean anything, so something is missing. The integral gave the CHANGE in volume, which is genuinely negative because the tank was draining.
The volume itself needs the starting amount added: 50 litres minus 25 gives 25 remaining. The theorem always produces a change, never an amount, so whenever the question asks how much there is rather than how much it changed, an initial value has to come from the problem.
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 rewriting, divide fractions term by term and expand brackets before looking for a formula. For net change, ask whether the question wants an amount or a change and add the initial value if the former. For motion, split at the velocity's zeros whenever a distance is wanted. For symmetry, substitute the negative input and compare. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the Net Change Theorem in a box, and beside it list four rate-and-total pairs from different fields with their units. Below, sketch a velocity curve that goes negative then positive, shade the two regions differently, and write the displacement and the distance travelled beneath with one line saying why they differ by more than a factor of two. In the middle of the page, draw an even function and an odd function over a symmetric interval and write each shortcut beside its picture, then add a mixed example and say what happens to each part. In the lower half, draw a rate curve and its running total side by side, with dashed lines connecting the rate's zero to the total's peak and the rate's peak to the total's steepest point. At the bottom, list the four standard rewrites that bring an integrand into the basic table, and write one sentence on what the table does not cover.
If your rate and total graphs put their peaks at the same place, look again — the total peaks where the rate crosses zero, and getting those two positions visibly different is the whole point of drawing them together.
Recap
Five things, and two of them are about asking the right question before computing.
| If you see | Then |
|---|---|
| A quotient with one term below | Divide term by term |
| A rate integrated | The answer is a net change |
| A question asking how much there is | Add the initial value |
| A distance wanted | Split at the velocity's zeros |
| A symmetric interval | Check the integrand's parity first |
| An odd integrand there | The answer is zero, with no work |
| A rate's peak | The total's steepest point, not its highest |
Section 5.5 removes the table's biggest limitation. Substitution reverses the chain rule, which brings compositions within reach and turns most of the integrals this section had to sidestep into routine ones.
OpenStax Calculus Volume 1, §5.4 Integration Formulas and the Net Change Theorem §5.4, pp. 488-505 — everything on these slides traces back here
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