A rate graph can determine change even when no single formula is supplied. Let g be the continuous piecewise-linear graph through (0,0), (2,2), (4,−2) and (6,0). Define H(x)=3+∫₀ˣg(t)dt for inputs x in [0,6]. Here t is a temporary integration variable, and the initial value H(0)=3 fixes the vertical position that a derivative alone cannot determine.
If g measures a rate in liters per minute and the horizontal coordinate is minutes, an area under this graph has units of liters. Positive area adds to H; negative area subtracts. The same mathematics applies without a physical interpretation. Sketch the three line segments and mark their crossings of the horizontal axis.
Predict
Predict whether H reaches its largest value at x=2, 3 or 4. The rate is greatest at 2, but does accumulation immediately stop increasing there? Also predict whether H(6) is less than, equal to or greater than its starting value. State which signed regions you expect to cancel before using the widget.
Experiment
Build accumulation from a supplied rate graph
This is a supplied exact teaching graph, not measured samples. Predict from its geometry before revealing calculations. Open circles exclude a point; filled circles include it. Lines and coordinates below define the entire graph.
Equivalent text: straight segments join (0,0), (2,2), (4,−2), (6,0), with all endpoints included. The horizontal axis is x and the vertical axis is g(x). No values outside [0,6] are supplied. Define H(x)=3+∫₀ˣg(t)dt. The initial value is H(0)=3.
Changing the probe hides old results. This temporary explorer saves no answers or completion credit. Record the geometric reasoning in the lesson practice.
Inspect the original rate graph and the accumulated result together. Treat the segments as exact mathematical data. The triangle from 0 to 2 has base 2 and height 2, so its signed area is (1/2)·2·2=2. On the middle segment, the rate reaches zero at 3. The two small triangles on either side have areas +1 and −1. The final triangle has area −2.
Interval added
Signed area
Accumulated endpoint value
[0,2]
+2
H(2)=3+2=5
[2,3]
+1/2·1·2=+1
H(3)=5+1=6
[3,4]
−1/2·1·2=−1
H(4)=6−1=5
[4,6]
−2
H(6)=5−2=3
Check the middle triangles carefully: each has height 2, not 1. A midpoint horizontal coordinate is not a triangle’s vertical height. Thus the exact maximum value is 6 for this stipulated graph. Explain which base and height you used before comparing the answer with your prediction.
Compute H(1) independently. On [0,1], g rises from 0 to 1, so the area is 1/2 and H(1)=7/2. This is a fresh checkpoint between the listed vertices; it does not rely on interpolating H linearly.
A supplied circular arc, not measured samples
Now use a different, exact graph q. On [0,4] it is the upper semicircle of radius 2 centered at (2,0); from (4,0) to (6,−2) it is a straight segment. Define A(x)=1+∫₀ˣq(t)dt on [0,6]. The circular arc and segment specify every intermediate value. Before reading the calculation, shade the positive and negative regions and predict where A is largest.
Exact circular and linear pieces on equal-scale axes. The dashed radius is 2. The initial value A(0)=1 belongs to the accumulation, not to this rate graph.
The positive area is half a disk: (1/2)π·2²=2π. The negative triangle has base 2 and height 2, so its signed area is −2. A quarter disk lies between 0 and 2. No inverse-trigonometric integration is needed to obtain these checkpoints.
Input x
Signed area from 0 to x
A(x)
0
0
1
2
π
1+π
4
2π
1+2π
6
2π−2
2π−1
The net change is 2π−2, whereas the integral of the rate’s absolute value is 2π+2. If the horizontal coordinate is seconds and q is velocity in meters per second, these are displacement and distance in meters, respectively. A radius on this mathematical graph is a coordinate length used to calculate area; the resulting integral carries the product of the two axis units.
Understand
The fundamental theorem of calculus applies because g is continuous. It gives H′(x)=g(x) for interior inputs. Therefore H increases up to 3 and decreases after 3. Continuity on the closed interval permits comparison with endpoints, so the global maximum is H(3)=6 and the minimum value is 3, attained at both endpoints. Endpoint derivatives, if discussed, are inward one-sided derivatives; an ordinary two-sided derivative is not asserted beyond the domain.
Do not join H’s listed values by straight lines. Integrating a linear rate gives a quadratic accumulation on each segment. For example, g(t)=t on [0,2], hence H(x)=3+x²/2 there. On [2,4], g(t)=6−2t. Writing u=x−2 gives H(x)=5+2u−u². Substituting u=1 recovers 6, while u=2 gives 5. This also shows why a simple average of H(2) and H(4) misses the peak.
Signed area measures net change: ∫₀⁶g(t)dt=0. Total accumulated magnitude instead uses ∫₀⁶∣g(t)∣dt=2+1+1+2=6. If g is velocity, these are displacement 0 and distance 6. They answer different questions. Reversing integration bounds negates the signed integral, so ∫₄²g(t)dt=0 here as well, whereas ∫₃²g(t)dt=−1.
The slope of g is 1, then −2, then 1. Consequently H is concave up, then down, then up. H remains differentiable at 2 and 4 because g is continuous there, although H″ is undefined at those corners. These conclusions depend on the full supplied segments. If only four measured data points were available, straight-line interpolation would be a modeling assumption, not proof of the true accumulated quantity.
For the circular example, continuity of q gives A′=q at every interior input, including x=4. Thus A′(2)=2 and A′(4)=0. The vertical tangent of the circular arc at its endpoint does not prevent this FTC conclusion: A′ requires continuity of q, not differentiability of q. The derivative of q itself is not finite there, so this argument does not supply A″(4).
The sign of q makes A strictly increasing on [0,4] and strictly decreasing on [4,6]. Therefore its absolute maximum is 1+2π at 4. For the absolute minimum, compare both endpoints: A(0)=1 and A(6)=2π−1>1. The minimum is 1 at 0. The highest rate, at 2, is where accumulation is increasing fastest, not where accumulation is highest. This distinction persists for curved rate graphs.
Apply
Change the initial value from 3 to −2 while retaining the rate. Predict the new maximum, both endpoint values and total net change without repeating every integral. The entire accumulation shifts down by 5: the maximum becomes 1 and both endpoints become −2, while net change remains zero.
Write a response to “the rate graph encloses area 6, therefore H(6)=9.” Identify the confusion between magnitude and signed change, then give both quantities with their interpretations. Include an interval-by-interval calculation so a reader can reproduce your result.
Try a new graph before opening the feedback
Let p be the lower semicircle of radius 1 centered at (1,0) on [0,2], followed by straight segments through (2,0), (3,2) and (4,0). Define B(x)=2+∫₀ˣp(t)dt. Use this supplied graph, not the earlier widget settings.
Changing the sign, radius and initial value creates a new accumulation problem. The shaded shapes describe signed integrals; neither shaded region is a graph of B.
Find B(1), B(2), B(4), the net change and the total absolute integral. Determine B′(1) and B′(2), then locate the absolute minimum and maximum of B, giving both inputs and values. Finally, explain why a corner of p at 3 does not prevent B′(3) from existing. Draw a qualitative accumulation sketch that matches your monotonicity and endpoint values rather than joining the listed values by straight lines.
Compare your reasoning with the worked feedback
The first half of the lower semicircle contributes −π/4, so B(1)=2−π/4. The whole semicircle contributes −π/2, giving B(2)=2−π/2. The positive triangle contributes (1/2)·2·2=2; hence B(4)=4−π/2. Net change is 2−π/2 and the total absolute integral is 2+π/2. Adding the initial value to distance would answer neither the final value nor the net-change question.
FTC gives B′(1)=p(1)=−1 and B′(2)=p(2)=0. The rate is negative before 2 and positive after 2, so the absolute minimum is B(2)=2−π/2. Compare endpoints for the maximum: B(4)=4−π/2>2=B(0), because π<4. Thus the maximum occurs at 4. Continuity of p at its corner gives B′(3)=2, although p′(3) and B″(3) do not exist. An accumulation sketch must fall until 2, then rise, and finish higher than it started.
Repair the sign-to-extremum inference
Let a tank’s stipulated net rate be Q′(t)=2−t litres per time unit on [0,4], with Q(0)=5 litres. The signed area from zero to two is a positive triangle of area two, so Q(2)=7. From two to four the rate is negative and its signed area is −2, returning the amount to five. Adding the absolute areas would give four litres of total variation in the amount, not four litres of net gain.
The sign change establishes the maximum: Q increases before t=2 and decreases afterward. The zero of Q′ alone would not classify the point. Endpoint amounts must be considered for an absolute-extremum claim on a closed interval. The explicit accumulation Q(t)=5+2t−t²/2 independently confirms all three candidate values.
Use the time control to distinguish three quantities at the same input: rate, signed change and amount. They have different units and different signs. After t=2 a negative derivative means the tank is losing liquid, not that it contains a negative amount.
The changed case reverses the crossing: Q′(t)=t−1 on [0,3], with Q(0)=4. The zero at one is now a minimum, because the rate changes from negative to positive. Compare both endpoints for the maximum instead of copying the previous conclusion. For written review, retain setup, bounds, sign justification and interpreted units as separate criteria. The interactive calculation is a bounded public retry, not independently observed calculator proficiency.
Separate sign, area and extrema
Review criterion: Use the sign of net rate for change and accumulation for amounts; compare endpoint values for global claims.
Predict, then change one input
Net rate Q′(t) (litres/time unit)
0 (rounded)
Signed change Q(t)−Q(0) (litres)
2
Amount Q(t) (litres)
7
The rate changes from positive to negative here, producing the maximum amount.
Explain the changed case: A triangular rate graph
Q′(t)=2−t on 0≤t≤4, and Q(0)=5 litres. At what time and value is Q maximal?
Rate crosses zero at t=2.
Signed area from 0 to 2 is 2 litres.
Signed area from 2 to 4 is −2 litres.
Write a reason before selecting a claim. Feedback is public practice, not a proof grader.
Inputs and choices are temporary. Nothing is saved, no lesson completion or score is changed, and repeating a public case is not unseen assessment.
Continue
Use the practice case to assemble areas before calculating an accumulation. Review the integral calculus unit if signed areas or the fundamental theorem are unfamiliar. Keep the initial value, rate units, endpoint convention and interpolation assumptions beside your final conclusion.
Apply · explain · revise
Practice and reasoning
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
For the supplied graph and H(x)=3+∫₀ˣg(t)dt, find H(5).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Use the supplied graph g to evaluate ∫₄¹g(t)dt.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Find the absolute maximum and minimum of H on [0,6], giving both values and all locations. Explain why a zero of g alone would not complete this argument.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Compute ∫₀⁶g and ∫₀⁶|g| from the supplied graph. If g were velocity in m/s, interpret both and state what initial position would add.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Define K(x)=−2+∫₂ˣg(t)dt. Compare K with H, find K(4), and decide whether their extrema and concavity occur at the same inputs.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
For 0<x<√6 define B(x)=∫ₓ^(x²)g(t)dt. Derive B′ and find B′(2). Explain why the formula also handles x²<x.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
