AP Calculus AB

1. Average height is not average slope

Learning goal: Choose the right quantity and units before calculating.

Average value asks how high a function is across an interval. Imagine redistributing its area into a rectangle of the same width. The rectangle height is the average value. For continuous f on [a,b], this is (1/(b−a))∫ₐᵇ f(t) dt. The integral has function-units multiplied by time-units; division restores the original function-units.

Average rate of change is a different question: how much did the endpoints change per unit time? It is [f(b)−f(a)]/(b−a), the slope of a secant line. This ignores the path between the endpoints. A function that rises and falls back to its starting value has zero average rate even if its average value is large.

For f(t)=t² over [0,2], the integral is 8/3, average value is 4/3 and average rate is 2. These quantities are not interchangeable. On a context question, first write what f represents, then label the units of the requested result. The 2025 Chief Reader report flags using the derivative as the integrand and using endpoint change instead of average value.

Try a second case with an invented rate. Let f(t)=3t on [0,4]. The integral is the area of a triangle with base 4 and height 12, which is 24. The average value is 24/4 = 6, so if f is in metres per second the average speed is 6 m/s. The average rate of change of f is (12−0)/4 = 3, and in this setting it has units of metres per second per second. Same function, two different questions, two different answers.

When you read a problem, underline the key phrase. "Average value of f" calls for an integral divided by the length of the interval. "Average rate of change of f" calls for endpoint outputs divided by the length. Writing the units of the answer next to your result is a quick check that you used the right formula.

Worked example

Find the average amount f(t)=t² over [0,3].

1. The interval length is 3.
2. The accumulated area is ∫₀³ t² dt = [t³/3]₀³ = 9.
3. Divide area by width: 9/3 = 3 amount-units.
4. The average rate is instead (9−0)/3 = 3 amount-units/time. The numbers coincide here, but their meanings and units still differ.

Mental model: Average value = area / width. Average rate = endpoint change / width.

Common trap: A numerical coincidence does not make two quantities the same.

2. From a rate to a total

Learning goal: Build an accumulation function with an initial condition.

If r(t) measures litres per minute, ∫ₐᵇ r(t) dt measures litres gained minus litres lost. It is net change, not automatically the final volume. The final amount is initial amount plus net change. Keep these two terms visible in your setup.

An accumulation function A(x)=A(a)+∫ₐˣ r(t) dt describes the amount at every endpoint x. Where r is continuous, A′(x)=r(x). This lets the rate graph tell you whether the accumulated amount rises or falls. A positive rate means A increases; it does not mean A itself is positive.

Total movement differs from net movement. For velocity v, displacement is ∫v and distance is ∫|v|. Split at sign changes before using the absolute value. A negative signed area is not a negative distance.

Work an invented tank problem. Water enters a tank at r(t)=2t litres per minute for 0 ≤ t ≤ 3, and the tank starts with 10 litres. The net change is ∫₀³ 2t dt = [t²]₀³ = 9 litres. The amount at t=3 is 10 + 9 = 19 litres. If you forget the initial 10, your answer describes only the water added, not the water in the tank.

Now suppose the rate becomes negative after some time, meaning water drains out. The accumulated amount then decreases, but it may still be positive. To find when the tank is fullest, look for where the rate changes from positive to negative. A graph of r tells you the slope of the amount function, while areas under r tell you changes in the amount. Keep these two readings separate.

Worked example

A tank begins with 10 litres. Its net rate is r(t)=2t−2 litres/minute from t=0 to t=3. Find final volume.

1. Use V(3)=10+∫₀³(2t−2)dt.
2. The antiderivative is t²−2t.
3. Net change is (9−6)−0=3 litres.
4. Final volume is 13 litres. The tank initially loses volume then gains it; endpoint change is still positive.

Mental model: Final amount = starting amount + signed accumulated rate.

Common trap: Integrating a rate does not erase the initial condition.

3. Read derivatives from meaning

Learning goal: Separate sign, size and rate of change.

Three functions can appear in one problem: position, velocity and acceleration; volume, net flow and flow change; cost, marginal cost and marginal-cost change. Name the level before reasoning. A graph labelled f′ is not a graph of f.

If f′>0, f increases. If f′<0, f decreases. If f″>0, f′ increases and f is concave up. None of these alone tells you whether f is positive. A car moving left has negative velocity, but it can still be gaining speed.

For motion in one dimension, speed is |v|. Away from v=0, speed increases when v and acceleration have the same sign, and decreases when signs differ. Use both signs, not acceleration alone. At v=0 investigate nearby behavior rather than mechanically applying a sign product.

Use an invented position function s(t)=t²−4t, with t in seconds and s in metres. Then v(t)=2t−4 and a(t)=2. At t=1, v=−2 and a=2. The signs differ, so the object is slowing down while moving left. At t=2, v=0 and the object is momentarily at rest. At t=3, v=2 and a=2, the signs match, so the speed is increasing while moving right.

A common slip is to say the object speeds up because the acceleration is positive. Check the sign of the velocity first. Another is to read a graph of v as a graph of position. If the graph is labelled v, a point above the axis means positive velocity, and the slope of that graph is the acceleration. Writing the name of each graph at the top of your scratch work prevents these mix-ups.

Worked example

A particle has v(2)=−3 and a(2)=−2. What can you conclude?

1. Velocity is negative, so position is decreasing.
2. Velocity and acceleration share a sign, so |v| increases.
3. The particle moves left and speeds up. This does not locate it on the number line.

Mental model: Name the function and derivative level before using signs.

Common trap: "Positive acceleration" does not mean "speeding up".

4. Local versus absolute extrema

Learning goal: Use candidate points and endpoints to justify a global answer.

A local maximum describes nearby points. An absolute maximum on an interval compares every point in that interval. A sign change at a critical input can show a local extremum. To establish an absolute extremum, cover the whole interval with the sign analysis, including endpoint behavior, or compare values at all candidates.

For a continuous function on [a,b], the Extreme Value Theorem guarantees absolute extrema. Candidates are endpoints and interior critical points where f′=0 or f′ does not exist. Evaluate f at all candidates and compare the values. If a point is outside the interval, discard it; if f is not continuous, do not cite this guarantee without checking.

Be precise about the requested output. A question may ask for the maximum value f(c), the location c, or both. When the given graph is f′ and the target is an accumulation function, compare accumulated values, not heights of f′. The 2025 calculus report flags incomplete candidates tests and mixing up the function being optimized.

Apply the candidate test to an invented function. Let f(x)=x³−3x on [−2,2]. Then f′(x)=3x²−3, which is zero at x=−1 and x=1. Evaluate at all candidates: f(−2)=−2, f(−1)=2, f(1)=−2 and f(2)=2. The absolute maximum value is 2, reached at x=−1 and at x=2. The absolute minimum value is −2, reached at x=−2 and at x=1.

Notice that the endpoints matter: x=2 ties with an interior point, and x=−2 ties with an interior one. If you had tested only interior critical points you would have missed half the locations. When you write a justification, list the candidates, show the values and state the comparison. A sign chart of f′ supports local claims, while the value table supports the absolute claim.

Worked example

Find absolute extrema of f(x)=x²−2x on [0,3].

1. f′(x)=2x−2, so x=1 is the interior critical point.
2. Candidates are x=0,1,3.
3. Values are f(0)=0, f(1)=−1, f(3)=3.
4. Absolute minimum is −1 at x=1. Absolute maximum is 3 at x=3.

Mental model: An absolute claim needs a complete candidate comparison.

Common trap: Answering a location when asked for a value.

5. Conditions are part of the theorem

Learning goal: Check continuity and differentiability before invoking a theorem.

The Intermediate Value Theorem requires continuity on a closed interval. It guarantees at least one input attaining any value between the endpoint outputs. It does not locate that input and does not guarantee uniqueness.

The Mean Value Theorem requires continuity on [a,b] and differentiability on (a,b). It guarantees at least one c with f′(c)=[f(b)−f(a)]/(b−a). A differentiable function is continuous at that point, but continuity alone does not imply differentiability.

In a written response, state why each hypothesis holds. "By MVT" without conditions may omit the argument. The 2025 Chief Reader report specifically flags failing to address continuity. For f(x)=|x| on [−1,1], continuity holds but differentiability fails at zero; the endpoint slope is zero, yet there is no interior derivative equal to zero.

Check two theorems on invented functions. For the Intermediate Value Theorem, let f(x)=x²−2 on [1,2]. It is continuous, f(1)=−1 and f(2)=2, so some c in (1,2) has f(c)=0. The theorem says such a c exists, and does not tell you its value or whether there is only one.

For the Mean Value Theorem, let f(x)=x² on [0,2]. It is continuous and differentiable everywhere. The average slope is (4−0)/(2−0)=2, and f′(c)=2c=2 gives c=1, which lies in (0,2). Compare this with the absolute value example above: the theorem fails there because differentiability does not hold at zero.

A written justification should have three parts: the hypothesis checked with a reason, the statement of the theorem's conclusion, and the specific claim you are making about the problem. Missing the first part is the most common gap.

Worked example

Use MVT for f(x)=x² on [1,3].

1. A polynomial is continuous and differentiable everywhere, so the hypotheses hold.
2. Secant slope is (9−1)/(3−1)=4.
3. Solve f′(c)=2c=4 to get c=2.
4. c=2 lies inside (1,3), so it satisfies the conclusion.

Mental model: Write hypotheses, theorem and conclusion as separate pieces.

Common trap: Treating an existence result as a uniqueness result.

6. Tables, approximations and units

Learning goal: Approximate an integral without confusing it with an endpoint difference.

A table gives discrete observations, not an exact continuous formula. A trapezoidal approximation models each interval with a straight segment. Its area is width × average of the two endpoint heights. Unequal widths must remain unequal in the calculation.

For rate data, each trapezoid gives an approximate amount. For amount data, the trapezoids give amount×time; divide by total time for average amount. State whether a result is approximate. A derivative estimate is instead a slope using nearby data values.

An error-direction claim needs a shape assumption. If the function is concave up across an interval, trapezoids lie above the curve and overestimate its integral. A few table points alone may not prove global concavity.

Use an invented table of a rate r(t) in litres per minute: at t=0, 2, 5, 6 minutes the values are 3, 5, 4, 6. The trapezoids give 2×(3+5)/2 = 8, then 3×(5+4)/2 = 13.5, then 1×(4+6)/2 = 5. The total is 26.5 litres, an approximation of the net amount that entered over six minutes. The widths differ (2, 3 and 1), and each was used as given.

Do not confuse this with an average rate of change, which would use only the endpoints. The approximation uses every row of data. If asked for the average rate over the six minutes, divide the amount by six, which is about 4.4 litres per minute. State the units and the word approximately so the reader knows the result comes from discrete data. A comment about overestimate or underestimate needs a stated assumption about concavity across the whole interval.

Worked example

r(0)=2, r(1)=4, r(3)=8 litres/minute. Approximate net change from 0 to 3.

1. First interval: (1−0)(2+4)/2 = 3 litres.
2. Second interval: (3−1)(4+8)/2 = 12 litres.
3. Add areas: approximately 15 litres.
4. The average value of r over [0,3] is approximately 15/3 = 5 litres/minute. The average rate of change of r is instead (8−2)/3 = 2 (litres/minute) per minute.

Mental model: Area weights by widths; slope divides change by width.

Common trap: Reporting rate units for an integral of rate.

7. Limits describe approach, not assignment

Learning goal: Separate a limiting value from a function’s value at a point.

A limit describes what f(x) approaches as x approaches a point, not necessarily what f equals at that point. Removing or redefining one point can change f(a) without changing the nearby limiting behavior.

A two-sided limit exists only when the left and right approaches agree. An infinite limit describes unbounded behavior; it is not a finite function value. Continuity at a requires f(a) to exist, the limit to exist and the two to match.

For f(x)=(x²−1)/(x−1), x≠1, factoring gives x+1 away from one. The limit at one is two even if the original formula is undefined there. Cancelling a factor does not restore the original domain automatically.

A piecewise function with left limit 1 and right limit 3 has no two-sided limit at the join. Assigning f(a)=2 does not fix this. By contrast, if both one-sided limits are 2 but f(a)=7, the limit exists and continuity fails only because of the assigned value. Diagnose the type of failure before using a continuity theorem.

Compare three invented situations at x=2. First, f(x)=(x²−4)/(x−2) for x≠2. Factoring gives x+2, so the limit as x approaches 2 is 4, even though f(2) is undefined. Second, define g(2)=9 and g(x)=x+2 elsewhere. The limit is still 4, but g(2)=9, so g is not continuous at 2. Third, h(x)=|x−2|/(x−2) has left limit −1 and right limit 1, so the two-sided limit does not exist.

These three cases show the three typical failures: a hole, a misplaced point and a jump. The method is the same for each: find the left and right behavior, compare, and then compare with the function value. Tables of values can suggest a limit, but a justification should use algebra or a graph with a clear statement of what the one-sided approaches show.

Worked example

Find limₓ→₁ (x²−1)/(x−1).

1. Factor x²−1=(x−1)(x+1).
2. For x≠1, the ratio equals x+1.
3. As x approaches one, x+1 approaches two.
4. The limit is two; the original expression remains undefined at x=1.

Mental model: Limit = approach; continuity also checks assignment.

Common trap: Cancelling a factor and silently changing the domain.

8. Tangent approximation is local

Learning goal: Use a derivative near its base point and state the approximation.

A tangent-line approximation uses f(a)+f′(a)(x−a). The derivative is a local rate at a; the expression predicts a nearby output, not an exact global formula.

Choose a base point where the function and derivative are easy to evaluate. The change is x−a, not x itself. Larger moves can increase approximation error; curvature determines whether the tangent is above or below the function when concavity is known throughout the relevant neighborhood.

The differential dy=f′(a)dx approximates the actual change Δy. It does not usually equal it. For f(x)=x² near a=2, f′(2)=4 and L(2.1)=4.4 while the true value is 4.41.

Estimate a square root with an invented base point. Let f(x)=√x and a=4. Then f(4)=2 and f′(x)=1/(2√x), so f′(4)=1/4. The tangent line is L(x)=2+(1/4)(x−4). At x=4.4, L(4.4)=2+0.1=2.1. The true value is about 2.0976, so the approximation is slightly high.

The direction of the error is explained by shape: √x is concave down, so tangent lines lie above the curve, which means the tangent approximation overestimates here. A different base point would be worse. Using a=1 to estimate √4.4 gives a poor result because 4.4 is far from 1. When you write the answer, show the base point, the derivative value, the linear formula and a short sentence saying that the estimate is approximate and local.

A good habit is to compute the error as well. Here the actual change is Δy = 2.0976 − 2 = 0.0976, while the differential dy = (1/4)(0.4) = 0.1. They are close but not equal. Doubling the step to 0.8 would make the gap grow faster than the step, which shows why the method belongs near the base point. State the base point, the linear formula and a sentence of reasoning about direction before you write the final number.

Worked example

Approximate √4.04 using a=4.

1. f(4)=2 and f′(x)=1/(2√x), so f′(4)=1/4.
2. Input change is 0.04.
3. L(4.04)=2+(1/4)(0.04)=2.01.
4. Report an approximation, not an exact square root.

Mental model: Linearization is anchored locally at a known point.

Common trap: Presenting a local approximation as an exact global identity.

9. Related rates keep the variables linked

Learning goal: Differentiate a relationship before substituting changing values.

A related-rates problem gives several quantities that change together over time. Write their governing equation, differentiate with respect to time and then evaluate at the specified instant.

For a circle A=πr², both A and r depend on t. Chain rule gives dA/dt=2πr dr/dt. Substituting a radius too early and treating it as permanently constant destroys the relationship.

Units check the result: radius×radius-rate gives area-rate. The sign describes expansion or contraction. A numerical radius is not a rate, and a rate is not the final quantity.

For a sphere V=(4/3)πr³, the linked rate is V′=4πr²r′. If volume is being supplied at a fixed rate, solve r′=V′/(4πr²): radius growth slows as the sphere gets larger. This dependence is visible before substituting any numbers and gives a useful reasonableness check.

Work an invented ladder problem. A 5 m ladder leans against a wall. The bottom slides away at 0.5 m/s. How fast does the top slide down when the bottom is 3 m from the wall? The relationship is x²+y²=25. Differentiating with respect to time gives 2x(dx/dt)+2y(dy/dt)=0. At x=3, y=4 (since 9+16=25). Substituting gives 3(0.5)+4(dy/dt)=0, so dy/dt=−0.375 m/s. The negative sign means the top is moving down.

Notice that the numbers 3, 4 and 0.5 appear only after differentiation. If you had put x=3 into the equation first, the 3 would be treated as a constant, and its derivative would have been zero. Write down the equation, differentiate, and only then plug in the instant's values with units attached.

Worked example

Radius is 3 cm and increasing at 2 cm/s. Find area rate.

1. Differentiate A=πr² to get A′=2πr r′.
2. Substitute r=3 and r′=2.
3. A′=12π cm²/s.
4. The positive sign matches expansion.

Mental model: Related rates: equation → derivative → instant.

Common trap: Dropping the time-dependent chain-rule factor.

10. Differential equations need an initial condition

Learning goal: Distinguish a family of solutions from the specified solution.

A differential equation specifies a relationship involving a derivative. Integrating often produces a family of functions because derivatives cannot determine an additive constant. An initial condition selects one member of the family.

For dy/dt=2t, the family is y=t²+C. If y(0)=5, then C=5. A slope field describes possible local directions at points; it is not itself the trajectory of every solution.

Exponential growth y′=ky has solutions y=Ceᵏᵗ. The growth constant k is a relative rate, not a fixed amount gained each unit time. Domain and initial conditions matter, and a biological growth model is an approximation, not a universal law.

Compare two invented cases. First, dy/dx=2x with y(1)=3. Integrating gives y=x²+C, and the condition gives 3=1+C, so C=2 and y=x²+2. Without the condition, every curve y=x²+C satisfies the equation. Second, an exponential model dy/dt=0.1y with y(0)=100, which has the solution y=100e^(0.1t). The constant 0.1 is a relative rate, so y grows by about 10% of its current size per unit of time, in this idealized model.

Slope fields show the direction at many points, and a solution curve must follow the little segments. Two solutions with different initial conditions never cross for these simple equations. In a context problem, say what the initial condition means, such as the starting amount, and mention that the model describes the situation only approximately.

Check a solution by substitution. For y=100e^(0.1t), y′ = 10e^(0.1t), which equals 0.1y, so it satisfies the equation. At t=0, y=100, matching the condition. Doing this two-line check catches sign and constant errors quickly, and it shows the reader the answer fits both parts of the problem.

Worked example

Solve y′=3t² with y(0)=4.

1. Integrate the rate: y=t³+C.
2. Apply y(0)=4 to obtain C=4.
3. The specified solution is y=t³+4.
4. Differentiate to check y′=3t² and substitute t=0 to check the condition.

Mental model: Check both the derivative equation and the initial condition.

Common trap: Mistaking one convenient antiderivative for the unique solution.

11. Speeding up needs two signs

Learning goal: Decide whether a particle speeds up or slows down by comparing velocity and acceleration.

Speed is the absolute value of velocity. A particle can have negative velocity and still be going fast, so the sign of velocity alone says only which way it moves. The sign of acceleration alone says only which way velocity is changing. Whether the speed grows or shrinks depends on both signs together.

If v(t) and a(t) have the same sign, the velocity is moving away from zero, so the speed increases. If they have opposite signs, the velocity is moving toward zero, so the speed decreases. If either is zero at a single instant, you cannot conclude anything from that instant alone. Look at the signs on each side.

The common error is to say "acceleration is positive, so the particle is speeding up". That is true only when velocity is also positive. A particle with v = −0.75 and a = 1 is slowing down: its velocity is rising toward zero, so its speed is falling. Write the two signs explicitly in your answer, and state the conclusion in terms of speed, not velocity.

Remember where each quantity comes from. a(t) = v′(t), and v(t) = x′(t). A sign chart for v and a on the same number line makes every interval easy to read. Mark the zeros of both functions, test one point in each interval, and then compare the signs interval by interval. Finish with a sentence that names the interval, the two signs and the conclusion about speed. Finally, check your answer against a quick picture. Imagine a ball thrown upward: on the way up its velocity is positive and its acceleration is negative, so it slows. At the top its velocity is zero. On the way down both are negative, so it speeds up. The same reading works for any particle, whatever the formula.

Worked example

A particle has velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 5. Is the speed increasing or decreasing at t = 0.5 and at t = 4?

1. Find acceleration: a(t) = v′(t) = 2t − 4.
2. At t = 0.5: v = 0.25 − 2 + 3 = 1.25 > 0 and a = 1 − 4 = −3 < 0.
3. The signs are opposite, so the speed is decreasing at t = 0.5.
4. At t = 4: v = 16 − 16 + 3 = 3 > 0 and a = 8 − 4 = 4 > 0.
5. The signs match, so the speed is increasing at t = 4.

Mental model: Same signs for velocity and acceleration mean speeding up; opposite signs mean slowing down.

Common trap: Acceleration being positive does not mean the particle is speeding up.

12. Concavity needs a sign change

Learning goal: Justify concavity, local extrema and inflection points with the second derivative.

The first derivative tells you whether f rises or falls. The second derivative tells you how the slope is changing. If f″ > 0 on an interval, the slope is increasing and the graph is concave up. If f″ < 0, the slope is decreasing and the graph is concave down.

The second derivative test uses this at a critical point. If f′(c) = 0 and f″(c) > 0, the graph curves up around a flat tangent, so f has a local minimum. If f′(c) = 0 and f″(c) < 0, it has a local maximum. If f″(c) = 0, the test says nothing and you must return to the sign of f′.

An inflection point is a point where concavity changes. Setting f″(x) = 0 only gives candidates. You must show that f″ changes sign across the candidate. The function g(x) = x⁴ has g″(x) = 12x², which is zero at x = 0 but never negative, so the graph is concave up on both sides and there is no inflection point.

For a written justification, name the function you are examining (f′ or f″), the interval or point, the sign and the conclusion. Statements such as "f″ is zero, so it is an inflection point" lose the key reason, which is the sign change. A sign line for f″ is a short and complete argument. Also keep the roles separate: f′ gives increasing and decreasing, f″ gives concavity, and the two are independent facts about a graph. As a final check, sketch the graph from your sign information. Rising and concave up looks like the right half of a bowl, falling and concave up looks like the left half, and so on. If your sketch disagrees with your sign chart, one of the derivatives has a sign error. Fix that before writing the justification.

Worked example

For f(x) = x³ − 6x² + 9x + 1, find the local extrema and the inflection point with justification.

1. f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), so the critical points are x = 1 and x = 3.
2. f″(x) = 6x − 12. Then f″(1) = −6 < 0, so f has a local maximum at x = 1 with f(1) = 5.
3. f″(3) = 6 > 0, so f has a local minimum at x = 3 with f(3) = 1.
4. f″(x) = 0 at x = 2. f″ is negative for x < 2 and positive for x > 2, so concavity changes.
5. The inflection point is (2, f(2)) = (2, 3).

Mental model: An inflection point needs f″ to change sign, not just to equal zero.

Common trap: Do not call every zero of f″ an inflection point.

13. Over or under: read the shape first

Learning goal: Decide whether left, right, trapezoid and midpoint approximations overestimate or underestimate.

A Riemann sum approximates an area using rectangles. Whether it lands above or below the true value depends on the shape of the graph, and you can often tell without calculating the exact integral. The shape facts you need are the sign of f′ (increasing or decreasing) and the sign of f″ (concave up or down).

For an increasing function, a left sum uses the lowest value on each subinterval, so it underestimates. A right sum uses the highest value, so it overestimates. For a decreasing function the roles reverse. These conclusions depend only on monotonicity.

Trapezoid and midpoint sums depend on concavity. For a concave up function, the chord of each trapezoid lies above the graph, so the trapezoid sum overestimates. The tangent line at a midpoint lies below the graph, so the midpoint sum underestimates. For a concave down function both conclusions reverse.

With a table of values, you can only use shape you can justify. A table alone does not prove concavity unless the problem gives f″ or a graph. State the given fact, such as "f is increasing", before drawing your conclusion. Also keep the units: the sum approximates the net accumulation of the function, so use the width of each subinterval and write the sum as a number with the units of the quantity. For unequal subintervals, compute each trapezoid separately: width times the average of the two end values. A quick sanity check helps. Sketch the curve and one rectangle or trapezoid. If the rectangle sticks out above the curve, the sum is too big; if it sits below, the sum is too small. This picture should always agree with the sign facts you quote, and it catches the most common slip, which is mixing up left and right on a decreasing function.

Worked example

Use a left sum with four equal subintervals to approximate ∫₀⁴ x² dx, and state whether it is an overestimate or underestimate.

1. Each subinterval has width 1, with left endpoints 0, 1, 2, 3.
2. Left sum = 1·(0 + 1 + 4 + 9) = 14.
3. f(x) = x² is increasing on [0, 4].
4. On an increasing function the left endpoint gives the smallest value on each subinterval.
5. So 14 is an underestimate; the exact value is 64/3, about 21.33.

Mental model: Monotonicity decides left and right sums; concavity decides trapezoid and midpoint sums.

Common trap: Do not claim concavity from a table unless the problem supplies it.

14. Calculator work still needs a setup

Learning goal: Write the equation, derivative or integral you evaluate before quoting a calculator result.

The official course description treats a graphing calculator as part of AP Calculus and says it is required on some portions of the exam. It lists four built-in capabilities that a calculator must have: plotting a graph in a chosen window, finding zeros of functions, numerically calculating a derivative and numerically calculating a definite integral.

The same document sets a rule for free-response answers. When a result comes from one of those capabilities, you must write the setup that leads to it. For a definite integral, that means writing the integral itself, with its limits and integrand. For a zero, it means writing the equation you solved. A bare decimal does not show reasoning, even when it is correct.

Other calculator features and programs are different: the course description says they require the mathematical steps that produce the result. So a calculator may find the number for ∫₀⁴ (6 + 4cos(t²/5)) dt, but it should not replace the explanation of why that integral gives the net change.

A practical routine is to answer in three lines. First, state what the quantity means and its units. Second, write the setup, such as 40 + ∫₀⁴ R(t) dt. Third, report the calculator value with its units and a short interpretation. Keep full precision inside the calculator until the final value, because rounding an intermediate result can shift the final digits. Also check that the calculator is in radian mode before evaluating trigonometric functions of t. When a question asks you to justify a claim, such as that a rate is zero or that two quantities are equal, a calculator value supports the claim only when the written equation or integral shows what was compared. Treat the setup as the part of the answer that a reader can check, and the number as the part they cannot.

Worked example

A tank holds 40 gallons at t = 0. Water enters at R(t) = 6 + 4cos(t²/5) gallons per minute. Find the volume at t = 4 using a calculator.

1. The volume at t = 4 is the initial amount plus net change: V(4) = 40 + ∫₀⁴ R(t) dt.
2. Write the setup: V(4) = 40 + ∫₀⁴ (6 + 4cos(t²/5)) dt, with the calculator in radian mode.
3. The calculator gives ∫₀⁴ R(t) dt ≈ 29.782 gallons.
4. V(4) ≈ 40 + 29.782 = 69.782 gallons.

Mental model: Write the setup, then use the calculator, then report the result with units.

Common trap: A calculator value without the integral or equation is not a complete answer.

15. Area and volume from a cross section

Learning goal: Set up area between curves and volume with known cross sections.

Both area and volume problems follow the same pattern. Find where the curves meet, decide which is on top over each interval, write the length of a representative slice, and integrate that length or area across the interval.

For area between two graphs, the length of a vertical slice is top minus bottom. The bounds come from the intersection points unless the problem gives them. If the curves cross inside the interval, split the integral at the crossing, because top and bottom swap. For y = x and y = x², the curves meet at x = 0 and x = 1, and the line is on top between them, so the area is ∫₀¹ (x − x²) dx = 1/6.

For a volume with known cross sections, each slice has a cross-sectional area A(x) that depends on the shape. If the cross sections are squares and the side is the length of the region at x, then A(x) = (side)². If they are semicircles, A(x) = (π/2)(side/2)². The volume is ∫ A(x) dx.

A frequent error is to integrate the side length instead of the area of the cross section. The integrand for volume must have units of area, because multiplying area by thickness gives volume. In the square example, the side is x − x², so the integrand is (x − x²)², which expands to x² − 2x³ + x⁴. Write the region, the slice and the integral before computing, and check the sign: area and volume cannot be negative.

Worked example

The base of a solid is the region between y = x and y = x² for 0 ≤ x ≤ 1. Cross sections perpendicular to the x-axis are squares. Find the volume.

1. The curves meet at x = 0 and x = 1, and y = x is above y = x² between them.
2. The side of the square at x is x − x².
3. The cross-section area is A(x) = (x − x²)² = x² − 2x³ + x⁴.
4. V = ∫₀¹ (x² − 2x³ + x⁴) dx = 1/3 − 1/2 + 1/5.
5. V = 1/30 cubic units.

Mental model: Find bounds, write the slice, make sure the integrand has the right units, then integrate.

Common trap: Integrating a length when the problem asks for a volume.