Antiderivative Stall: Six Steps Out

You already know that a definite integral needs an antiderivative first. What you don't yet have is a reliable way to find that antiderivative when the integrand isn't a single clean symbol like x² or sin x. You name the technique, u-substitution or integration by parts, but somewhere between "pick u" and "rewrite du" the work freezes. You recognize Newton-Leibniz but can't always tell whether the answer you just wrote is right, and when a physics problem asks you to set up an integral from scratch, you're not sure which variable to integrate, where the limits should come from, or whether the differential you're carrying through the calculation is doing real work.

This chapter is built around that bottleneck. It is organized as six short steps, each of which closes one specific hole in the antiderivative-finding reflex. The destination is the same in every step: when you see an integral, you should be able to look at the integrand, recognize its shape, write down an antiderivative in one or two lines, and verify the answer with a quick sign, unit, or magnitude check. Physics scenarios appear in step six as the final test, not as the main course. Integration is the destination; physics is the vehicle.

1. Why you keep getting stuck at the antiderivative step

The bottleneck is not a missing tool. It is a missing reflex. You learned to differentiate by memorizing the chain rule, the product rule, the power rule, and the standard derivatives of e^x, sin x, cos x, and ln x. Integration is the inverse operation, but no one ever showed you the derivative table running backwards. That is the whole missing piece.

MathWorld's Indefinite Integral entry makes the inverse-operation connection formal. An antiderivative F of a function f is any function on an interval satisfying F'(x) = f(x). The integral sign with no limits, ∫f(x) dx, simply names the family of all such antiderivatives, and because the derivative of any constant is zero, the family is non-unique: any two antiderivatives of the same integrand differ by a constant. The conventional notation captures this as F(x) + C, where C is the constant of integration.¹

That non-uniqueness is what the trailing +C is doing in every antiderivative you write down, and it is also the reason the constant quietly disappears from definite integrals. If you evaluate a definite integral ∫_a^b f(x) dx via an antiderivative F, the operation is F(b) − F(a). Any C added to F cancels at the subtraction step, so writing +C in an antiderivative is required for indefinite integrals (where the constant identifies the whole family) and irrelevant for definite integrals (where the constant collapses). Chapter 5 returns to this in operational form; the only thing worth holding in mind now is that the +C is not a typo or a formality, it is the algebraic record of the family.

Once the inverse-operation lens is in place, the integration table that sits at the end of MathWorld's entry stops looking like a memorization exercise. Every row of the table is a derivative rule, read in the opposite direction. Six pairings carry the whole toolbox the rest of this chapter uses:

The right-hand column is not new material. It is the left-hand column, played backwards.

Caption: Left column shows the derivative rules you already know. Right column shows the same rules read in reverse: that is the integration toolbox. The arrows say "the integrand in the middle column is the output of the derivative rule on the left, so the antiderivative is the function that was differentiated."²

The typical stall happens not at the recognition step, but between the recognition step and the write-down step. You see that ∫(3x²+2x)·e^(x³+x²) dx contains a factor (3x²+2x) that is the derivative of (x³+x²), and you correctly pick u = x³+x². The next move is to rewrite (3x²+2x) dx as du, and that is where the work freezes for most learners. dx and du get treated as part of the integral operator rather than as variables you can manipulate, so the rewrite step feels illegal even when it is correct.

That stall is the same stall that appears in every technique this chapter covers. The fix is mechanical: re-read the integrand line by line, confirm the inner expression and the derivative factor separately, then do the substitution one substitution at a time. Chapter 3 walks through this in detail; for now, the takeaway is that the stall is a workflow problem, not a knowledge problem. You already have the derivative rules. The reflex of reading them backwards, and the line-by-line rewrite habit that breaks the freeze, are what the next five chapters build.³

2. What's actually in the integration toolbox, and how to read it backwards

The full integration toolbox is wider than you need right now. Stewart's main text and the Chinese bilingual counterpart focus on four families whose antiderivatives you should be able to write down on sight: power, exponential, trigonometric, and the explicit 1/x exception. Internalize these four rows before touching any technique.

Power family. For any n except −1,

$$\int x^n , dx = \frac{x^{n+1}}{n+1} + C$$

The rule covers integer exponents (∫ x³ dx = x⁴/4 + C), rational exponents (∫ √x dx = (2/3) x^(3/2) + C), and negative exponents (∫ x^(−2) dx = −1/x + C). The same formula works for sums: integrate term by term, distribute +C at the end. OpenStax §4.10 exercises in this range include practice items that span integer, rational, and negative exponent cases; verify each item against your edition before working them.⁴

Exponential family. The fixed point of differentiation is e^x, so

$$\int e^x , dx = e^x + C$$

For a general base a, ∫ a^x dx = a^x / ln(a) + C, valid for a > 0 and a ≠ 1. In practice this matters for ∫ 2^x dx = 2^x / ln 2 + C and similar; everything else reduces to e^x once you rewrite a^x as e^(x ln a).¹

Trigonometric family. Three identities carry most of the load:

$$\int \sin x , dx = -\cos x + C \qquad \int \cos x , dx = \sin x + C \qquad \int \sec^2 x , dx = \tan x + C$$

The first one is the highest-frequency sign trap in elementary integration. The derivative of cos x is −sin x, not sin x, so the antiderivative of sin x is −cos x, not +cos x. The trap is asymmetric: ∫ cos x dx = sin x + C has no minus sign. A quick reference card for this trap is worth keeping nearby.⁵

Caption: Goal ∫ sin x dx. Wrong answer +cos x + C misses the minus sign that comes from d/dx cos x = −sin x. Correct answer −cos x + C. Both rows include +C; the trap is the sign on cos, not the constant of integration.⁶

1/x is its own row, not the power rule at n = −1. The power rule would require dividing by n+1 = 0, which is undefined. The correct row is

$$\int \frac{1}{x} , dx = \ln|x| + C$$

The absolute value makes the antiderivative defined wherever the integrand is defined (both sides of x = 0, never crossing). Verify by differentiation: d/dx ln|x| = 1/x on both branches.¹

A condensed cheatsheet for the four rows looks like this:

Caption: Four-row reference card. Power row uses the n+1 over n+1 formula with n ≠ −1. Exponential row uses e^x fixed point and a^x / ln a. Trigonometric row lists the three basic identities plus the ∫ sec x tan x dx = sec x + C bonus. The 1/x row is explicitly flagged as the exception where the power rule breaks (n+1 = 0).⁷

After every row in the table, run the reverse-check reflex: differentiate your answer and confirm you get the integrand back. d/dx [x⁴/4] = x³, so ∫ x³ dx = x⁴/4 + C is correct. d/dx [−cos x] = sin x, so ∫ sin x dx = −cos x + C is correct. d/dx [ln|x|] = 1/x, so ∫ (1/x) dx = ln|x| + C is correct. The reverse check catches every sign error and every "I dropped the constant" mistake at once, and it costs about five seconds per problem.⁴

3. When the integrand is a composition, how do you unwind it

The next time you stall on finding an antiderivative, the most likely cause is that the integrand is a composition, not a single clean symbol. The composition is wrapped around something, like e^(x³+x²) or sin(x²), and the bare toolbox rows don't apply directly. The technique that handles this is u-substitution, and it is the reverse of the chain rule.

The chain rule says d/dx f(g(x)) = f'(g(x))·g'(x). Run that backwards and you get the recognition heuristic for u-substitution: scan the integrand for a composition f(g(x)) whose derivative factor g'(x) is sitting in the integrand as a multiplicative factor. If you find it, that g(x) is your u.³

Walk through the canonical example. Take

$$\int (3x^2 + 2x) \cdot e^{x^3 + x^2} , dx$$

Scan the integrand. The exponent x³ + x² has derivative 3x² + 2x, which is the factor multiplying the exponential. That is exactly the f(g(x))·g'(x) shape. So u = x³ + x², du = (3x² + 2x) dx, and the integral becomes

$$\int e^u , du = e^u + C = e^{x^3 + x^2} + C$$

One line of algebra. That is what the textbook means by "u-sub solves it instantly."³

A second worked example makes the rewrite step explicit. Take

$$\int (2x + 1) \sqrt{x^2 + x} , dx$$

The factor (2x + 1) is exactly the derivative of the expression inside the radical, so u = x² + x, du = (2x + 1) dx. The integral becomes

$$\int u^{1/2} , du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} (x^2 + x)^{3/2} + C$$

Again, one line.³

The four-step reflex that drives every u-substitution is:

1. Scan the integrand for an expression whose derivative (up to a constant) already multiplies the rest. That expression is u.
2. Differentiate u to get du = g'(x) dx. Treat du as an algebraic object, not as part of the integral operator.
3. Rewrite the integrand in terms of u. If u appears once with no leftover x's, the rewrite is done; if some x's survive, stop and reconsider.
4. Reduce to a toolbox entry. The result should look like ∫ f(u) du for some f you recognize.

The canonical stall is between steps 2 and 3. You have named u correctly, you have computed du, and the rewrite step feels illegal because dx and du "shouldn't be variables." They are variables for the duration of the substitution, and treating them as such is the entire point of the technique. The fix is the line-by-line check built into step 3: re-read the integrand and the differential separately, confirm that everything on the left matches everything on the right, and the freeze breaks.³

A geometric picture of what u-substitution unwinds is shown below.

Caption: The chain rule composes an inner function g(x) with an outer function f. Its derivative is f'(g(x))·g'(x). U-substitution reverses this: identify the inner function and its derivative factor inside the integrand, then change variable from x to u = g(x), reducing the problem to a single toolbox entry.⁸

A worked example with a definite integral finishes the picture. Take

$$\int_0^1 x^2 \cdot 2^{x^3} , dx$$

Set u = x³, du = 3x² dx, so x² dx = du/3. Then

$$\int_0^1 x^2 \cdot 2^{x^3} , dx = \frac{1}{3} \int_0^1 2^u , du = \frac{1}{3} \cdot \frac{2^u}{\ln 2} \Big|_0^1 = \frac{2 - 1}{3 \ln 2} = \frac{1}{3 \ln 2} = \frac{1}{\ln 8}$$

The lesson at the end: u-substitution preserves the original variable during evaluation (back-substitute u = x³ after integrating), which keeps the +C cancellation visible. The alternative is to convert the limits to u-values at the substitution step and skip back-substitution; both routes give the same answer, and the back-substitution route is the one the standard textbook uses.³

The boundary with integration by parts is sharp. U-substitution is the reverse of the chain rule, and it applies when you can spot an inner function whose derivative sits in the integrand. Integration by parts is the reverse of the product rule, and it handles product-shaped integrands where no such inner function exists. If the integrand is a product of two unlike families, like x·sin x or x·e^x, parts is the next move.

4. When the integrand is a product of unlike families, how do you split it

If u-substitution does not apply because no inner-function-plus-derivative-factor is sitting in the integrand, the integrand is probably a product of two unlike families. Polynomial-times-exponential (x·e^x), polynomial-times-trig (x·sin x), and log-times-polynomial (x·ln x) are the canonical cases. The technique is integration by parts, which is the reverse of the product rule.

The derivation is three lines. Start with the product rule:

$$(uv)' = u'v + uv'$$

Integrate both sides. The left side gives uv plus a constant, which is absorbed:

$$uv = \int u' v , dx + \int u v' , dx$$

Rearrange to isolate the integral you do not yet have, and you get the integration-by-parts formula:

$$\int u , dv = uv - \int v , du$$

In words: the original integral ∫u·v' dx equals u times the antiderivative of v' minus the new integral ∫v·u' dx. The technique exchanges one integral for another, and the whole point is that the new integral is easier because u' is "simpler" than u (a polynomial derivative that eventually drives the degree to zero) and v is a function with a known antiderivative.⁹

A geometric reading is in the figure below.

Caption: Parametric-area decomposition of integration by parts. A rectangle of dimensions u(b) × v(b) − u(a) × v(a) is split by a parameterized curve into two regions A₁ and A₂. The formula ∫u dv = uv − ∫v du is the algebraic statement that the two region areas add up to the rectangle area minus a small correction.¹⁰

The decision process is: which factor in the product should be u, and which should be dv? The empirical answer is captured by the LIATE ordering, which assigns the role of u to whichever factor comes earlier in the list:⁹ LIATE LIATE

• L — Logarithmic (ln x, log_b x)
• I — Inverse trigonometric (arctan x, arcsin x)
• A — Algebraic (x, x², x³, …)
• T — Trigonometric (sin x, cos x, tan x)
• E — Exponential (e^x, a^x)

The rule says: pick u from earlier in the list, dv from later. The reason is that functions lower in the list generally have simpler antiderivatives than the ones above them, so the new integral ∫v·u' dx ends up easier than the original.

Work the canonical example. Take

$$\int x \cos x , dx$$

LIATE gives u = x (A, algebraic) and dv = cos x dx (T, trigonometric). Then du = dx and v = sin x. Apply the formula:

$$\int x \cos x , dx = x \sin x - \int \sin x , dx = x \sin x + \cos x + C$$

The reverse-check: d/dx [x sin x + cos x] = sin x + x cos x − sin x = x cos x. Correct.⁹

The contrast case makes the rule's value visible. If you had instead picked u = cos x (T) and dv = x dx (A), then du = −sin x dx and v = x²/2. Apply the formula:

$$\int x \cos x , dx = \frac{x^2}{2} \cos x - \int \frac{x^2}{2} (-\sin x) , dx = \frac{x^2}{2} \cos x + \frac{1}{2} \int x^2 \sin x , dx$$

The new integral is harder than the original. A second application of parts moves to x³·cos x − ∫ x³·sin x dx, and the recursion never terminates. LIATE was constructed precisely to prevent this trap.⁹

When the polynomial factor is higher degree, the same parts step has to be applied multiple times in sequence, and the bookkeeping of "which coefficient carries over from the previous step" becomes its own trap. The tabular method, also called the "tic-tac-toe method" in some texts, organizes the bookkeeping by writing the derivatives of u down one column, the integrals of dv up another column, and a sign column alternating +/−/+/−/+:

The answer is read by multiplying each sign with the diagonal entries: 2x⁴ e^(x/2) − 16x³ e^(x/2) + 96x² e^(x/2) − 384x e^(x/2) + 768 e^(x/2) + C. The method terminates when the u-derivative column reaches zero, which always happens for a polynomial factor. (If the function factors produce a circle, like ∫e^θ cos θ dθ, parts twice returns the original integral on both sides, and algebra solves for the answer as (1/2)(e^θ cos θ + e^θ sin θ) + C.)¹¹

A second boundary case closes the loop with u-substitution. If after one or more applications of parts the remaining integral has the form ∫ x·e^(x²) dx or similar, that is the signal to switch to u-substitution. The two techniques layer; they do not compete.

A historical note: the formula is named after Brook Taylor, who published the integration-by-parts technique in his 1715 Methodus Incrementorum. This attribution is widely cited but rests on secondary biographical sources rather than a primary text reading, so treat it as a useful historical pointer rather than a sourced fact.⁹

5. How an antiderivative becomes a number, and how not to lose the answer on the way

You have an antiderivative F. The definite integral ∫_a^b f(x) dx is a number, not a function, and the bridge between the two is the Newton-Leibniz formula, also called the second fundamental theorem of calculus:

$$\int_a^b f(x) , dx = F(b) - F(a)$$

where F is any antiderivative of f and a, b are the bounds.¹² FTC

The five operational steps are:

1. Set up the integral ∫_a^b f(x) dx.
2. Find F such that F'(x) = f(x).
3. Plug in the limits: compute F(b) and F(a) as numbers.
4. Subtract: F(b) − F(a). The constant +C cancels at this step, because (F(b)+C) − (F(a)+C) = F(b) − F(a).
5. Sanity-check sign, units, and magnitude.

The order of steps can vary; some textbooks evaluate F(a) first. What matters is that both F(b) and F(a) appear and are subtracted in the right order, and that +C is dropped at step 4.¹²

The diagram below turns the abstract identity into a procedure card.

Caption: Five-step operational flow for evaluating a definite integral. Box 1 sets up the integral; box 2 finds the antiderivative F; box 3 plugs in the limits; box 4 subtracts F(b) − F(a), with +C cancelling automatically; box 5 runs the sign, units, and magnitude sanity check. The order is necessary but not rigid; the two non-negotiable ingredients are that both F(b) and F(a) appear and are subtracted in that order.¹³

The "signed area" reading of the same formula is what gives the procedure its sign-check. A region where f(x) is negative (the graph sits below the x-axis) contributes a negative amount to the total. The worked velocity example makes this concrete. With v(t) = 4 − t over [0, 8]:

• Positive triangle from t = 0 to t = 4 (where v > 0): contributes +8 m to net displacement.
• Negative triangle from t = 4 to t = 8 (where v < 0): contributes −8 m.
• Net displacement: F(8) − F(0) = (4·8 − 8²/2) − (4·0 − 0²/2) = 0 m.¹²

Caption: Velocity vs. time for v(t) = 4 − t over [0, 8] s. The blue triangle above the axis (positive velocity region, t ∈ [0, 4]) adds +8 m; the orange triangle below the axis (negative velocity region, t ∈ [4, 8]) adds −8 m. Net displacement is 0 m. The negative lobe subtracts from the positive lobe; the integral records signed contribution, not unsigned distance traveled.¹⁴

The worked wrong/right pair below makes the two most common evaluation errors visible in one figure.

Caption: Goal ∫₀² e^(−x) dx. Mistake 1 writes the antiderivative as e^(−x) (missing the minus sign, since d/dx[−e^(−x)] = e^(−x)) and evaluates to e^(−2) − 1, the wrong sign on the antiderivative. Mistake 2 writes −e^(−x) correctly but evaluates as −e^(−2) + (−e⁰) = −e^(−2) − 1, which applies F(a) − F(b) instead of F(b) − F(a) and drops the minus sign on the lower-limit term. Correct: 1 − e^(−2) ≈ 0.865. The trap pattern is the lesson; the numeric answer is incidental.¹⁵

The unit check closes the chapter. Whatever physical quantity the integral represents, the units must come out right. Integrate force (N) over distance (m) and the result must be in J (N·m), which is the unit-level proof that the work integral is set up correctly. Integrate velocity (m/s) over time (s) and the result must be in m, which is the unit-level proof that displacement is the right reading. Integrate linear density (kg/m) over length (m) and the result must be in kg. If the units don't multiply to the right thing, the integrand is wrong or the limits are wrong or the differential is missing. The unit check is the cheap universal test that catches setup errors before they propagate to the answer.¹²

6. How to translate a physics scenario into the right integral

The remaining gap is the one you named in the diagnostic: turning a physics description ("a variable force, a non-uniform velocity, a rod with non-uniform density") into the integral expression itself. The mechanism is one procedure applied to three different products.

Take a constant-coefficient identity you already know: A = l·w (area), d = r·t (distance), M = D·V (mass). When one factor varies, replace the constant factor with a function and the product becomes an integral:

• Area under y = f(x): A = ∫_a^b f(x) dx.
• Distance with non-uniform velocity: s = ∫_a^b v(t) dt.
• Mass of a non-uniform rod: m = ∫_a^b ρ(x) dx.

Three settings, one procedure. Stewart 9e Figure 6.14 places them side by side as the unifying frame for the chapter.¹⁶

Caption: Three settings for one accumulation pattern. Left panel: area under a curve y = f(x). Middle panel: distance traveled by an object with velocity y = v(t). Right panel: mass of a horizontal bar with linear density y = ρ(x). The same procedure (slice the interval, write the integrand, sum the slices in the limit) produces three physically different quantities.¹⁷

The same procedure in each case reduces to four meta-questions:

1. Activate: is there a simpler route? For variable-force work with a polynomial force, sometimes yes (an algebraic identity gives the answer without integration). For non-uniform density, usually no. The activation check is your first sanity gate.¹⁸
2. Construct C1 — variable: which quantity plays the role of x (or t)? The variable is the one that "varies" along the physical extent of the motion or distribution.
3. Construct C2 — differential: what does dW = F(x) dx, dm = ρ(x) dx, or v(t) dt physically represent? The differential is the slice thickness; without it, the integral loses its units.¹⁸
4. Construct C3 — limits: where does the physical extent start and end? The limits come from the geometry of the problem (natural length to stretched length, start time to end time, rod endpoint to endpoint), not from the domain of the integrand.
5. Reflect: do the units multiply correctly? Does the sign make physical sense? Does the magnitude match a sanity estimate?¹⁸

The ACER framework that organizes these questions comes from a 2019 Physics Education Research study (Wilcox and Corsiglia, "Cross-context look at upper-division student difficulties with integration"). The original study surveyed upper-division physics courses at CU Boulder; its findings generalize directionally but not as exact population statistics to introductory single-variable work. The framework's value here is the named list of failure modes, not the frequency numbers.¹⁸ ACER

The diagnostic grid for chapter 6's three scenarios is below.

Caption: Seven rows, one per ACER stage, listing the named pitfall, a concrete example, the frequency range from the paper's Tables I–III, and whether the difficulty fades with practice or persists. The frequencies cover upper-division multivariable contexts and should not be treated as exact population statistics for introductory single-variable work. The persistence labels are the diagnostic that drives chapter 7's "if stuck at X, re-read chapter Y" pointers.¹⁹

Worked example A: variable-force work. A spring follows Hooke's law F(x) = kx, with x the distance from the natural length. A 10 N force in the negative direction compresses the spring 0.2 m, so F(−0.2) = −10 gives k = 50 N/m and F(x) = 50x. How much work does it take to stretch the spring from natural length (x = 0) to x = 0.5 m?

• Variable: x, the extension from natural length.
• Limits: 0 to 0.5 m. These come from the physical extent of the motion, not from the domain of F.
• Differential: dW = F(x) dx. Each slice of thickness dx contributes force times thickness; units check: N · m = J.
• Setup: W = ∫₀^0.5 50x dx.
• Execute: Antiderivative 25x². Evaluate: 25(0.5)² − 25(0)² = 6.25 J.
• Reflect: Units check N·m = J ✓. Sign check: F(x) > 0 on [0, 0.5], so W > 0 ✓. Magnitude check: a 25 N average force over 0.5 m is 12.5 J expected if force were constant; 6.25 J is half that, which is the correct shape for a linear ramp from 0 to 25 N. ✓.¹⁶

Worked example B: displacement under non-uniform velocity. A particle moves with v(t) = 3t − 5 m/s from t = 0 to t = 3 s. Find the net displacement.

• Variable: t, time.
• Limits: 0 to 3 s, the time window of the motion.
• Differential: ds = v(t) dt. Units: (m/s)·s = m.
• Setup: s = ∫₀^3 (3t − 5) dt.
• Execute: Antiderivative (3/2)t² − 5t. Evaluate: (3/2)(9) − 15 − 0 = −3/2 m.
• Reflect: Units m ✓. Sign check: v(t) is negative on [0, 5/3] and positive on [5/3, 3], so the net displacement can plausibly be negative; physically, the particle spends the first 5/3 s moving backward before reversing. The negative sign is meaningful, not an error.¹⁶

Worked example C: mass of a non-uniform rod. A thin rod lies along the x-axis from x = a to x = b, with linear density ρ(x) kg/m. Find the mass.

• Variable: x, position along the rod.
• Limits: a to b, the endpoints of the rod.
• Differential: dm = ρ(x) dx. Units: (kg/m)·m = kg.
• Setup: m = ∫_a^b ρ(x) dx.
• Execute: depends on the form of ρ(x). For ρ(x) = kx, antiderivative (k/2)x², evaluated as (k/2)(b² − a²). For ρ(x) = 10e^(−0.1x), antiderivative −100 e^(−0.1x), evaluated as 100(e^(−0.1a) − e^(−0.1b)).¹⁶
• Reflect: units check kg/m · m = kg ✓. Sign check: ρ(x) > 0 for any physical rod, so m > 0 ✓. Limit check: if ρ(x) ≡ ρ₀ (constant density), the formula must reduce to m = ρ₀(b − a), which it does.

The differential is the single most common setup error across these three scenarios. Students drop the dx or attach it to the wrong variable and the unit check immediately fails. Wilcox and Corsiglia report that across their upper-division sample, 14–30% of physics-integral solutions involved dropping or mis-attaching the differential element, and the difficulty partially fades with practice for line distributions but not for the density side. The Nguyen-Rebello 2011 study, as paraphrased in Wilcox and Corsiglia (2019), reaches a similar conclusion for introductory contexts: students often drop the differential or attach it incorrectly, losing the physical meaning of the integrand.¹⁸

The activation step is the most under-appreciated of the four. About 20–30% of students in the upper-division sample substituted a simpler method (treating the rod as a point source, computing potential energy first, etc.) before realizing that direct integration was required. The persist-vs-fade classification from Wilcox and Corsiglia labels activation, limits, and reflection as persistent (they do not go away with more exposure) and the line-element form of the differential as one that partially fades. That pattern drives the next-and-final chapter's "if you got stuck at X, re-read chapter Y" pointer table.

A closing thought on the practice rhythm. The ACER rows that persist (activation, limits, difference-vector, reflection) need explicit re-reads of the corresponding chapter, not more practice problems. The rows that fade (line-element form, basic execution) need more problems. Stewart's main text and the 人大社《微积分》bilingual counterpart agree on chapter ordering for the three scenarios: Stewart 9e §5.4 (or §6.4 in Early Transcendentals) for variable-force work and the velocity → displacement example, §8.3 for the non-uniform-density rod. 人大社《微积分》第 6 版 ch. 5 covers the Net Change Theorem and ch. 8 covers the rod-mass application. If you got stuck on the activation step (tried a simpler method), re-read the opening of whichever Stewart section you were working in, and check that direct integration is genuinely required. If you got stuck on limits, re-read the worked example's limit-conversion paragraph. If you got stuck on the differential, re-read the micro-element discussion. If the answer came out numerically plausible but the units don't multiply to the right thing, the differential is the first place to look.

The reflex that closes the loop is: every definite integral is an antiderivative sandwiched between two limits, with a sanity check on the way out. Find F by reading the derivative table backwards; if F is a single clean symbol, the toolbox row handles it; if F is a composition, u-substitution handles it; if F is a product of unlike families, integration by parts handles it. Plug F into F(b) − F(a), let +C cancel, then check sign, units, and magnitude against the physics. That sequence is the path out of the stall.

References

The core sources cited in this tutorial; for deeper study, read the originals:

1. Weisstein, Eric W. "Indefinite Integral." MathWorld. — https://mathworld.wolfram.com/IndefiniteIntegral.html
2. Weisstein, Eric W. "Fundamental Theorems of Calculus." MathWorld. — https://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html
3. Khan Academy — AP® Calculus AB, Unit 7, Lesson 11: "Integrate using substitution." — https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-9
4. Wikipedia — "Integration by parts." — https://en.wikipedia.org/wiki/Integration_by_parts
5. Paul Dawkins — "Calculus II — Integration By Parts," Lamar University course notes. — https://tutorial.math.lamar.edu/classes/calcii/integrationbyparts.aspx
6. Sanderson, Grant — "Essence of Calculus, Chapter 8/11: Integration and the Fundamental Theorem of Calculus," 3Blue1Brown. — https://www.3blue1brown.com/lessons/integration/
7. mathmistakes.info — "Common Calculus Mistakes: Definite Integral" (Example/40) and "Indefinite Integral" (Example/37). — https://www.mathmistakes.info/mistakes/calculus/index.html
8. OpenStax — Calculus, Volume 1, Chapter 4 (Antiderivatives), §4.10. — https://openstax.org/books/calculus-volume-1/pages/chapter-4
9. Stewart, James; Clegg, Daniel; Watson, Saleem — Calculus, 9th Edition (Cengage, 2021), §5.4 / §6.4 / §8.3. LibreTexts mirror. — https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcendentals_(Stewart)
10. Wilcox, Bethany R.; Corsiglia, Vincent P. — "Cross-context look at upper-division student difficulties with integration," Phys. Rev. Phys. Educ. Res. 15, 020136 (2019). — https://journals.aps.org/prper/abstract/10.1103/PhysRevPhysEducRes.15.020136