Why does the integral of cos(3x+2) need a division by three? Differentiating sin(3x+2) produces 3cos(3x+2), so simply writing sin(3x+2) as the antiderivative is three times too large. Integration reverses differentiation, including the inside derivative. The most reliable habit is to differentiate the proposed answer and check that it returns the original integrand.
This independent eduKateSG Ministry of Education V3.0 Reference 391 develops the composite-function integration strand of C1 Differentiation and Integration in the 2027 SEC G3 Additional Mathematics K341 syllabus. It focuses on antiderivatives of (ax+b)^n, sin(ax+b), cos(ax+b) and e^(ax+b), the special form sec²x, rational powers, integration constants, and definite-evaluation checks. It complements the previously published signed-area chapter rather than repeating area-below-axis or two-curve questions.
All calculations below are invented educational examples, not official examination questions. The official C1 excludes finding area between two curves, and the power antiderivative formula needs an exception at exponent −1; this guide explicitly states the mathematical restrictions. It follows the immutable Clementi first-principles three-pupil standard without altering that article and is independent of MOE and SEAB.
1. The official C1 syllabus and the niche of this guide
The 2027 K341 C1 topic requires integration as the reverse of differentiation, power and basic trigonometric/exponential antiderivatives, and composite linear arguments including (ax+b)^n, sin(ax+b), cos(ax+b) and e^(ax+b). It also covers definite integrals and area or one-dimensional motion in separate applications. This article concentrates on how the inner linear coefficient changes an antiderivative and on checking indefinite answers. It is not a duplicate of a general signed-area guide, nor a claim that the syllabus requires substitution techniques for arbitrary advanced functions.
2. Integration reverses a derivative operation
If dF/dx=f(x), then an indefinite integral of f(x) is F(x)+C on an appropriate interval. The constant C appears because differentiation removes additive constants. For instance the derivative of x² is 2x, and so is the derivative of x²+7; both produce the same integrand. Omitting +C from an indefinite integration answer loses information about the full family of antiderivatives. A definite integral has numerical bounds and is evaluated as an endpoint difference instead, so a free constant cancels there.
3. The basic power rule carries a denominator
For n≠−1, ∫xⁿ dx=x^(n+1)/(n+1)+C on a suitable domain. The rule raises the exponent by one, then divides by the new exponent. For ∫x³ dx, answer x⁴/4+C, because its derivative is x³. Simply writing x⁴+C gives a derivative 4x³ and fails the reverse check. Rational exponents are also within C1, but real-domain restrictions matter when the result contains roots or negative powers. Exponent −1 must be treated separately.
4. Why the exponent −1 is exceptional
When n=−1, the ordinary power formula would divide by zero. Mathematically ∫(1/x)dx=ln|x|+C on intervals not crossing zero. This is a special logarithmic antiderivative, not a valid result of plugging −1 into a denominator n+1. A learner should first recognise the exceptional form rather than claim every rational power has the same formula. If x is restricted positive, ln x can be written; if negative values are possible, the absolute value is important. Such domain care also supports correct differentiation checks.
5. Constant factors can be taken outside
The integral of 5x² is 5×x³/3+C. A numerical coefficient outside the function is not the same as a coefficient multiplying x inside a composite argument. For ∫5(2x+1)³ dx, the factor five stays outside while the inside derivative two also has to be reversed. Students who handle every visible number identically may either omit the division by two or divide by five unnecessarily. Mark the outside multiplier and inside linear coefficient as different roles before starting.
6. Linear arguments require a reverse chain adjustment
Differentiating (ax+b)^(n+1) gives (n+1)a(ax+b)^n. Therefore, for a≠0 and n≠−1, the antiderivative of (ax+b)^n is (ax+b)^(n+1)/[a(n+1)]+C on a suitable domain. The factor a is not optional: it cancels the inside derivative produced by the chain rule. For (3x+1)², integration gives (3x+1)³/9+C. Differentiate to check: 3(3x+1)²×3/9=(3x+1)².
7. A negative inner coefficient changes the sign
The integral of (4−2x)³ is (4−2x)⁴/[−2×4]+C=−(4−2x)⁴/8+C. The negative sign reverses the derivative of the inside linear expression. A student who writes +(4−2x)⁴/8 has an antiderivative whose derivative is the negative of the desired integrand. One derivative check is more persuasive than a remembered slogan. The rule works with a negative a, provided a is nonzero and the real power is defined on the relevant interval.
8. Sinusoidal antiderivatives preserve their inside argument
Since d/dx[sin(ax+b)]=a cos(ax+b), ∫cos(ax+b)dx=sin(ax+b)/a+C when a≠0. Conversely d/dx[−cos(ax+b)]=a sin(ax+b), so ∫sin(ax+b)dx=−cos(ax+b)/a+C. The sine integral includes a minus sign, while the cosine integral does not, apart from any sign from a. Read whether the integrand is sine or cosine before adjusting its inner multiplier. The final derivative must recreate both the correct function and the exact original coefficient.
9. Exponentials with a linear argument behave similarly
Because d/dx[e^(ax+b)]=a e^(ax+b), its antiderivative is e^(ax+b)/a+C for a≠0. Thus ∫e^(5x−1)dx=e^(5x−1)/5+C. In a question with an outside factor 10, the result becomes 2e^(5x−1)+C. A learner should not confuse integrating e^x, where a=1, with e^(5x−1), where the linear chain contributes five. This simple difference is a central reason the reverse-chain method deserves focused practice.
10. The integral of sec²x is tan x
The derivative of tan x is sec²x on intervals where tangent is defined. Therefore ∫sec²x dx=tan x+C on any such interval. A common student error is to write sec³x/3 by treating the function name sec as an algebraic variable raised to a power; that ignores how trigonometric differentiation actually works. This syllabus-specific basic antiderivative can be verified directly by differentiation. The expression sec²(ax+b) also has a reverse-chain form tan(ax+b)/a under the corresponding mathematical conditions, but pupils should distinguish that extension from the exact elementary entry.
11. Sums and differences can be integrated term by term
Integration is linear: ∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx, with the independent constants combined into one final C. For 3x²+2cosx, an antiderivative is x³+2sinx+C. Terms may require different rules within the same problem. Do not place a separate C after every term and treat those constants as independent unknown functions. One combined arbitrary constant is sufficient for a single indefinite integral on a connected domain. Differentiate the entire result as a final check.
12. Reverse chain rule is not arbitrary substitution
A composite antiderivative with linear inside argument has a constant derivative a that can be cancelled by dividing by a. For more complicated inside functions, blindly dividing by the coefficient of x may not work. As an illustration, ∫cos(x²)dx is not sin(x²)/(2x)+C, because differentiating that quotient introduces extra terms. The 2027 C1 named forms include linear arguments, which is why this simple method works. A pupil should identify the inner expression and confirm its derivative is constant before applying the shortcut.
13. A definite integral evaluates endpoints, not +C
For a definite integral ∫_a^b f(x)dx, find an antiderivative F and compute F(b)−F(a). Adding an arbitrary constant to F makes no difference because it cancels. This guide uses definite examples to check reverse-chain calculations, while detailed signed-area and motion interpretations belong to the companion Reference 362. A student should label the upper and lower bounds and calculate them in the correct order. Reversing the bounds changes the sign, even when the integrand is nonnegative.
14. A constant of integration can be determined from a condition
If dy/dx=6x and a question gives y(1)=5, integration yields y=3x²+C. Substitute x=1 to obtain 5=3+C, so C=2. The resulting function is y=3x²+2. A student who forgets C cannot use the point condition properly and may assume that every antiderivative passes through the origin. In a composite example, the same method applies after reversing the chain rule. The condition supplies one particular function from a family.
15. A fraction outside is not automatically the inside derivative
For ∫(1/2)cos(3x)dx, the outside multiplier is one half, while the inside derivative of 3x is three. The antiderivative is (1/6)sin(3x)+C. A common error combines the numbers by guessing that a half and a three should produce three-halves. The reliable route writes the outside coefficient first and divides it by the nonzero inside coefficient. Differentiate the result: (1/6)×3cos3x=(1/2)cos3x. This short check proves the choice of numerical factor.
16. Negative powers and domain restrictions require care
For x>0, ∫x^(−2)dx=−x^(−1)+C, because differentiating −1/x yields +1/x². A power such as x^(−1/2) integrates to 2√x+C for x>0. The restrictions matter: expressions involving reciprocal powers are undefined at zero, and some real powers require positive or nonnegative bases. A student who applies a rule across a point where the integrand is undefined may incorrectly claim one antiderivative family works on both sides with one common constant. State the relevant interval when the algebra demands it.
17. Trigonometric integration retains radians in the standard calculus rules
The familiar derivatives d/dx(sinx)=cosx and d/dx(cosx)=−sinx use x as a radian angle. Calculus questions should be read using the conventional mathematical variable units, not a calculator’s degree-mode output. A student who treats sin(30x) as a degree expression without conversion may obtain an inconsistent derivative constant. The 2027 course includes trigonometric derivatives and integrals, so the argument’s linear coefficient and radian convention must be understood. Symbolic differentiation provides a check that does not depend on choosing arbitrary angle values.
18. The reverse-chain scale is determined by differentiation
Suppose the proposed answer for ∫sin(4x)dx is −Acos(4x)+C. Differentiating gives 4A sin(4x), so matching the original integrand requires 4A=1 and A=1/4. This reasoning makes the division by four inevitable rather than a memorised rule. For ∫7sin(4x)dx, matching requires 4A=7 and the antiderivative is −(7/4)cos(4x)+C. Students who construct the unknown coefficient and solve by matching derivatives can recover the technique even when they forget the formula.
19. Exponential integration with a negative coefficient
For ∫e^(2−3x)dx, the inside derivative is −3, so an antiderivative is −(1/3)e^(2−3x)+C. Differentiating it produces −1/3×−3×e^(2−3x)=e^(2−3x), as required. The negative sign is a consequence of the decreasing linear argument. A student who uses +1/3 simply because exponentials are positive produces an antiderivative with the wrong derivative. Positivity of the function does not determine the sign of the antiderivative coefficient; the chain derivative does.
20. Integration of (ax+b)^n needs the correct combined denominator
The denominator in ∫(ax+b)^n dx is a(n+1) for n≠−1, not merely a or n+1 alone. For ∫(5x−2)³dx, the answer is (5x−2)⁴/20+C because a=5 and n+1=4. Check: differentiating multiplies by four and then five, cancelling twenty. If a pupil divides only by four, the derivative returns five times the integrand. A factor-by-factor derivative check is particularly helpful when the powers or coefficients are fractional.
21. Indefinite and definite answers have different final forms
An indefinite result such as ∫2x dx=x²+C describes a family of functions. A definite result such as ∫_0^3 2x dx=[x²]_0^3=9 is a number with its interpretation depending on context. Adding +C to the final numerical value is not meaningful; C cancels in the endpoint subtraction. Some pupils wrongly remove C from every indefinite answer after practising definite integrals. The question’s bounds are the decisive clue. If limits are present, evaluate F(b)−F(a); if they are absent, give the antiderivative family.
22. Numerical differentiation can supplement symbolic checking
The exact check is to differentiate the answer algebraically, but a calculator can also compare numerical slopes and integrand values at a few valid points as a coarse diagnostic. This does not prove a general identity; different functions can agree at isolated points. A symbolic check is stronger because it establishes equality of derivatives on the stated interval. If a pupil finds a numerical mismatch, revisit the inside coefficient, exponent increment or trigonometric sign. The calculator should help detect mistakes, not replace a valid antiderivative argument.
23. A two-term integral can require different inverse rules
For ∫[4(3x+1)²+6cos(2x)]dx, integrate each term independently: 4(3x+1)³/9+3sin(2x)+C. Differentiating the first returns 4(3x+1)² and the second returns 6cos2x. Combining them into one antiderivative is valid because integration is linear. A student who tries to treat the whole sum as one composite function with one shared inside derivative will usually make an unjustified manipulation. Identify each summand’s structure before integrating.
24. Independent transfer changes the form and the inside sign
After solving ∫(3x+1)²dx, a pupil should handle ∫(1−4x)^(1/2)dx, ∫sin(−2x+3)dx and ∫e^(7−x)dx. These require the same reverse-chain idea but different outer functions and possible negative inside coefficients. The transferable routine is to identify the outer antiderivative, compute the derivative of the inside linear argument, divide by that factor, include +C and differentiate back. A learner who explains those steps on an unseen example has mastered the concept rather than a single repeated template.
25. Twelve-stage Reverse–Scale–Differentiate method
Stage 1 — Identify whether bounds are present
Indefinite integrals need a constant C; definite integrals need endpoint evaluation instead.
Stage 2 — Recognise the outer function
A power, sine, cosine, exponential or secant squared has a different antiderivative pattern.
Stage 3 — Identify the full inner expression
Use the whole linear argument ax+b, including a negative coefficient.
Stage 4 — Check the inner derivative
For the named composite forms, its derivative is the constant a.
Stage 5 — Apply the correct base rule
For powers n≠−1, increase exponent and divide; for sine, use negative cosine; for cosine, use sine.
Stage 6 — Cancel the inner derivative
Divide by a once, provided a≠0, so differentiation can reconstruct the integrand.
Stage 7 — Retain outside numerical factors
Multipliers outside the integrand remain and combine with the reverse-chain scale.
Stage 8 — Treat special cases separately
The n=−1 power case leads to a logarithm; sec²x integrates to tanx where defined.
Stage 9 — Preserve suitable domain restrictions
Avoid claiming a real antiderivative across points at which the function is undefined.
Stage 10 — Add C or evaluate limits
Keep the correct final answer form according to bounds and any given condition.
Stage 11 — Differentiate to verify
Check that the full derivative, including the inner chain factor, matches the original integrand.
Stage 12 — Transfer without a template
Solve a new linear argument with changed sign, power and trigonometric or exponential form.
26. Fictional three-pupil calculus tutorial
Three imaginary G3 learners examine ∫4(3x+1)²dx. Pupil A recognises the outer power and proposes a cube, pupil B computes the inside derivative three and the correct denominator, and pupil C differentiates 4(3x+1)³/9+C to verify the integrand. The tutor then changes the problem to ∫sin(2−5x)dx, requiring attention to the negative inner coefficient. After each learner rotates roles, all three solve a different unfamiliar exponential or fractional-power question independently. The teacher diagnoses the first wrong choice—outer rule, exponent exception, inner coefficient, integration constant or domain—then retests the repaired idea with different numbers.
27. Twenty-six original worked examples
Worked Example 1 — Basic fourth power
∫x⁴dx=x⁵/5+C. Differentiating x⁵/5 gives x⁴ because the factor five from differentiation cancels the denominator. The exponent increases before the division, not afterward.
Worked Example 2 — A coefficient outside a power
∫6x²dx=2x³+C. The derivative of 2x³ is 6x². The outside six is handled by ordinary linearity and is not a reverse-chain adjustment.
Worked Example 3 — Negative second power
∫x⁻²dx=−x⁻¹+C on an interval excluding zero. Increasing exponent −2 by one gives −1, so division by −1 creates the minus sign. Differentiation checks the result.
Worked Example 4 — The exceptional reciprocal
∫(1/x)dx=ln|x|+C for x≠0 on a connected domain interval. The usual power denominator n+1 would be zero at n=−1, so a separate logarithmic rule is needed.
Worked Example 5 — Composite cube with positive a
∫(2x+1)³dx=(2x+1)⁴/8+C. The denominator is inside derivative two multiplied by new exponent four. Differentiation gives 4×2/8 times the original cube, exactly one.
Worked Example 6 — Composite square with a=3
∫(3x−2)²dx=(3x−2)³/9+C. The new exponent three and inside derivative three create the denominator nine. A quick derivative verifies the cancellation.
Worked Example 7 — Negative inner coefficient with square
∫(4−x)²dx=−(4−x)³/3+C. The inside derivative is −1. The negative antiderivative coefficient cancels it when differentiated, returning the required positive square.
Worked Example 8 — Composite cube with negative a
∫(2−3x)³dx=−(2−3x)⁴/12+C. The denominator is four times −3; differentiation contributes those two factors and restores the original integrand.
Worked Example 9 — Outside five and inside two
∫5(2x+3)²dx=5(2x+3)³/6+C. The outside factor five remains while the denominator combines the new exponent three with inside derivative two.
Worked Example 10 — Outside four and inside three
∫4(3x+1)²dx=4(3x+1)³/9+C. Differentiating gives (4/9)×3×3(3x+1)², which equals 4(3x+1)² as required.
Worked Example 11 — Negative composite power
∫(2x+1)⁻²dx=−1/[2(2x+1)]+C on intervals excluding x=−1/2. The power increases to −1 and the denominator includes both −1 and inside derivative two.
Worked Example 12 — Square-root composite power
∫√(1+4x)dx=(1+4x)^(3/2)/6+C on a suitable real domain. The new exponent is 3/2 and inside derivative four, giving denominator six.
Worked Example 13 — Inverse-square-root with negative a
∫(5−2x)⁻¹ᐟ²dx=−√(5−2x)+C on a domain with positive radicand. The new exponent is 1/2, multiplied by inside derivative −2, giving −1.
Worked Example 14 — Cosine with a=3
∫cos(3x)dx=sin(3x)/3+C. Differentiation of sine brings down a factor three, which the denominator cancels. Forgetting it makes the answer three times too large.
Worked Example 15 — Outside six on cosine
∫6cos(2x)dx=3sin(2x)+C. The outside six is divided by inside derivative two. Differentiate the proposed answer: 3×2cos(2x)=6cos(2x).
Worked Example 16 — Sine with a=4
∫sin(4x)dx=−cos(4x)/4+C. The negative is part of the sine antiderivative and the division by four reverses the chain derivative.
Worked Example 17 — Outside seven on sine
∫7sin(4x)dx=−7cos(4x)/4+C. Differentiating yields −7/4×−4sin(4x)=7sin(4x). Both minus sign and scale matter.
Worked Example 18 — Sine with negative inner a
∫sin(2−5x)dx=cos(2−5x)/5+C. The ordinary sine antiderivative contributes negative cosine, and division by inner derivative −5 cancels the minus.
Worked Example 19 — Cosine with negative inner a
∫cos(1−2x)dx=−sin(1−2x)/2+C. Differentiating gives −1/2×−2cos(1−2x), which returns the original cosine.
Worked Example 20 — Exponential with a=5
∫e^(5x−1)dx=e^(5x−1)/5+C. The exponential function remains the outer form, while inside derivative five must be reversed.
Worked Example 21 — Exponential with negative a
∫e^(2−3x)dx=−e^(2−3x)/3+C. The negative coefficient is necessary because differentiating the exponential creates a factor −3.
Worked Example 22 — Outside ten on an exponential
∫10e^(5x−1)dx=2e^(5x−1)+C. Outside ten divided by inner derivative five gives two, and differentiating confirms the original factor ten.
Worked Example 23 — Secant-squared basic form
∫sec²x dx=tan x+C wherever tan x is defined. It is a named trigonometric antiderivative and should not be confused with a polynomial power rule applied to the word sec.
Worked Example 24 — Sum of power and cosine
∫(3x²+2cosx)dx=x³+2sinx+C. Each summand uses its own antiderivative, and the two arbitrary constants combine into one final constant C.
Worked Example 25 — A given point determines C
If dy/dx=6x and y(1)=5, integrate to y=3x²+C. Then 5=3+C gives C=2, so the particular function is y=3x²+2.
Worked Example 26 — Definite reverse-chain check
∫₀²3(2x+1)²dx=[(2x+1)³/2]₀²=(125−1)/2=62. The constant cancels between endpoints. Differentiating the antiderivative recreates 3(2x+1)².
28. Ten mistakes and targeted repairs
1. Forgetting the chain factor
For ∫cos(3x)dx, the correct antiderivative is sin(3x)/3+C rather than sin(3x)+C. Differentiate to see the missing factor three. This diagnosis is about the inner linear argument, not the ordinary cosine integral.
2. Applying the power rule without dividing by the new exponent
The integral of x⁴ is x⁵/5+C, not x⁵+C. A correct derivative of the proposed answer must recreate the original power. Show the derivative explicitly when debugging.
3. Using the power rule at exponent minus one
For ∫1/x dx, n+1 would be zero. Use ln|x|+C on a valid domain interval rather than dividing by zero. This is an exceptional antiderivative, not a reason to abandon all rational powers.
4. Losing the sine integral’s negative sign
The antiderivative of sinx is −cosx+C because d/dx(−cosx)=sinx. With a composite linear argument, also divide by its derivative. A positive cosine alone gives the wrong sign.
5. Ignoring a negative inside coefficient
The integral of e^(2−3x) contains −1/3 as a multiplier. If the answer is +e^(2−3x)/3, differentiation gives the negative integrand, revealing the missing sign.
6. Dividing by the outer coefficient rather than the inside derivative
For ∫5(2x+3)²dx, keep five outside and divide by inside two and new exponent three. The answer 5(2x+3)³/6+C retains the correct factor.
7. Adding C to the final definite integral
An indefinite antiderivative includes a family constant. A definite integral’s numerical endpoint difference cancels that constant. Do not append +C to a final evaluated number.
8. Omitting C from indefinite integration
If there are no bounds and no special condition selecting one solution, include +C. Differentiation cannot recover additive constants, so a single function is not the complete antiderivative family.
9. Ignoring the real domain of reciprocal or root forms
An integrand containing 1/x is undefined at zero, and one involving square roots can restrict the linear argument. State an interval where the real expression and its antiderivative are valid.
10. Applying reverse chain to an arbitrary nonlinear inside expression
The simple division by a constant works for specified linear arguments ax+b. For an inside expression x², its derivative is 2x, so an unrelated division by two cannot justify the same shortcut. Check the chain derivative before using a formula.
29. A ninety-minute small-group sequence
Begin with a retrieval test on derivatives of powers, sine, cosine and exponentials. Give the three students a simple composite power and ask each to explain a distinct operation: choose the outer antiderivative, calculate inside derivative and verify by differentiating the answer. Rotate roles and introduce negative inner coefficients, then the exceptional reciprocal power and sec². The final segment uses a two-term integral and one definite example. The tutor records whether a pupil’s first mistake was the outer rule, new exponent, inside scale, missing constant or domain, and assigns a fresh independent example to retest the repaired point. A short mixed set the following lesson is more effective than memorising one repeated form.
30. Depth–Load–Transfer diagnosis
Depth means understanding why antiderivatives are checked through differentiation, including a chain factor and arbitrary constant. Load asks a pupil to work with a fractional or negative power, a negative linear argument, a trigonometric term and a choice between definite and indefinite forms. Transfer requires choosing the appropriate rule when no formula label is provided. The student should be able to explain why the denominator is a(n+1) in a power case and a alone in a cosine or exponential case. An accurate final answer is only one part of mastery; the reverse check is the proof that the method worked.
31. Independent practice prompts
Integrate (2x−1)⁴, cos(5x), sin(3−2x), 7e^(4x+1), (1+3x)^(−1/2) on a suitable domain and 4sec²x. For each, state the inside derivative, the antiderivative and the result of differentiating back. Next solve dy/dx=4x with y(1)=6 to find the particular constant, and evaluate ∫₀¹(3x+2)²dx using the appropriate antiderivative and endpoint subtraction. Mixed practice should require the pupil to identify the method without a topic heading telling them which rule to use.
32. Frequently asked questions
What does integration reverse?
Differentiation. An antiderivative F satisfies F′=f, with an additive constant for indefinite integration.
Why divide by a in (ax+b)?
Differentiation of the composite form multiplies by a; integration must cancel that factor.
How do I integrate sin(ax+b)?
Use −cos(ax+b)/a+C when a≠0.
How do I integrate cos(ax+b)?
Use sin(ax+b)/a+C when a≠0.
How do I integrate e^(ax+b)?
Use e^(ax+b)/a+C when a≠0.
What about (ax+b)^n?
Use (ax+b)^(n+1)/[a(n+1)]+C for n≠−1 and valid real domains.
What if n=−1?
Use the logarithmic exception on intervals where the linear argument is nonzero.
What is ∫sec²x dx?
tanx+C where defined.
Why is +C needed?
Differentiation eliminates arbitrary constants, so indefinite integration has a family of answers.
What is the most reliable check?
Differentiate the entire answer and verify that it equals the original integrand on the relevant domain.
33. Closing principle
Integration of a named composite function is reverse chain rule in action. The inside linear derivative is what determines the scaling, not a number guessed from the appearance of the problem. A strong learner can identify the original function, select the correct antiderivative, reverse the inside multiplier, include the appropriate constant or bounds, and prove the result by differentiating. This sequence is compact, exact and transferable across power, trigonometric and exponential forms.
Official syllabus and connected references
Official SEAB 2027 G3 Additional Mathematics K341 C1 · Reference 362 — Signed Area and Motion · Reference 363 — Differentiation · MOE V3.0 reference library · Immutable Clementi first-principles benchmark.
34. Final diagnostic: differentiate before accepting the answer
A robust audit begins before the final line is submitted. If a student proposes F(x) as the integral of a composite expression, ask them to calculate F′(x) without looking at the original working. For F(x)=cos(2−5x)/5, differentiation produces [−sin(2−5x)×(−5)]/5=sin(2−5x); notice that this positive cosine expression works for the integrand, whereas −cos(2−5x)/5 would differentiate to −sin(2−5x). For an exponential, check the linear coefficient inside the exponent. For a power, check both the new exponent and the derivative of the linear argument. Finally ask whether the original question had bounds, a given point or neither. That determines whether the final answer is a number, a particular function or a family with +C. Requiring one full reverse check trains learners to locate errors rather than guess a different coefficient whenever a solution feels uncertain.