To find a curve when its gradient function is given, integrate first, then use a point or another stated condition to determine the constant. Differentiation loses constant information. Integration restores a family of possible functions, and the extra condition selects the member that answers the question.
Integration as the reverse of differentiation is included in the 2027 SEC G2 K232 and G3 K341 Additional Mathematics syllabuses. Their permitted function types differ; the polynomial core below serves both, while trigonometric and exponential examples are marked G3. See K232 page 8 and K341 page 8. The reconstructed-curve problems are original applications of that reverse relationship.
This Punggol guide answers a specific question that differs from finding area under a curve: how do we recover the unknown function itself? It explains gradient versus height, one constant versus two, compatible versus contradictory conditions, and the final checks that make a solution convincing.
Quick example: if dy/dx=6x−4 and the curve passes through (2,5), integration gives y=3x²−4x+C. Substituting the point gives 5=12−8+C, so C=1. The curve is y=3x²−4x+1. Differentiate it and test the point to verify both pieces of information.
Find the missing piece of the calculus problem
- Why a derivative does not determine one unique curve
- Recovering a curve using one point
- Second derivatives and two conditions
- Gradient, normal and area conditions
- G3 trigonometric, exponential and motion examples
- Error diagnosis, independent practice and parent questions
Many curves can have the same gradient function
The functions y=x², y=x²+3 and y=x²−7 all have derivative 2x. Adding a constant moves the graph vertically without changing its tangent gradient at any fixed x. Therefore knowing dy/dx=2x is not enough to decide which of those curves was intended.
The general antiderivative y=x²+C records that missing vertical position. C can be any real constant unless more information is supplied. Writing C is not a ritual added for an examiner; it states honestly that the given derivative alone leaves a degree of freedom.
A point on the curve can fix that freedom. If y(0)=3, then C=3. If y(1)=5, then C=4. Different extra conditions select different curves while leaving the same derivative. That is the central idea behind every example in this chapter.
Gradient is not the same quantity as height
For y=x²+1 at x=3, the function value is ten and the gradient is six. A point (3,10) tells us where the graph is. The derivative value six tells us how steeply it is changing there. These two pieces of information answer different questions.
If a problem says the curve passes through (3,10), substitute x=3 and y=10 into the recovered function, not into dy/dx. If it says the tangent gradient at x=3 is six, substitute into the derivative. The notation tells the learner which relationship should receive the condition.
A useful readiness check offers one point condition and one gradient condition and asks the student to identify the correct equation for each before calculating. This isolates an interpretation error that might otherwise appear to be an integration difficulty.
Reverse the power rule and check it immediately
If the derivative is 6x−4, an antiderivative is 3x²−4x+C. The exponent of x increases by one when integrating the first term, and the coefficient is divided by that new power. A constant term such as −4 integrates to −4x, because differentiating −4x recovers −4.
Students sometimes integrate 6x as 6x² or leave −4 unchanged. Differentiate the proposed answer to expose the mistake. The first gives 12x; the second would contribute zero rather than the required negative constant.
Do this check before substituting a point. Otherwise an incorrect antiderivative can be made to pass through the given point by choosing a constant, disguising the fact that it still has the wrong gradient everywhere else.
Worked example 1: one gradient function and one point
Given dy/dx=6x−4 and y=5 when x=2, integrate to obtain y=3x²−4x+C. Now apply the point condition: 5=3(2)²−4(2)+C=4+C. Hence C=1 and y=3x²−4x+1.
The derivative check returns 6x−4. The point check returns y(2)=12−8+1=5. Both pass. These are independent obligations: the first tests the reverse-calculus step, while the second tests the chosen member of the antiderivative family.
The final answer should be the equation of the curve, not merely C=1. The constant was an intermediate unknown. When the command asks for y in terms of x, provide that complete expression and retain any stated domain.
Worked example 2: a point with negative coordinates
Suppose f′(x)=4x+3 and the curve passes through (−1,2). Integration gives f(x)=2x²+3x+C. Substitute the whole negative input with brackets: 2=2(−1)²+3(−1)+C=−1+C.
Therefore C=3 and f(x)=2x²+3x+3. The derivative is 4x+3, and f(−1)=2−3+3=2. A sign error in the substituted square would produce the wrong constant even if the antiderivative was otherwise correct.
This is a good place to distinguish (−1)² from −1². The point supplies a negative number as the complete input, so the square applies to that whole number. Brackets communicate exactly which quantity is being raised to a power.
Worked example 3: several polynomial terms
Given f′(x)=12x²−6x+2 and f(1)=7, integrate term by term: f(x)=4x³−3x²+2x+C. At x=1, the nonconstant terms total 4−3+2=3. Therefore C=4.
The curve is f(x)=4x³−3x²+2x+4. Differentiate every term to recover 12x²−6x+2. Then evaluate at one to obtain seven. The final constant has no effect on the derivative, but it is essential to the point condition.
A common incomplete response integrates only the highest-power term and forgets the others. The problem asks for an antiderivative of the entire gradient function. Matching term by term is a useful organisational check before any numerical substitution.
Worked example 4: a reciprocal power on a stated interval
Suppose f′(x)=2−3/x² for x>0, and f(1)=6. Rewrite the derivative as 2−3x⁻². Its antiderivative is f(x)=2x+3x⁻¹+C, or 2x+3/x+C, on the given interval.
At x=1, 6=2+3+C, so C=1. The result is f(x)=2x+3/x+1 for x>0. Differentiating 3/x gives −3/x², confirming the sign that students often reverse.
The original interval should remain with the answer. The expression is undefined at zero, so a point condition there could not be used. Do not treat the appearance of a formula as permission to apply it outside the domain stated in the problem.
Worked example 5: a fractional power
Given f′(x)=3√x for x>0 and f(1)=5, write √x=x^(1/2). Integration gives f(x)=2x^(3/2)+C because three divided by three-halves is two. Using the point gives 5=2+C, hence C=3.
The function is f(x)=2x^(3/2)+3 on the stated interval. Differentiation returns 2×(3/2)x^(1/2)=3√x. Substitution at one returns five. Both exact checks avoid unnecessary decimal approximations.
When fractional indices cause the difficulty, practise the arithmetic of dividing by a fraction separately. A pupil who understands the reverse-calculus idea may still need that earlier number operation repaired before the coefficient becomes reliable.
Worked example 6: an inner linear expression
Suppose f′(x)=6(2x+1)² and f(0)=4. A convenient antiderivative is f(x)=(2x+1)³+C. Differentiating that cube gives 3(2x+1)²×2=6(2x+1)², exactly the given derivative.
At x=0, 4=1+C, so C=3. The recovered curve is f(x)=(2x+1)³+3. Expanding first and integrating the resulting polynomial would also work, but the compact form makes the derivative check particularly clear.
The coefficient six was not ignored. It is accounted for by the new power three and inner derivative two. In a changed example, the multiplier may not fit so neatly, so the learner should verify rather than copy this particular coefficient pattern.
A second derivative loses two pieces of information
Knowing f″(x) requires two integrations to recover f(x). The first produces a constant in f′, and the second produces another constant in f. Those constants play different roles and should be given different names.
For f″(x)=6x, the first integration gives f′(x)=3x²+A. The second gives f(x)=x³+Ax+B. A affects the slope function and contributes a linear term to the original curve; B shifts the curve vertically without affecting its derivative.
This explains why a single point generally cannot identify the whole curve from a second derivative. It gives one equation relating A and B, leaving a family of possibilities unless another independent condition is provided.
Worked example 7: second derivative, slope and point
Given f″(x)=6x, f′(1)=2 and f(1)=4, integrate once: f′(x)=3x²+A. The slope condition gives 2=3+A, so A=−1. Therefore f′(x)=3x²−1.
Integrate again: f(x)=x³−x+B. The point condition gives 4=1−1+B, so B=4. The answer is f(x)=x³−x+4.
Check all three conditions, not only the last one. The second derivative is 6x, the first derivative at one is two, and the function value at one is four. A solution that passes one of these while failing another has not reconstructed the required curve.
Worked example 8: second derivative and two points
Suppose f″(x)=2, f(0)=1 and f(2)=9. Integrating twice gives f′(x)=2x+A and f(x)=x²+Ax+B. The first point immediately gives B=1.
The second point gives 9=4+2A+1, so A=2. Hence f(x)=x²+2x+1. The curve has second derivative two and passes through both specified points.
This example links calculus to simultaneous-equation thinking. The unknown constants are determined by independent conditions. If the same point were merely repeated twice, it would not provide two independent equations and would not fix both constants.
Worked example 9: a stationary point supplies two conditions
Let f″(x)=2 and suppose the curve has a stationary point at (3,5). The phrase “at (3,5)” gives f(3)=5, while “stationary” gives f′(3)=0. These are two different conditions, both needed.
From f′(x)=2x+A, the stationary condition gives A=−6. Then f(x)=x²−6x+B. Using f(3)=5 gives 9−18+B=5, so B=14. The curve is f(x)=x²−6x+14=(x−3)²+5.
The completed-square form supplies an independent geometric check: the vertex is indeed (3,5), and the curve opens upward. The second derivative is positive, so the stationary point is a minimum. A student who used only the coordinate or only the zero slope would have left information unused.
Worked example 10: determine a coefficient before the constant
Suppose f′(x)=kx+2, the tangent gradient at x=2 is six, and f(2)=9. The gradient condition belongs in f′: 6=2k+2, so k=2. We now know f′(x)=2x+2.
Integrating gives f(x)=x²+2x+C. The point gives 9=4+4+C, hence C=1. The recovered curve is f(x)=x²+2x+1.
The given point and slope did different jobs. The slope fixed an unknown coefficient in the derivative, while the point fixed the integration constant. Reversing those roles would confuse the function with its rate of change and produce an incorrect system for the unknowns.
Worked example 11: a normal gradient must be converted first
Suppose the same derivative f′(x)=kx+2 is given, but the normal at x=2 has gradient −1/6. A normal is perpendicular to the tangent. Since the stated normal slope is finite and nonzero, the tangent slope is six.
Therefore f′(2)=6, not −1/6. With a point condition f(2)=9, the remaining calculation is the preceding example: k=2, C=1 and f(x)=x²+2x+1.
Check the geometry after recovering the curve. At x=2 its derivative is six, so a normal through (2,9) has slope −1/6. The two gradients multiply to −1. A learner who substitutes the normal slope directly into f′ has made a geometric interpretation error before integration begins.
Worked example 12: a line can encode point and gradient information
Suppose a curve has derivative f′(x)=2x+k, and its tangent at x=1 is y=5x−2. The line’s gradient is five, so f′(1)=5 and k=3. The contact point also lies on the line, so f(1)=5−2=3.
Integration gives f(x)=x²+3x+C. Substituting the contact point gives 3=1+3+C, so C=−1. Thus f(x)=x²+3x−1.
Verify that f(1)=3 and f′(1)=5. The tangent formula y−3=5(x−1) simplifies to y=5x−2, reproducing the supplied line. This reverse check tests the whole problem, not just one algebraic coefficient.
Worked example 13: an integral condition fixes a vertical shift
Suppose f′(x)=2x and the definite integral of f(x) from zero to two is 14/3. First reconstruct the family f(x)=x²+C. The integral condition then gives [x³/3+Cx] from zero to two =8/3+2C=14/3.
Solving gives 2C=2, so C=1 and f(x)=x²+1. Differentiate to verify the gradient and integrate over the stated interval to verify the total. The condition fixes C because shifting the curve vertically changes its accumulation over a nonzero-width interval.
This is a mathematical reconstruction problem, not a claim that every such combination is a separate named syllabus topic. Its ingredients are the reverse derivative and a definite integral. Read whether the condition states a signed integral or a geometric area before interpreting it.
What insufficient information looks like
If all we know is f′(x)=2x, the correct conclusion is the family f(x)=x²+C. Picking C=0 without a condition would invent information. The graph might pass through the origin, but the problem has not said so.
Likewise, f″(x)=2 and f(0)=1 give f(x)=x²+Ax+1 with A still free. A second point, a slope or another independent condition is needed to identify one curve. A long calculation cannot create a condition that was never supplied.
A student should learn to state this limitation positively and precisely: “The given information determines this family; one constant remains unspecified.” That is a complete mathematical conclusion when uniqueness is not justified.
What contradictory information looks like
Suppose a problem states f′(x)=2x, f(0)=3 and f(1)=5 on an interval containing both points. The first two conditions give f(x)=x²+3. That function has f(1)=4, not five. No differentiable function on that interval satisfies all three stated conditions.
The correct response is to identify the inconsistency, not adjust the constant separately for each point. C is constant throughout the connected interval being considered. Using C=3 at zero and C=4 at one would describe different curves.
This example is useful for testing whether the learner understands the role of a constant. It also encourages verification of supplied or copied data before assuming every worksheet must contain a consistent unique solution.
One interval and a fixed constant
The statement that antiderivatives differ by a constant applies on an interval where the functions are differentiable. If an expression is undefined at a point splitting the real line, conditions on one side do not automatically determine a constant on the other side.
For routine school questions, a stated domain such as x>0 often supplies the needed interval. Keep that domain instead of silently extending the recovered function across an excluded point. This is especially relevant for reciprocal powers and other expressions with zero denominators.
The aim is not to turn every problem into advanced analysis. It is to preserve the domain already given and avoid claiming a unique global function when the information only establishes a curve on a particular interval.
G3 example: recover a trigonometric curve
Suppose f′(x)=4cos(2x), with x in radians, and f(0)=3. A suitable antiderivative is f(x)=2sin(2x)+C, because differentiation gives 2×2cos(2x). The point condition gives C=3.
The curve is f(x)=2sin(2x)+3. Its derivative matches the given gradient, and its value at zero is three. The constant changes the midline but not the derivative, amplitude or period of the sine component.
The new outer function does not change the reconstruction logic: integrate, use the condition and verify. A learner already secure with polynomial recovery can transfer that reasoning once the relevant G3 trigonometric antiderivative is understood.
G3 example: recover an exponential curve
Given f′(x)=3e^(3x) and f(0)=5, integration gives f(x)=e^(3x)+C. The factor three is accounted for by the inner derivative of 3x when checking. At zero, 5=1+C, so C=4.
The recovered function is f(x)=e^(3x)+4. It is not 3e^(3x)+4: differentiating that incorrect expression would produce 9e^(3x). The point alone cannot detect every coefficient mistake, so the derivative check remains essential.
Keep e^(3x) exact unless a numerical evaluation is requested. Rounding e or an intermediate exponential value is unnecessary when the question asks for an equation of a curve.
G3 motion example: acceleration, velocity and position
In an original one-dimensional model, let a(t)=6t metres per second squared, with v(0)=2 metres per second and s(0)=5 metres. Integrating acceleration gives v(t)=3t²+A. The initial velocity fixes A=2.
Integrating velocity gives s(t)=t³+2t+B. The initial position fixes B=5. Thus v(t)=3t²+2 and s(t)=t³+2t+5 in the model. At t=1, velocity is five metres per second and position is eight metres.
The displacement from t=0 to t=1 is three metres, not eight. Eight is the final position relative to the chosen origin. This example connects two integrations with two physical conditions; it does not describe any measured Punggol journey.
Initial does not always mean at zero
A condition can be given at any permitted input. If v(t)=2t+1 and position s=10 when t=2, integration gives s(t)=t²+t+C. Substituting t=2 gives 10=6+C, so C=4.
Calling C the “starting value” without checking the function can be misleading. C equals s(0) in this particular polynomial because the other terms vanish at zero, but the given reference condition was at t=2. In a function containing e^t, for example, the nonconstant term does not vanish at zero.
The reliable method is always substitution into the recovered function. Do not assign C directly from a y-coordinate merely because the word “initial” or “passes through” appears somewhere in the question.
A curve can pass the point check and still be wrong
Suppose a student incorrectly integrates dy/dx=6x as y=6x²+C. If the point is (1,5), they choose C=−1 and obtain y=6x²−1. The point check passes: y(1)=5. But differentiating gives 12x, not the required 6x.
The correct integration is y=3x²+C, and the point then gives C=2. Therefore y=3x²+2. Both proposed curves pass through the point, but only one has the stated gradient function.
This example explains why “I substituted the point and it worked” is not a complete verification. Every independent item of given information creates an obligation that must be checked.
A second integration must integrate the first constant
From f″(x)=6x, the first integration is f′(x)=3x²+A. Integrating again gives f(x)=x³+Ax+B. A student who writes x³+A+B has failed to integrate the constant A with respect to x.
Differentiate x³+A+B and the A contribution disappears, giving only 3x² rather than 3x²+A. Differentiate x³+Ax+B and A is correctly recovered. The reverse check makes the role of Ax visible without requiring another memorised rule.
Keep two distinct constant names and record which condition determines each. Reusing C for both can conceal an accidental merger of two independent unknowns into one.
When the student cannot decide what to do first
Ask the learner to underline the given function and circle the requested output. If the given quantity is f′ and the target is f, integration is needed once. If the given quantity is f″ and the target is f, integration is needed twice. If the question already gives f and asks for a tangent gradient, the direction is differentiation instead.
Next classify every condition: point, slope, second derivative, integral total or geometric relation. Place each condition beside the equation to which it belongs. This reduces the temptation to substitute all numbers into the first expression on the page.
The diagnosis may reveal a notation problem rather than a calculation problem. A pupil who does not distinguish f from f′ will benefit from a graph and a simple numerical comparison before being asked to integrate more complicated functions.
Independent checkpoint with complete answer checks
Task A: f′(x)=8x+1 and f(1)=7. Integration gives f(x)=4x²+x+C, so C=2 and f(x)=4x²+x+2. Differentiate and evaluate at one. Task B: f′(x)=3x²−2 and f(0)=6. The curve is f(x)=x³−2x+6.
Task C: f″(x)=4, f′(0)=−3, f(0)=2. Two integrations give f′=4x−3 and f=2x²−3x+2. Task D: f′(x)=2x and f(0)=1. Can the same curve also satisfy f(2)=6? No: the first two conditions imply f(2)=5.
These tasks test one constant, a simple initial value, two constants and inconsistent conditions. A learner should explain why each answer has the stated level of certainty. Four correct final expressions without any derivative or condition checks provide less diagnostic information.
A teaching session that tests reconstruction, not copying
An illustrative 90-minute session can use ten minutes distinguishing values from gradients, twenty minutes deriving one complete curve, fifteen minutes on guided variants and twenty minutes on independent problems. Reserve fifteen minutes for two-condition questions and ten for a final error explanation and follow-up task.
In a three-student group, one pupil can differentiate a proposed answer, another test the point and a third explain which condition fixes which constant. Each then needs a different unaided problem. The purpose of discussion is to make reasoning visible, not to produce three identical copies of the board.
A later check should change the function or move the reference point away from zero. This reveals whether the student learned the reconstruction method or merely the convenient pattern in the first demonstration. The schedule is a teaching example, not a guaranteed course-completion period.
Parent questions answered
Why is there a constant after integration?
All vertical translations of a curve have the same derivative. The constant records the information differentiation cannot recover on its own. A point or another independent condition is needed to select one curve from the family.
Can the constant be set to zero for convenience?
Only when a given condition actually implies zero. Otherwise that choice invents information and may exclude the intended curve. If no suitable condition is supplied, state the family with its undetermined constant rather than guessing.
Why do two integrations produce two constants?
Each integration reverses a derivative that has lost constant information. After the first integration, the constant becomes part of the first derivative; after the second, it contributes a linear term while a new vertical-shift constant appears. Independent conditions are needed to determine both.
Is a normal gradient the derivative?
No. The derivative gives the tangent gradient. Convert the stated normal information through perpendicularity before applying it to f′, taking care with horizontal and vertical special cases. Then use the actual point condition in f.
What is the strongest final check?
Differentiate the recovered function to reproduce the given derivative, then verify every supplied point, slope or total separately. Passing one check does not guarantee the others. A later fresh problem solved without prompts is the best practical test of whether the method has been retained.
Official sources and related Punggol guides
Curriculum check: 21 October 2026, using the official SEAB 2027 K232 and K341 documents linked above. The worked curves and motion models are original illustrations. The guide does not predict examination questions, school subject offers or grades.
