01 · At zero
F′(x) = 4x − 3 and F(0) = 2. Find F.
Hint
Integrate to 2x² − 3x + C.
Worked solution
F(x) = 2x² − 3x + 2.
Understand · explore · practise
Finding the constant of integration
Recover a curve from its derivative and a known point. Find the integration constant, use initial conditions and interpret motion models and repeated integration.
Before you startIntegrating powers and substituting coordinates into equations
01 / A family of curves
The derivative gives the shape; a point fixes the vertical position.
If F′(x) = 2x + 2, then F(x) = x² + 2x + C. Every choice of C has the same derivative. The curves are vertical translations of one another.
Use the model to move the curve. The fixed point supplies an extra condition: choose the constant that makes the curve pass through it. At a fixed x, changing C changes the height but leaves the tangent gradient alone.
Choose the curve through the pointMove at your pace
F(x) = x² + 2x + C, with C = 0. At x = 1, F(1) = 3 and its gradient is 4. The required value is 7. This curve does not yet satisfy the point condition. Changing C moves the curve vertically and leaves its derivative unchanged. The green ring marks the required point.
Watch an initial point select one curve from the family
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Use a known point
Substitute into the integrated function.
Find F when F′(x) = 2x + 2 and F(1) = 7Worked example
F(x) = x² + 2x + C
Integrate first.
7 = 1² + 2(1) + C
Use both coordinates of the point (1,7).
C = 4
Solve the resulting equation.
F(x) = x² + 2x + 4
State the complete function and check F(1) = 7.
Substituting into F′ would test a gradient, not the height of the curve. A point condition belongs in F.
03 / A point away from zero
The constant need not equal the supplied y-coordinate.
Suppose F′(x) = 3x² − 4x + 2 and F(2) = 9. Then:
F(x) = x³ − 2x² + 2x + C
9 = 8 − 8 + 4 + C ⇒ C = 5
F(x) = x³ − 2x² + 2x + 5
Only when the non-constant part is zero at the specified input does C equal the given output. Use brackets for negative inputs.
A negative input
If F′(x) = 4x³ − 2x and F(−1) = 3, then F(x) = x⁴ − x² + C. At −1, the two powers cancel, giving C = 3.
04 / Roots, fractions and the domain
Keep exact values until C is found.
For F′(x) = 3√x − 2/x² on x > 0, with F(4) = 10:
F(x) = 2x3/2 + 2/x + C
10 = 2(8) + 1/2 + C
C = −13/2
F(x) = 2x3/2 + 2/x − 13/2
The point x = 4 lies in the domain. A condition at x = 0 would be invalid for this derivative because of its reciprocal term. An initial point on one side of a domain gap does not fix an independent constant on the other side.
05 / Initial values in a model
A rate and a starting value determine the quantity.
A particle has velocity v(t) = 12 − 4t m/s for 0 ≤ t ≤ 5 s, and position s(0) = 5 m. Since s′ = v:
s(t) = 12t − 2t² + C
C = 5
s(4) = 21 m
The displacement from t = 0 to 4 is 21 − 5 = 16 m. It is not the total distance: the velocity changes sign at t = 3. The area lesson will split the journey at that turning time.
A falling-height model
A model gives h′(t) = −10t m/s and h(0) = 45 m. Integration gives h(t) = 45 − 5t². At t = 2 s the height is 25 m; setting h = 0 gives t = 3 s after rejecting negative time.
This model assumes a constant downward acceleration of 10 m/s² and ignores air resistance. Its falling-height interpretation ends at impact; negative predicted heights are not a continuation of the same physical motion.
06 / An initial time that is not zero
Use the time you are actually given.
Suppose a position function satisfies s′(t) = (t + 2)² and s(1) = 2. Expand and integrate:
s(t) = t³/3 + 2t² + 4t + C
2 = 1/3 + 2 + 4 + C
C = −13/3
s(3) = 104/3
The change from t = 1 to 3 is 104/3 − 2 = 98/3. Do not set C = 2 just because 2 is the initial position in the question.
07 / Integrating more than once
Each integration introduces its own constant.
If F″(x) = 6x, F′(0) = −2 and F(0) = 3:
F′(x) = 3x² + A ⇒ A = −2
F(x) = x³ − 2x + B ⇒ B = 3
F(x) = x³ − 2x + 3
The first condition fixes A, and the second fixes B. A single point condition would not determine both constants.
Extension: a sequence of antiderivatives
Let F₁(x) = 2x³ and, for n ≥ 2, let F′ₙ = Fₙ₋₁ with every Fₙ(0) = 0. Then F₂ = x⁴/2, F₃ = x⁵/10, and:
Fₙ(x) = 12xn+2/(n + 2)!
Differentiating this expression recovers the previous one, and its value at zero is zero. The factorial collects the successive divisors.
Extension: non-zero initial values in a sequence
Let G₀ = 2, with G′ₙ = Gₙ₋₁ and Gₙ(0) = 2. Then G₁ = 2 + 2x, G₂ = 2 + 2x + x² and G₃ = 2 + 2x + x² + x³/3. In general Gₙ = 2∑k=0nxᵏ/k!. Each new constant restores the required value 2 at zero.
08 / Recover a cubic from turning points
The stationary inputs determine the derivative’s factors.
A cubic has turning points at (−1,8) and (2,−19). Its derivative is a quadratic with zeros at −1 and 2, so for some non-zero constant k:
F′(x) = k(x + 1)(x − 2)
F(x) = k(x³/3 − x²/2 − 2x) + C
Both supplied heights belong in F. Subtract the point equations to eliminate C:
F(2) − F(−1) = k[−10/3 − 7/6]
−19 − 8 = −9k/2 ⇒ k = 6
8 = 6(7/6) + C ⇒ C = 1
F(x) = 2x³ − 3x² − 12x + 1
Check the heights and that F′ vanishes at both inputs. Here F″(−1) = −18 and F″(2) = 18, confirming a maximum then a minimum. This method combines stationary-point information with integration; assuming the derivative’s leading coefficient is 1 would lose a necessary unknown.
09 / Your turn
Check the derivative and the point condition.
Write down the complete function once you have found C.
02 · A non-zero input
F′(x) = 3x² + 2 and F(1) = 6. Find F.
Hint
6 = 1 + 2 + C.
Worked solution
F(x) = x³ + 2x + 3.
03 · A root derivative
dy/dx = 2/√x for x > 0, and y = 5 at x = 4. Find y.
Hint
The antiderivative is 4√x + C.
Worked solution
5 = 8 + C ⇒ C = −3
y = 4√x − 3.
04 · A negative-domain condition
F′(x) = 6/x² on x < 0, and F(−2) = 1. Find F on that interval.
Hint
The antiderivative is −6/x + C.
Worked solution
1 = 3 + C ⇒ C = −2
F(x) = −6/x − 2, x < 0.
05 · An exact constant
dy/dx = 2x² − x and y(2) = 4. Find y.
Hint
At x = 2, the non-constant part is 16/3 − 2.
Worked solution
y = (2/3)x³ − x²/2 + 2/3.
06 · A starting time
s′(t) = 6 + 2t m/s and s(1) = 4 m. Find s(3) and the displacement from t = 1 to 3.
Hint
s(t) = 6t + t² + C.
Worked solution
C = −3; s(3) = 24 m.
Displacement = 24 − 4 = 20 m.
07 · Time to reach the ground
A height model has h′(t) = −8t and h(0) = 64, using metres and seconds. When does it first reach zero for t ≥ 0?
Hint
Integrate to 64 − 4t².
Worked solution
64 − 4t² = 0 ⇒ t = 4 s.
The other algebraic root is negative time; the physical height model is used only until impact.
08 · Two integrations
F″(x) = 4, F′(0) = −1 and F(0) = 3. Find F.
Hint
Use separate constants for F′ and F.
Worked solution
F′(x) = 4x − 1
F(x) = 2x² − x + 3.
09 · Is the condition enough?
F′(x) = 2x + 2 and F′(1) = 4. Does this determine C in F(x) = x² + 2x + C?
Hint
The second statement is about the derivative too.
Worked solution
No. It already follows from the derivative formula and holds for every C. A value of F at a point would determine its vertical position.
10 · Repeated antiderivatives
Let H₁ = 3x², H′ₙ = Hₙ₋₁ for n ≥ 2, and Hₙ(0) = 0. Find H₂ and H₃.
Hint
Each constant is zero because of the value at the origin.
Worked solution
H₂(x) = x³
H₃(x) = x⁴/4.
11 · A cubic from its turning points
A cubic has turning points at (0,5) and (2,1). Find its equation.
Hint
Write F′(x) = kx(x − 2), then integrate and use both heights.
Worked solution
F(x) = k(x³/3 − x²) + C
F(0) = 5 ⇒ C = 5
F(2) = 1 ⇒ −4k/3 + 5 = 1 ⇒ k = 3
F(x) = x³ − 3x² + 5.
10 / Recap
The extra information fixes what differentiation removed.
- Integrate before substituting a point condition.
- Use the known x and y values in the antiderivative.
- Keep exact fractions and check the point is in the domain.
- A gradient condition may add no information about C.
- Repeated integration needs a new constant each time.
- Interpret rates, positions and physical validity separately.
Section 1 of 10 · A family of curves