To recover a function, integrate its derivative term by term and add a constant c. Then substitute the given point into the result and solve for c.
This skill opens integration methods, and it returns whenever a question gives a rate of change together with a starting value.
Why is there a constant at all?
Differentiation removes constants. The functions y = x² + 1, y = x² and y = x² − 7 all have gradient 2x, so the gradient alone cannot tell you which one you started from.
When you integrate, you write the whole family: y = x² + c. A point such as (2, 9) then says which member you need. Substituting it gives 9 = 4 + c, so c = 5.
How do you recover the function, step by step?
- Integrate each term: raise the power of x by one and divide by the new power.
- Write + c at the end, before you do anything else.
- Substitute the point into your integrated expression, with x and y in the right places.
- Solve for c, then rewrite the equation with that value.
- Check by substituting the point again and by differentiating.
Worked example
A curve has gradient dy/dx = 6x² − 4x + 3 and passes through (1, 5). Find its equation.
Step 1, integrate: y = 6x³/3 − 4x²/2 + 3x + c = 2x³ − 2x² + 3x + c.
Step 2, substitute x = 1, y = 5: 5 = 2 − 2 + 3 + c. So 5 = 3 + c and c = 2.
Step 3, write the equation:
y = 2x³ − 2x² + 3x + 2
Step 4, check: at x = 1 the value is 2 − 2 + 3 + 2 = 5, which matches the point. Differentiating gives 6x² − 4x + 3, which matches the gradient.
The mistake to watch for
A common slip is to leave out the constant and then substitute the point anyway.
Mistaken answer: y = 2x³ − 2x² + 3x
At x = 1 this gives 3, not 5. The point is not satisfied because the constant was never added.
The correction is to write + c the moment you finish integrating. If the point is given, the constant is never optional. A quick test is to put the point back into your final equation and see whether it is true.
Check yourself
Try these on paper, then open each answer.
1. A curve has dy/dx = 4x − 3 and passes through (2, 5). Find y in terms of x.
Show answer
y = 2x² − 3x + c. Substitute (2, 5): 5 = 8 − 6 + c, so c = 3.
y = 2x² − 3x + 3
2. Given f′(x) = 3x² + 2 and f(−1) = 0, find f(x).
Show answer
f(x) = x³ + 2x + c. Then f(−1) = −1 − 2 + c = 0, so c = 3.
f(x) = x³ + 2x + 3
3. A curve has gradient 8x³ − 6x and passes through (1, 1). Find its equation.
Show answer
y = 2x⁴ − 3x² + c. Substitute (1, 1): 1 = 2 − 3 + c, so c = 2.
y = 2x⁴ − 3x² + 2
Where this leads next
Next, handle harder terms in integrating a power with the correct constant, then test yourself with the integration methods practice set. The function composition and inverse explorer and the non-calculator working trainer are useful for checking values and arithmetic.
If the set-up of these questions is where you stall, our teachers can work on it in online one-to-one Additional Mathematics tuition.