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If you know a curve’s gradient function and one point it passes through, you can recover the original function exactly — by integrating and then using the point to pin down the constant of integration.
What you'll be able to do
Given , integrating gives — but with an unknown . You need extra information to find .
Substitute the coordinates of a known point on the curve into your integrated expression and solve for .
Tip — Integrate first (keep the + c), THEN substitute the point — not the other way round.
The same idea recovers a quantity from its rate of change plus an initial value — for example, finding displacement from a velocity model and a starting position.
Formula recap
Common mistakes to avoid
Key takeaways
Test yourself
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