Why do you add + C when you integrate?
Updated
The short answer
An indefinite integral asks for every function whose derivative is the integrand, and those functions differ only by a constant. So ∫ 2x dx = x² + C, and leaving off the + C gives just one of them.
What it looks like
| Wrong | Right |
|---|---|
| ∫ 2x dx = x² | ∫ 2x dx = x² + C |
| ∫ cos x dx = sin x | ∫ cos x dx = sin x + C |
The missing C matters most when a problem gives you a starting value. Say f′(x) = 2x and f(0) = 5. With f(x) = x², you get f(0) = 0, and there's nothing left to adjust. With f(x) = x² + C, f(0) = C, so C = 5 and f(x) = x² + 5.
In alai, an iPad app that reads your handwriting as you work, you can hold to talk and ask the tutor, Ally, why a step works.
Why it happens
Integration is usually practiced as undoing a derivative, and undoing the derivative of x² seems to give one answer. But x² + 5 and x² − 1 have the same derivative, 2x, because the derivative of any constant is 0. Differentiating wipes out the constant, so integrating can't tell what it was. The + C stands for that unknown constant. On a graph, x² + C is the same parabola shifted up or down, and at every x all of those curves have the same slope, 2x.
It's also easy to forget because the usual check doesn't catch it. If you differentiate x², you get 2x, so the answer looks fine. Paul Dawkins counts the dropped constant among the biggest errors students make in integration. He describes two versions: leaving it out entirely, and dropping it from the middle steps and adding it back at the end. In differential equations, he adds, either one can leave you with a very different function from the right one.
How to catch it yourself
Look at the integral sign. If it has no limits on it, your answer is a whole family of functions, and it needs a + C. Then point at your answer and ask yourself:
Is x² the only function whose derivative is 2x?
When a problem gives a starting value, find C right after you integrate, before you do anything else with the function.
With definite integrals, the ones with limits, the C cancels out, so you can leave it off there: ∫₀³ 2x dx = (9 + C) − (0 + C) = 9.
Try these
- ∫ 3x² dx
- ∫ (4x + 1) dx
- Find f(x) if f′(x) = 6x and f(1) = 10.
Answers
- x³ + C
- 2x² + x + C
- f(x) = 3x² + 7, because f(x) = 3x² + C and f(1) = 3 + C = 10
How alai helps
If you're not sure why the + C matters, hold to talk and ask, and Ally answers out loud with a short reply. Ally also remembers what tripped you up in past sessions, so a later session can start from there.