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A friend and I were doing some math together when we had to integrate 2(100 + t).

I just multiplied it out and integrated 200 + 2t, which should be 200t + t^2.

He did u-substitution and got (100 + t)^2

When I take the derivative of both of ours, I seem to get the same thing. But (100 + t)^2 = 10,000 + 200t + t^2, which is not equal to just 200t + t^2.

What is going on here?

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Like the answers say, you and your friend have found two different primitives $G$ and $H$ for the same function. Two such primitives necessarily differ by a constant $c$, in this case $c = 10000$. However, when you calculate the integral of $2(100+t)$ over an interval $[a,b]$ by either of your primitives (i.e. when you calculate $\int_a^b f(t) = G(b)-G(a)$), you get the same thing no matter which primitive $G$ or $H$ you use. The reason is that the constant $c$ cancels out. Try writing the equations out, you'll see how it works. –  Gunnar Þór Magnússon Dec 5 '10 at 22:07
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Like the waitress told the mathematician, "...plus a constant!" simonsingh.net/Joke_Competition.html –  Arturo Magidin Dec 6 '10 at 0:13

3 Answers 3

up vote 4 down vote accepted

The difference is just the constant of integration. Both your solutions are correct, as is any one of the form C+200t+t^2 for any constant C.

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When you integrate indefinitely, you must add an arbitrary constant. So in essence, your answers are the same, up to addition of a constant (10,000 in this case). In other words, his constant will be 10,000 less than yours.

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Failure to recognize that an indefinite integral is only defined up to a constant can lead to strange results:

$$\begin{align}\int \frac{1}{x} dx & = \frac{1}{x} x - \int x \frac{(-1)}{x^2} dx \\ \int \frac{1}{x} dx & = 1 + \int \frac{1}{x} dx \end{align}$$

Or, subtracting the integral from both sides, we obtain $0=1$. As troll face would say:"Math will go bankrupt!".

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Actually, subtracting the integral from each side includes subtracting its constant of integration in the context in which you're using it. I.e. you will end up with $0 = 0$ unless you arbitrarily change your context in the middle, which makes any following answer meaningless. –  jnm2 Dec 13 '11 at 23:20

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