1
$\begingroup$

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 $(100 + t)^2 = 10000 + 200t + t^2$, which is not equal to just $200t + t^2$.

What is going on here?

  • 3
    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.2010-12-05
  • 2
    Like the waitress told the mathematician, **"...plus a constant!"** http://www.simonsingh.net/Joke_Competition.html2010-12-06

3 Answers 3

5

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.

1

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.

1

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!".

  • 0
    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.2011-12-13