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Let $f$ be a function. If one finds $\displaystyle \frac{\mathrm d}{\mathrm dx}f$ and computes it at $x=a$, then one gets the rate of change of $f$ at $a$. That can be useful in some situations. But if one finds $\int f \space \mathrm dx$ and computes it at $x=a$, what information can we obtain about the function $f$?

Peter Woolfitt
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  • @Kaster I think they mean "find the antiderivative and evaluate it at a point," which makes sense. – icurays1 Jun 19 '13 at 17:59
  • Oh, @icurays1 I think you're right. – Kaster Jun 19 '13 at 18:00
  • I believe part of the problem is, the integral is a global answer, meaning what's a net effect in some region, whereas the derivative is very local. But I'm sure someone can put that in better words. Even the family of curves mentioned below, should still fall under the same idea. – Brady Trainor Jun 19 '13 at 18:05

2 Answers2

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You use the indefinite-integral tag, so I'll assume you are talking about indefinite integrals. In that case, recall that when integrating an integrable function f(x): $\displaystyle \int f \,dx$, we obtain a family of functions: $F(x) + C$, where $C$ can be any constant.

So we can not evaluate $F(x) + C$ at $a$ and obtain a distinct value for $a$, without knowing $C$.

amWhy
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  • For instance, let $C=0$. Given $f(x)=x$, then $\frac{ \mathrm d}{\mathrm dx} f=1$, and $\int f \mathrm dx=\frac{x^2}{2}$. At $x=2$ the derivative tells that the rate of change is 1 and if I compute it the antiderivative at $x=2$ one obtain $2$. What's the meaning of this $2$? Thanks –  Jun 19 '13 at 18:13
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    It really gives us no information, apart from knowing the function and what it represents. – amWhy Jun 19 '13 at 18:18
  • The value at $x=2$ of a function with linear ($x$) rate of change that passes through the origin. :) – Yoni Rozenshein Jun 19 '13 at 18:20
  • @Amzoti One of the (many) things that impresses me about you is your enthusiasm, and your willingness to display it/share it! Some folks are either too stodgy and stiff, or else they are "secretly" enthusiastic, but "too cool" to display it! – amWhy Jun 20 '13 at 01:24
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Because antiderivatives are only unique up to a constant, the answer is: not much, since evaluating the antiderivative can, in fact, give you any number (see @amWhy's answer).

However, we could reinterpret your question in the following way:

Given a function $f(x)$, an antiderivative is a function $F(x)$ such that $F^\prime(x)=f(x)$. In other words, $F(x)$ is a function which solves the equation

$$ F^\prime(x)=f(x) $$

As previously mentioned, this equation typically has infinitely many solutions - all varying by a constant. For example, consider

$$ F^\prime(x)=e^x $$ What type of function has derivative $e^x$? Well, it must be the exponential function plus a constant:

$$ F(x)=e^x+C $$ How might we restrict this collection of solutions? Well, suppose we knew the value of $F(x)$ at a point - say we know $F(0)=1$. Then, as it turns out, we only get one solution to this equation - we choose the constant of integration so that $F(0)=1$. In our example above, we would choose $C=0$ since $e^0=1$. This type of process is just the beginning of the (vast and fascinating) study of Differential Equations - determining a function from knowledge of its derivatives.

So what is the meaning of, say, $F(1)$ in the example above? It would be the value of a function whose derivative is $e^x$, and whose value at zero is $1$.

icurays1
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