How come the inverse of a derivative is an integral?

381 views

I can’t conceptually grasp this the same why I can grasp addition/subtraction (the opposite of adding stuff is to take it away), but if the derivative of a function is the slope at any given point, then why would the area under the curve be the opposite of that?

In: 2

8 Answers

Anonymous 0 Comments

Let’s first consider the discrete analogue of some operators on sequences.

For a given sequence

>s = s_1, s_2, s_3, . . .

we can form a sequence *P(s)* of partial sums of the first one, two, three etc. terms, like this

>P(s) = s_1, s_1+s_2, s_1+s_2+s_3, . . .

and a sequence *D(s)* of differences between each term and the preceding term, like this

>D(s) = s_1, s_2-s_1, s_3-s_2, . . .

Now let’s compute *D(P(s)),* the result of applying both to some sequence *s.* We have

>P(s) = s_1, s_1+s_2, s_1+s_2+s_3, . . .

and for this sequence we list the sequence of differences between terms by definition of *D*

>D(P(s)) = s_1, (s_1+s_2)-(s_1), (s_1+s_2+s_3)-(s_1+s_2), . . .

but that is the original sequence *s = s_1, s_2, s_3, . . .* , so we observe that *D(P(s))=s.*

Similarly, for *P(D(s))* we have

>D(s) = s_1, s_2-s_1, s_3-s_2, . . .

so that

>P(D(s)) = s_1, s_1+(s_2-s_1), s_1+(s_2-s_1)+(s_3-s_2)

is again the original sequence. Thus also *P(D(s))=s*.

The operations *P* and *D* are therefore inverses of each other.

Finally, integrals and derivatives are similar operators but on *functions* instead of sequences. The difference is that while sequences have a clear “unit step”, there is no such thing for (real) functions, that’s what the *dx* in the integral and 𝛥*x* in the definition of the derivative represent and the rest is just some rigor, but the basic idea is the same.

Integral is kind of a “continuous sum” like the operator *P* and derivative is kind of a “continuous consecutive difference” like the operator *D.*

You are viewing 1 out of 8 answers, click here to view all answers.