I know that if we start with an original function and take one derivative, we get another function. If we take the integral of that new function we get the original function back. So I see how they are opposite operations but I don't see how they are opposite in the sense that the integral is supposed to be the area under the curve and the derivative is supposed to be the instantaneous slope. It seems like they are separate; deriving one function is the opposite of integrating it but when we look at it graphically it doesn't make sense. Also, if I start with some function, differentiate it and then take the integral I get the original function so how is that the area under the curve? Thanks.

  • $\begingroup$ The two operations are indeed different and not obviously "inverse". That's why the fundamental theorem of calculus is so profound! $\endgroup$
    – user98602
    Aug 25, 2014 at 18:34
  • $\begingroup$ but what about when you start with one function, differentiate it and then integrate it. you end up with the original. So in that sense they are inverse? $\endgroup$ Aug 25, 2014 at 18:36
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    $\begingroup$ @KingSquirrel This isn't true. You get a constant of integration when you integrate. For example $$\int \frac{\mathrm{d}x}{x^2} = -\frac{1}{x}+k \equiv \frac{kx-1}{x}$$ $\endgroup$ Aug 25, 2014 at 18:41

3 Answers 3


Well...it's not obvious, as you point out, why "finding a slope" and "finding area under a curve" are opposites. That non-obviousness is why it's called a "theorem" (indeed, more formally stated, it's called "the fundamental theorem of calculus").

But maybe I can help out with the intuition a little. Instead of thinking of $f'(b)$ as the slope of the graph of $f$ at the point $(b, f(b))$, think of it this way: if we moved a little to the right of $b$, say, to $b + h$, where $h$ is a small number, what would we expect $f(b+h)$ to be? Well, if you draw a picture, you'll see that we expect it to be

$$ f(b+h) \approx f(b) + h * slope $$ and since the slope is $f'(b)$, this is

$$ f(b + h) \approx f(b) + h f'(b). $$

That's true for any "nice" (smooth, etc.) function, for small values of $h$. Let me define such a function, the "accumulated area" function:

$$ F(x) = \int_0^x f(t) ~dt. $$

That's the area under the graph of $f$ between $0$ and $x$. In particular, we have

$$ F(b) = \int_0^b f(t) ~dt $$ is the area under the graph of $f$ from $0$ to $b$. What's the value of $F(b + h)$ for a small value of $h$?

We have two possible answers: the first is from the definition: it's just

$$ F(b+h) = \int_0^{b+h} f(t) ~dt = \int_0^{b} f(t) ~dt + \int_b^{b+h} f(t) ~dt. $$

Hold that thought.

The second answer comes from that general stuff I wrote above, namely: $$ F(b+h) \approx F(b) + h F'(b). $$ Setting these two equal (since they both represent the value of $F(b+h)$, more or less), we get $$ F(b) + h F'(b) \approx \int_0^{b} f(t) ~dt + \int_b^{b+h} f(t) ~dt $$ Since the $F(b)$ on the left hand side is the same as the first integral on the right hand side, this simplifies to $$ h F'(b) \approx \int_b^{b+h} f(t) ~dt $$ and dividing through by $h$, becomes $$ F'(b) \approx \frac{1}{h} \int_b^{b+h} f(t) ~dt $$

Now when $h$ is very small, the integrand, $f(t)$ is roughly constant ... and its value is approximately $f(b)$, So the integral on the right is roughly the integral of the constant $f(b)$ over an interval of width $h$; that value is just $h f(b)$. So we get

$$ F'(b) \approx \frac{1}{h} \int_b^{b+h} f(t) ~dt \approx \frac{1}{h} [h f(b) ] = f(b). $$

So the derivative of the "accumulated area" function $F$ is the original function $f$.

As for derivative and integral being "opposites", you might want to look at

$$ f(x) = x^2 + 1. $$ Try taking its derivative, to get a new function $g$, and then write down the accumulated area function $$ G(x) = \int_0^x g(t) dt. $$ Evaluate the integral, and see whether it equals $f$. (Hint: it doesn't!). So "differentiate then integrate" doesn't necessarily bring you back to the original function. On the other hand, for nice enough functions (e.g., continuous), "integrate then differentiate" does bring you back to the original.


The derivative is the instantaneous rate of change. The best way to think about the derivative is: \begin{equation} f(x + \Delta x) \approx f(x) + f'(x) \Delta x. \end{equation}

The (second) fundamental theorem of Calculus says, intuitively, that "the total change is the sum of all the little changes".

Suppose we want to know how much a function $f$ changes across an interval $[a,b]$. Chop up $[a,b]$ into tiny subintervals $[x_i,x_{i+1}]$ of width $\Delta x_i$. Then \begin{align} \underbrace{f(b) - f(a)}_{\text{total change}} &= \sum_i \underbrace{f(x_{i+1}) - f(x_i)}_{\text{little change}} \\ &\approx \sum_i f'(x_i) \Delta x_i \\ &\approx \int_a^b f'(x) \, dx. \end{align}

It seems plausible that, by chopping up $[a,b]$ into extremely tiny subintervals, we can make this approximation as close as we like. It follows that the two quantities must be equal: \begin{equation*} f(b) - f(a) = \int_a^b f'(x) \, dx. \end{equation*}

The geometric interpretations of the derivative and the integral, are not always the best way to think about them.

  • $\begingroup$ This is the best answer I've seen so far. The integral from a to b is the two derivatives in reverse order, b - a. That's why they're opposite. $\endgroup$
    – DrZ214
    Jun 12, 2018 at 7:15
  • $\begingroup$ Why we need to chop into $[a,b]$ into tiny subintervals if always the $\Delta x_i$ is present both on numerator and denominator so always cancels out? $\endgroup$
    – user599310
    Feb 9, 2020 at 22:05
  • $\begingroup$ @user599310 There's no $\Delta x_i$ in $f'(x_i).$ Sometimes the definition of the derivative is written with a $\Delta x$ in the denominator, but by the time you get to $f'(x_i)$ that denominator has already been taken to zero in a limit; it is not the $\Delta x_i$ described in this answer. $\endgroup$
    – David K
    Oct 25, 2021 at 2:39

enter image description here

Suppose that $A(x)$ represents the area under the curve $f$ between $a$ and $x$. Now try to find A'(x) with limits. $$A'(x)=\lim_{h\to 0} \frac{A(x+h)-A(x)}{h}$$

Now try graphing $A(x)$ and $A(x+h)-A(x)$ as shaded regions under the graph of $f$, where $A(x+h)$ is the union of the $A(x)$ and $A(x+h)-A(x)$ regions. enter image description here

I think that you will find that $$A(x+h)-A(x)\approx hf(x+h)\approx hf(x)$$ where $hf(x+h)$ is the area of an approximating rectangle. The limit then becomes $$A'(x)=\lim_{h\to 0} \frac{hf(x+h)}{h}=f(x)$$

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    $\begingroup$ Was that made in Paint? $\endgroup$
    – Gahawar
    Aug 25, 2014 at 21:23
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    $\begingroup$ aww, did I forget the MSPaint NOW! logo? $\endgroup$
    – John Joy
    Aug 25, 2014 at 22:58
  • $\begingroup$ there! logo added $\endgroup$
    – John Joy
    Aug 26, 2014 at 13:36
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    $\begingroup$ Oh no! I preferred it before! (MS pawn? Not I!) But +1 anyway. $\endgroup$
    – TonyK
    Aug 26, 2014 at 15:18
  • $\begingroup$ Re: resisting MS. If at first you don't succeed, give up. Ain't no use in being a damn fool. $\endgroup$
    – John Joy
    Aug 26, 2014 at 17:11

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