I have learnt integration as well as differentiation. In the early days I learnt a very simple proof for why the derivative of $\sin(x)$ is $\cos(x)$ and that of $\tan(x^2)$ is $2x\sec(x^2)$. This basically involved the use of the limit theorem which goes something like $f'(x) = \lim_{h\rightarrow0} \frac{f(x+h)-f(x)}h$. Now I can analytically solve this limit and derive the derivatives of all everyday functions. This me "see" how differentiation works!

Now my question here is about the integral. There is something called Riemann Sum in which we keep dividing a shape into smaller and smaller strips and we can get its area as the limit. OK that makes sense. But, how is the area under a curve equal i.e the Riemann Sum actually equal to the anti-derivative? We just reverse the differentiation to get the anti-derivative, nothing new about that.

My teacher has NEVER explained this nor have I come across a single book that sheds light on this! So,... why is area between $f(x) = x^2 + 2x$ and the $x$-axis equal to $g(x) = \frac{x^3}3 + x^2$. WHY??? Just like I can "see" how derivative works, can someone explain to me how the anti-derivative gives the area???

OK, this time can you tell me of a book that explains this concept?

  • $\begingroup$ Have a look at James Stewart's Calculus book! $\endgroup$
    – Jeb
    May 2, 2014 at 18:17
  • $\begingroup$ I think this website explains how a sum and anti-derivative are related: mathsisfun.com/calculus/integration-introduction.html $\endgroup$
    – stackErr
    May 2, 2014 at 18:25
  • $\begingroup$ Hmm... The definition of an integral? Did you never go over that in class? $\endgroup$
    – Shahar
    May 2, 2014 at 19:25

1 Answer 1


Intuitively, the formula $\int_a^b f'(x) \, dx = f(b) - f(a)$ says that "the total change is the sum of all the little changes".

Suppose you want to know the total change in $f$ as its input changes from $a$ to $b$.

Strategy: chop up the interval $[a,b]$ into tiny pieces, and add up the changes in $f$ across each of the subintervals.

So we pick points $a = x_0 < x_1 < x_2 < \cdots < x_M = b$. The "little change" in $f$ as its input changes from $x_i$ to $x_{i+1}$ is approximately $f'(x_i) (x_{i+1} - x_i)$. In fact, this is nothing more than the definition of $f'(x_i)$. Instead of saying $f'(x_i) \approx \frac{f(x_{i+1}) - f(x_i)}{x_{i+1} - x_i}$, we are saying

\begin{equation} f(x_{i+1}) - f(x_i) \approx f'(x_i)(x_{i+1} - x_i). \end{equation}

Now, just add up all the little changes (across all the tiny intervals) to get the total change: \begin{equation} f(b) - f(a) \approx \sum_{i=0}^{M-1} f'(x_i) \underbrace{(x_{i+1} - x_i)}_{\Delta x_i}. \end{equation}

On the right we see, for the first time, a Riemann sum. To get a better approximation to the total change, we could repeat this calculation but this time chop up $[a,b]$ into even tinier subintervals. By repeating this process again and again, using smaller and smaller subintervals each time, we get better and better approximations to the total change, and we see that the total change $f(b) - f(a)$ is exactly equal to a limit of Riemann sums.

With this strategy for computing the total change, we simultaneously invent the integral and discover that \begin{equation} f(b) - f(a) = \int_a^b f'(x) \, dx . \end{equation}

(Note that we could also approximate $f(x_{i+1}) - f(x_i)$ as $f'(\xi_i)(x_{i+1} - x_i)$, where $\xi_i$ is any point in $[x_i,x_{i+1}]$. For example, choosing $\xi_i$ to be the midpoint rather than an endpoint of $[x_i,x_{i+1}]$ seems likely to give a better approximation. By the Mean Value Theorem this approximation is exact for some choice of $\xi_i \in (x_i,x_{i+1})$, which suggests a way of turning our intuitive argument into a rigorous proof.)


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