Integration finds the area under any curve, even a shape with no area formula of its own. Differentiation finds a rate of change. They sound like completely different jobs — so why is each one the exact reverse of the other? Watching how the area builds up gives the answer.

Concept Accumulating area under a constant

Start with $y = 5$ , a flat line at height 5. Its integral is the accumulation of area between the line and the $x$ -axis, measured from the left. Sweep to the right one unit at a time and watch the total:

5,\; 10,\; 15,\; 20,\; 25,\; \ldots

The accumulated-area function is $F(x) = 5x$ . Every step it climbs by the same $5$ , because a strip of width 1 under a height of 5 has area 5. The rate at which area arrives is 5 — and $5$ is exactly what you get by differentiating $F(x) = 5x$ .

Concept When the incoming area is not constant

Now take $y = x$ . Its integral is $\dfrac{x^{2}}{2}$ . The area still accumulates from the left, but each new strip is larger than the one before it.

The strips come in as

\tfrac{1}{2},\; 1\tfrac{1}{2},\; 2\tfrac{1}{2},\; 3\tfrac{1}{2},\; \ldots

— each one larger than the last by exactly 1. That step of 1 is no accident: along the

x

-axis the value goes up by 1 each time, so the height of the strip does too.

The rate at which area arrives is therefore $y = x$ itself — and $\dfrac{d}{dx}\!\left(\dfrac{x^{2}}{2}\right) = x$ .

Note Two sides of one idea

Integration accumulates area. Differentiation measures how fast that area is accumulating. The speed at which each new strip enters is just the height of the original curve at that point — so differentiating the accumulated-area function hands the original function straight back.

\frac{d}{dx}\int_{0}^{x} f(t)\,dt = f(x)

Area and rate of change are not two different subjects. They are one process looked at from opposite ends, which is why each operation undoes the other.

Summary

Integration is the accumulation of area under a curve; differentiation is a rate of change.
Under $y = 5$ the area grows by a constant 5 each step, so the accumulated area is $5x$ and its derivative is $5$ .
Under $y = x$ the incoming strips are $\tfrac{1}{2}, 1\tfrac{1}{2}, 2\tfrac{1}{2}, \ldots$ , rising by 1 each step; the integral is $\dfrac{x^{2}}{2}$ and its derivative is $x$ .
The rate at which area arrives equals the height of the original function.
That is why $\dfrac{d}{dx}\displaystyle\int_{0}^{x} f(t)\,dt = f(x)$ : the two operations are inverses.

Integration vs. Differentiation