Before the method: the idea behind it
Why differentiation and integration are opposites
You have been using \(\int_a^b f(x)\,dx=F(b)-F(a)\) as a recipe. This lesson is about why that recipe is true. The answer is one of the most important results in mathematics: if you build a function by accumulating area under a curve, then differentiating it gives you the original curve back. Area and gradient, which look nothing like each other, turn out to be exact inverses.
Part one
A function made out of area
Fix a starting point a. For any x, let F(x) be the signed area under the curve from a up to x.
That definition turns area into a function: feed it an endpoint, get back a number. Written properly it is \(F(x)=\displaystyle\int_a^x f(t)\,dt\). Notice the letter inside changed to t. That is not decoration. The x is the moving endpoint, the thing F depends on; t is just the name of the position sweeping along inside. Using x for both would be like using the same variable for a box and for what’s in it.
Drag the endpoint below and watch two things at once: the shaded area on the left, and the point being plotted on the right. The right-hand curve is F, drawn one value at a time.
The left panel shows the region being swept out; the right panel plots how much area has accumulated. For f(t) = t the area is a triangle, so F(x) = x²/2, a parabola.
Part two
The sliver argument
Now ask: how fast does F grow as the endpoint moves?
Push the endpoint from x to x + h, a small step. The extra area you sweep out is a thin sliver: its width is h and its height is roughly f(x), because over such a narrow strip the curve barely changes. So the extra area is about \(h\cdot f(x)\).
Shrink h and watch the middle readout. The extra area divided by h closes in on the height of the curve.
Divide that extra area by h and you get the average rate at which F is growing:
That left-hand side is exactly the difference quotient from first principles: as h shrinks to zero it becomes F′(x). And the approximation gets better the smaller h is, because a shorter strip means less room for the curve to change. In the limit it is exact.
This is the fundamental theorem of calculus. Read it slowly: accumulating area, then differentiating, gets you back exactly where you started. The two operations undo each other. That is why anti-differentiation, which was defined as reversing a derivative, turns out to compute areas, something it had no obvious right to do.
Part three
From the theorem to the recipe
The evaluation rule you have been using follows in three lines.
Suppose G is any anti-derivative of f, so G′ = f. Part two says F is also an anti-derivative of f. Two functions with the same derivative can only differ by a constant, so \(F(x)=G(x)+c\) for some fixed c.
Now use the fact that F accumulates from a, so it has swept out no area at all when x = a: \(F(a)=0\). That forces \(c=-G(a)\), so \(F(x)=G(x)-G(a)\). Putting x = b gives the recipe.
The second form is worth staring at on its own. It says: integrating a rate of change over an interval gives the total change in the quantity. Integrate a velocity and you get the change in position. Integrate a flow rate and you get the change in volume. Integrate a marginal cost and you get the change in total cost. Every “rate to total” question you have done is this one line.
Why this is a landmark
It made areas computable
Before this theorem, finding the area under a curve meant summing ever-thinner rectangles, a limit calculation done from scratch for every new curve. Afterwards it became: find an anti-derivative, subtract two numbers. Problems that had taken mathematicians months became exercises for students.
It links every rate to its total
Any quantity you can only measure as a rate, speed, flow, current, infection rate, can be turned into a total by integration. That single move underlies physics, engineering and statistics, including the probability calculations you will meet in Term 3.
Before you start
The three statements to know
| Statement | What it means |
|---|---|
| \(F(x)=\int_a^x f(t)\,dt\) | The signed area function: area accumulated from a to x |
| \(F'(x)=f(x)\) | Differentiating an accumulated area returns the original function |
| \(\int_a^b f'(x)\,dx=f(b)-f(a)\) | Integrating a rate of change gives the total change |
Next: the working method for each, twenty practice questions, ten harder ones, and the ClassPad guide.