Is an integral an area?

A definite integral measures signed area between a curve and the horizontal axis. For a curve above the axis, that is the ordinary area underneath it.

Start with the blue curve $f(x)=x^2$. Imagine coloring the region underneath it green, from 0 to a movable endpoint $x$. As you move the endpoint to the right, the green region grows. Call its area $A(x)$.

There are two quantities to keep track of: the curve’s height, $f(x)$, and the area accumulated so far, $A(x)$.

Build the area from rectangles

Divide the interval from 0 to $x$ into equal pieces. On each piece, draw a rectangle whose height is the curve’s height at the left edge. Each rectangle contributes:

$$\text{rectangle area}=\text{height}\times\text{width}$$

Add those areas to estimate the green region. Try 4 rectangles, then 16 rectangles, then 128 rectangles.

Build the area under x²Reset

Blue: curve heightGreen: area and rectangles

Endpoint x

3.0

04

4 rectangles16 rectangles64 rectangles128 rectangles

Rectangle sum5.9064 left-edge rectangles

Exact area A(x)9.000x³ / 3

Missing area3.094exact area − rectangle sum

Narrower rectangles leave less area uncovered. Their sum approaches the integral.

Because $x^2$ increases on this interval, these left-edge rectangles underestimate the area. Increasing the rectangle count makes them narrower and brings the sum closer to the exact area.

The integral is the limit of these sums as the widths approach zero:

$$A(x)=\int_0^x t^2\,dt=\frac{x^3}{3}\qquad(x\geq 0)$$

The lower limit, 0, says where accumulation starts. The upper limit, $x$, says where it stops. The letter $t$ is a temporary position inside the interval; using it keeps that position separate from the endpoint $x$. The $dt$ indicates that we integrate with respect to $t$.

At $x=3$:

$$A(3)=\frac{3^3}{3}=9$$

The green area is 9 square units. The curve’s height is also 9 here, but that numerical match is a coincidence: at $x=2$, the height is 4 while the area is $8/3$.

Move the endpoint: one extra strip

What happens when we move from $x$ to $x+\Delta x$? We add a thin strip on the right. Its exact area is:

$$\Delta A=A(x+\Delta x)-A(x)$$

For a small positive $\Delta x$, the height barely changes across the strip. Treating it as a rectangle gives:

$$\Delta A\approx f(x)\,\Delta x=x^2\,\Delta x$$

Keep the endpoint at 3 in the explorer below. Try Δx = 0.1, then Δx = 0.001. Watch the area gained per unit of horizontal movement approach 9.

One extra strip of areaReset

Blue: curve heightGreen: accumulated areaOrange: added strip

Endpoint x

3.0

04

Δx = 0.5Δx = 0.1Δx = 0.01Δx = 0.001

Strip enlarged horizontally; height scaled separately. Dashed rectangle: the estimate x² Δx.

Added area ΔA5.291666667A(x + Δx) − A(x)

Average height ΔA / Δx10.583333over this finite strip

Instantaneous rate A′(x)9.000equals the curve height x²

The average height is 1.583333 above x². Shrinking Δx makes that difference approach zero, even though the added area also shrinks.

At $x=3$ and $\Delta x=0.001$, the rectangle estimate is:

$$\Delta A\approx 9\times 0.001=0.009$$

The exact added area is $0.009003000333\ldots$. The small extra amount comes from the curve rising across the strip.

We use $\Delta x$ and $\Delta A$ for these finite changes. Derivative notation, $dA/dx$, names the limiting rate as the horizontal change shrinks to zero.

Why divide the added area by the width?

The ratio $\Delta A/\Delta x$ asks: how much area do we gain per unit of horizontal movement?

For our narrow rectangle:

$$\frac{\Delta A}{\Delta x}\approx\frac{x^2\,\Delta x}{\Delta x}=x^2$$

Dividing an area by its width leaves a height. For a finite strip, the ratio is the average height across that strip. As the width approaches zero, that average approaches the curve’s height at the endpoint:

$$A'(x)=\lim_{\Delta x\to 0}\frac{A(x+\Delta x)-A(x)}{\Delta x}=x^2=f(x)$$

You can also see the approximation become exact by expanding the cubes:

$$\frac{(x+\Delta x)^3/3-x^3/3}{\Delta x} =x^2+x\,\Delta x+\frac{(\Delta x)^2}{3}$$

The last two terms vanish as $\Delta x$ approaches zero, leaving $x^2$.

The Fundamental Theorem of Calculus

The same relationship holds for any continuous function $f$:

$$A(x)=\int_a^x f(t)\,dt \quad\Longrightarrow\quad A'(x)=f(x)$$

The rate at which the accumulated area changes equals the height of the original curve. This is the accumulation part of the Fundamental Theorem of Calculus.

It also gives a practical way to calculate definite integrals. If $F'(x)=f(x)$, then:

$$\int_a^b f(x)\,dx=F(b)-F(a)$$

For $f(x)=x^2$, an antiderivative is $F(x)=x^3/3$. That is why the area from 0 to 3 is $F(3)-F(0)=9$.

Height and slope are different

There are two different derivatives in this example:

So $A'(3)=9$ is the slope of the area function $A$, and it equals the height of the blue curve $f$. The slope of the blue curve itself is $f'(3)=6$.

Think of water flowing into a tank. If $f(t)$ is the flow rate in liters per second, then $\int_0^x f(t)\,dt$ is the amount added over $x$ seconds, in liters. Differentiating that accumulated amount recovers the flow rate.

What if the curve goes below the axis?

With limits running from left to right, regions below the horizontal axis contribute negatively. For example:

$$\int_0^2 (t-1)\,dt=0$$

The negative triangle from 0 to 1 cancels the positive triangle from 1 to 2. Their total geometric area is 1 square unit, but their signed area is 0. To measure total geometric area, integrate the absolute value of the function.

Beyond geometric area

Integration accumulates continuously varying quantities. For a continuous random variable with probability density $p$:

$$P(a\leq X\leq b)=\int_a^b p(t)\,dt$$

Here, an area under the density curve represents a probability. An expected value accumulates each possible value weighted by its density:

$$\mathbb E[X]=\int_{-\infty}^{\infty} t\,p(t)\,dt$$

This formula assumes the expectation exists. Probability and expectations are two ways this same idea of continuous accumulation appears in machine learning.

Check your understanding

Try five questions about area, thin strips, and rates of change. Select an answer, then use Check answer to see the reasoning. You can revisit a question and try again.

Question 1 of 50 checked · 0 correct

For f(x) = x², at x = 2, which pair gives the curve height and the area accumulated from 0?

Height = 4; area = 8/3.Height = 4; area = 4.Height = 8/3; area = 4.

Check answer

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Keep learning

To explore derivatives as slopes and see how they guide neural-network training, continue with The derivative: slope at a point.