Is an integral an area?

Can you estimate the area under a curve, then predict how much extra area a small move of the endpoint adds? You need rectangle areas, powers, and the idea of a rate as change divided by change. We will introduce integral and derivative notation after the experiments.

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, $A(x)$.

Keep a record of your investigation

Use a notebook or text file. For each comparison, record the reference settings, your prediction, and the observed values. First keep the curve and endpoint fixed while changing only the rectangle count. Later keep the curve and endpoint fixed while changing only the extra strip width. Explain each result from the recorded evidence before opening the comparison.

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:

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

Keep $f(t)=t^2$ and $x=3$ fixed. Use 4 left-edge rectangles as the reference. Before trying 16 rectangles and 128 rectangles, write whether the estimate will rise or fall, and whether it will lie above or below the curved region. Record the rectangle sum and missing area for each count.

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)9.000

Missing area3.094exact area − rectangle sum

Compare each result with the four-rectangle reference. Did the green region change, or only the estimate? Explain the direction of the error using the rectangle edges before opening the comparison.

Compare your explanation

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.

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.

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:

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

Keep $x=3$ fixed and use $\Delta x=0.5$ as the reference. Before choosing Δx = 0.1 and Δx = 0.001, write predictions for both the added area and the added area divided by the strip width. Must both approach zero? Change only the strip width and record both readouts each time.

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

Compare your observations with the reference. The strip is getting smaller: does its average height behave the same way as its area? Write which number the ratio seems to approach and use the curve’s height to explain why.

Compare your explanation

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

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

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.

Design your own comparison

Choose a new positive endpoint within the explorer’s range, different from 3. Predict the number that $\Delta A/\Delta x$ should approach. Keep that endpoint fixed and test two strip widths of your choice. Record both the added area and the ratio, then explain what supports or contradicts your prediction.

For a separate comparison, keep your endpoint and rectangle count fixed. On paper, replace left-edge rectangles with right-edge rectangles. Predict whether the estimate is now too low or too high; calculate a small case or sketch the rectangles to justify the claim. Finish by comparing your evidence with your original predictions.

Check the reasoning in your investigation

At your chosen endpoint $x$, the strip ratio is $x^2+x\Delta x+(\Delta x)^2/3$, so it approaches $x^2$ as the positive width shrinks, while the added area approaches zero. For $t^2$ on a positive interval, right-edge rectangles overestimate and left-edge rectangles underestimate. The increasing curve explains the inequality; a few numerical trials alone do not prove the limit.

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.