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MATHEMATICS / ACCUMULATION

The integral: make a curved area countable

We know how to find the area of a rectangle. A curved edge does not have a single height. Cut the interval into strips, approximate each strip by a rectangle, and see what happens when the strips become thinner.

01How much area lies beneath a curve?

Start with f(x) = x², the x-axis and the boundaries a = 0 and b = 3. The teal region is the area we want. The gold rectangles are an approximation, not the answer. Click a strip to inspect it, or use the strip selector.

f(x)=x2f(x)=x^2Equal-width partition · n = 4
012345678910a = 0b = 3xf(x)Δxf(xᵢ*)
Rectangle width Δx0.7500
Rectangle sum Sₙ5.9063
Exact integral (reference)9.0000
The curve increases from left to right. A left-endpoint rectangle misses area; a right-endpoint rectangle includes extra area. As n doubles, each width halves, so the rectangles follow the curve more closely.

1. Divide the width

Split the total width b − a into n equal pieces. The Greek letter Δ means a change or interval width.

Δx=ban\Delta x=\frac{b-a}{n}

2. Choose a height

Pick a point xᵢ* in strip i. The curve’s height at that point becomes the rectangle’s height.

Ai=f(xi)ΔxA_i=f(x_i^*)\Delta x

3. Add the pieces

Σ means “add all these terms”. This finite sum is called a Riemann sum.

Sn=i=1nf(xi)ΔxS_n=\sum_{i=1}^{n}f(x_i^*)\Delta x

02Same width. Three ways to choose the height.

Compare the same curve using eight rectangles. A sampling point must lie in its own strip. Play the guide to highlight left, midpoint and right sampling in turn; the three results stay visible for comparison.

Left endpoint

0246810a = 0b = 3xf(x)
Riemann sum S₈7.3828

xi=a+(i1)Δxx_i^*=a+(i-1)\Delta x

Midpoint

0246810a = 0b = 3xf(x)
Riemann sum S₈8.9648

xi=a+(i12)Δxx_i^*=a+(i-\frac12)\Delta x

Right endpoint

0246810a = 0b = 3xf(x)
Riemann sum S₈10.7578

xi=a+iΔxx_i^*=a+i\Delta x

Left endpointUse the left edge of each strip. For an increasing function, this gives the lowest height in that strip and hence an underestimate.

03What if the maximum width tends to zero?

For these continuous functions, all three sampling choices approach the same limit as the equal-width strips become arbitrarily thin. We call that limit the definite integral. The animation stops at 256 rectangles; it illustrates convergence and does not literally reach infinity.

abf(x)dx=limni=1nf(xi)ban\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\frac{b-a}{n}

An elongated S for summation. It denotes the limiting accumulation, not a new area formula unrelated to rectangles.

a · b · f(x)

a and b are the lower and upper bounds on x. f(x) supplies the signed height. When f(x) ≥ 0 throughout, the integral is the geometric area between the curve and the x-axis.

dx

dx identifies x as the integration variable. In this construction it comes from strip widths tending to zero; we never set every finite rectangle’s width equal to zero.

See an exact limit for x² on [0, 3]

With right endpoints, xᵢ = 3i/n. Substitute into the sum and use the sum of squares formula. The terms containing 1/n and 1/n² vanish in the limit. This establishes the exact value beyond the finite animation.

Sn=27n3i=1ni2=9+13.5n+4.5n29S_n=\frac{27}{n^3}\sum_{i=1}^{n}i^2=9+\frac{13.5}{n}+\frac{4.5}{n^2}\longrightarrow9
Can the integral be zero while the total area is not?

No. For sin x from 0 to 2π, the positive contribution is 2 and the negative contribution is −2. The integral is 0, but the total geometric area is 4. To add areas without cancellation, integrate |f(x)|. Midpoint sampling happens to cancel exactly here even with finitely many symmetric strips; that is not proof that its rectangles fit the curve exactly.

We used equal-width partitions for clarity. General Riemann integration allows unequal widths, provided the largest strip width tends to zero and the sums approach a common value independent of sample choices.