The Definite Integral

Rectangles under a curve

Take the region under $y=x^2$y = x², above the x-axis, from x = 0 to x = 3. Its top is curved, so no area formula fits it. Cut it into three strips, each 1 wide, and replace each strip by a rectangle.

If each rectangle is as tall as the curve at the right edge of its strip, the heights are 1, 4 and 9, and the rectangles add to 14. Every one of them pokes above the curve, so 14 is too much. If each is as tall as the curve at the left edge, the heights are 0, 1 and 4, which add to 5, too little, since each sits under the curve. The area is somewhere between 5 and 14: close enough to bracket it, too coarse to name it.

Three rectangles over the region under $y=x^2$y = x² from x = 0 to x = 3, each 1 wide and as tall as the curve at its right edge: heights 1, 4 and 9, adding to 14.

Height times width

Each rectangle has area $f\left(x\right)\times \Delta x$f(x) × Δx: the height of the curve at its chosen point, times the strip width $\Delta x$Δx. The estimate is all of these added up.

With six strips, $\Delta x=0.5$Δx = 0.5. The right-edge heights are 0.25, 1, 2.25, 4, 6.25 and 9, which add to 22.75, so the right sum is 22.75 × 0.5 = 11.375. The left-edge heights are 0, 0.25, 1, 2.25, 4 and 6.25, which add to 13.75, so the left sum is 6.875. The bracket has narrowed from 5 to 14 down to 6.875 to 11.375.

Six rectangles under $y=x^2$y = x² from x = 0 to x = 3, each 0.5 wide and as tall as the curve at its right edge. Their areas add to 11.375, closer than the 14 from three strips.

Four rectangles under $y=x^2$y = x² from x = 0 to x = 2, each as tall as the curve at its left edge. They add to 1.75, below the area under the curve, $\frac{8}{3}\approx 2.667$8/3 ≈ 2.667. Drag the number of rectangles up and the shortfall shrinks; switch to right edges, which give 3.75 with four rectangles, and they close in from above.

The limit

Keep cutting. With n strips on $y=x^2$y = x² from 0 to 3, the right sum is always above the area and the left sum below it, and the two differ by the last right rectangle less the first left one: $9\times \frac{3}{n}-0=\frac{27}{n}$9 × 3/n − 0 = 27/n. That gap goes to 0 as n grows, so both sums are squeezed onto one number.

With 30 strips the sums are 8.555 and 9.455. With 300 they are 8.95505 and 9.04505. With 3000 they are 8.9955005 and 9.0045005. Both close in on 9, and 9 is the area of the region.

That limit is the definite integral. ${\int }_a^{b}f\left(x\right)dx$∫ₐᵇ f(x) dx is the number that the sum of $f\left(x\right)\times \Delta x$f(x) × Δx, over strips from x = a to x = b, approaches as every strip width shrinks toward 0. For this region, ${\int }_0^3{x}^{2}dx=9$∫₀³ x² dx = 9.

The region under $y=x^2$y = x² from x = 0 to x = 3, shaded. Its area, the limit of the rectangle sums, is 9.

Areas that are already known

Where the region is a familiar shape, the integral is its area. Under y = 2x from 0 to 4 the region is a triangle with base 4 and height 8, so ${\int }_0^{4}2xdx=\frac{1}{2}\times 4\times 8=16$∫₀⁴ 2x dx = ½ × 4 × 8 = 16. With four strips 1 wide, the left sum is 0 + 2 + 4 + 6 = 12 and the right sum is 2 + 4 + 6 + 8 = 20, either side of 16.

Under y = x + 1 from 1 to 3 the region is a trapezium with parallel sides 2 and 4, 2 apart, so $\underset{1}{\overset{3}{\int }}\left(x+1\right)dx=\frac{2+4}{2}\times 2=6$∫ from 1 to 3 of (x + 1) dx = (2 + 4)/2 × 2 = 6. Under y = 3 from 1 to 5 it is a rectangle, so ${\int }_1^{5}3dx=3\times 4=12$∫₁⁵ 3 dx = 3 × 4 = 12.

Strips below the axis

The sum adds $f\left(x\right)\times \Delta x$f(x) × Δx, and where the curve is below the x-axis, f(x) is negative, so those strips add negative amounts. ${\int }_0^3\left(-2\right)dx$∫₀³ (−2) dx is a rectangle 2 deep and 3 wide below the axis, and its value is −6.

For y = x − 1 from 0 to 2, the line is below the axis from 0 to 1, a triangle of area ½ counting as −½, and above it from 1 to 2, a triangle counting as +½. So ${\int }_0^2\left(x-1\right)dx=0$∫₀² (x − 1) dx = 0, even though the two triangles cover a total area of 1. In the same way, $\underset{-1}{\overset{2}{\int }}xdx=-\frac{1}{2}+2=\frac{3}{2}$∫ from −1 to 2 of x dx = −½ + 2 = 3/2. The definite integral is a signed area; the lesson Area Below the Axis takes this further.

The line y = x − 1 from x = 0 to x = 2. The triangle below the axis, from 0 to 1, counts as −½, and the triangle above it, from 1 to 2, as +½, so the integral is 0.

The usual mistakes

Taking a rectangle sum for the integral. Six right rectangles give 11.375, not 9. Only the limit is exact.

Expecting more strips to make the error worse. Each extra cut leaves smaller gaps between the flat tops and the curve; the gap between the left and right sums here is $\frac{27}{n}$27/n, which only shrinks.

Reading a signed integral as a total area. ${\int }_0^2\left(x-1\right)dx=0$∫₀² (x − 1) dx = 0, but the region between the line and the axis has area 1.

Swapping the limits. The strips run from the lower limit at the bottom of the sign to the upper limit at the top.