TL;DR:
An integral adds up a rate times small steps. Under a speed-time graph, the area is distance.
Rectangles approximate the area, and more thinner rectangles squeeze toward the exact value.
Area below the axis counts as negative, so the integral gives signed area.
A car travels at a steady 20 m/s for 5 seconds. How far does it go? You multiply: 20 × 5 = 100 m. On a graph of speed against time, that is the area of a rectangle 5 wide and 20 tall. An integral is what you do when the speed is not steady, but the idea of area as distance still holds.
Area as accumulated change
Speed times time gives distance. Rate times input gives accumulated output. On a speed-time graph, the width is time, the height is speed, so area (height times width) has units of (m/s) × s = m. This works for any rate: litres per minute times minutes gives litres, and watts times hours gives watt-hours. The integral of a rate is the total change.
Worked example: area under a speed-time graph equals distance
A cyclist's speed grows steadily from 0 to 10 m/s over 5 seconds, a straight line on the graph. The area under that line is a triangle with base 5 s and height 10 m/s.
Area of a triangle:
½ × base × height.Substitute:
½ × 5 × 10 = 25.Units:
s × m/s = m.
The cyclist travels 25 m. Check with the average speed: the speed goes from 0 to 10 evenly, so the average is 5 m/s, and 5 × 5 = 25 m. The two methods agree.
What if the curve is not straight? Riemann sums
Take a speed v = t² m/s over 3 seconds. The graph curves, so no simple shape fits. Cut the time into slices and replace each by a rectangle. With 3 slices of width 1, using the right end of each slice for the height: 1 × (1 + 4 + 9) = 14. Using the left end gives 1 × (0 + 1 + 4) = 5. The true area lies between 5 and 14.
Use 6 slices of width 0.5. Right ends: 0.5 × (0.25 + 1 + 2.25 + 4 + 6.25 + 9) = 0.5 × 22.75 = 11.375. Left ends: 0.5 × (0 + 0.25 + 1 + 2.25 + 4 + 6.25) = 0.5 × 13.75 = 6.875. The window has narrowed to 6.875 up to 11.375. Keep going and the two estimates squeeze toward one number, which turns out to be 9. That limit is the definite integral of t² from 0 to 3, written ∫₀³ t² dt.
Slices | Left sum | Right sum |
|---|---|---|
3 | 5 | 14 |
6 | 6.875 | 11.375 |
How do we know it is 9? In the next article, an antiderivative of t² is t³/3, and 27/3 − 0 = 9. The average of our two six-slice sums is (6.875 + 11.375) / 2 = 9.125, already close.
A rate that changes in steps
Water flows into a tank at 3 litres per minute for the first 4 minutes, then at 5 litres per minute for the next 6 minutes (a made-up schedule). The graph is two rectangles. The first has area 3 × 4 = 12 L and the second 5 × 6 = 30 L. The total is 12 + 30 = 42 L, which is the integral of the flow rate over 10 minutes. Real flows change smoothly, and the integral is the same sum with infinitely many thin steps.
How fast do the estimates close in?
For a rising graph, the right sum overshoots and the left sum undershoots, and their gap is exactly the width of a slice times the total rise. With 3 slices that is 1 × 9 = 9 (and indeed 14 − 5 = 9). With 6 slices it is 0.5 × 9 = 4.5 (and 11.375 − 6.875 = 4.5). Halving the slice width halves the gap. With 1,000 slices the gap would be 0.003 × 9 = 0.027, so the estimate is within about 0.03 of 9.
Signed area
When the graph dips below the horizontal axis, the height is negative, so that part of the area counts as negative. Take a velocity v = 4 − 2t over 4 seconds. From 0 to 2 s the velocity is positive and the area is a triangle: ½ × 2 × 4 = 4. From 2 to 4 s it is negative, with a triangle of area 4 below the axis. The integral is 4 − 4 = 0: the object ends where it started. The total distance travelled is 4 + 4 = 8 m. This is why you must separate displacement (signed) from distance (total).
How to check your answer
Units. Output times input: m/s times s gives m.
Bounds. Left and right sums should bracket the answer for a steadily increasing function.
Shape check. A triangle or rectangle shortcut should match the sum when the graph is straight.
Sign. Part of the graph below the axis subtracts.
Common mistakes
Forgetting that the integral is a total, not a rate.
Counting area below the axis as positive when asked for displacement.
Using the bottom of the rectangle instead of a height on the curve.
Thinking more rectangles means a different answer. Narrower slices give a better estimate of the same number.
Ignoring the units of the graph.
Where you will use it next
The integral is linked to slope by one of the central results in maths, given in the fundamental theorem of calculus in plain words. Before that, optimization: the biggest open box from a square sheet shows the other half of calculus at work.
Frequently asked questions
Is an integral always an area?
It is area when the function is a height. When the function is a rate, the integral is an accumulated amount, such as distance, volume or energy.
Why do narrower rectangles help?
Each rectangle then hugs the curve more closely, so the leftover sliver above or below the curve is smaller.
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