Skip to content

Beyond: Areas under curves, and the big link

Change and motion · step 5 of 5

A question to start

A car's odometer (the meter that counts total distance) is broken, but its speedometer works, and a camera filmed it for a whole trip. You can see the speed at every moment. Can you work out how far the car went?

If the speed never changed, it would be easy: distance = speed × time. But the speed changes all the time. This page shows that the answer is an area, and that finding areas is the exact reverse of finding derivatives. That link is called the fundamental theorem of calculus, and it is one of the most important facts in all of mathematics.

The idea

Distance is the area under a speed-time graph

Draw a speed-time graph: time along the bottom, speed up the side.

  • If a car goes at a steady 20 m/s for 30 seconds, the graph is a flat line at height 20. The region under it is a rectangle, 30 wide and 20 tall. Its area is \(30 \times 20 = 600\), and the distance is \(20 \times 30 = 600\) m. The area is the distance.
  • If a stone is dropped, its speed is \(v(t) = 9.8t\) (from the Master step). The graph is a sloping straight line. Up to time \(t\), the region under it is a triangle with base \(t\) and height \(9.8t\), so its area is \(\dfrac{1}{2} \times t \times 9.8t = 4.9t^2\). That is exactly the distance \(d(t) = 4.9t^2\) the stone has fallen.

This is no accident. In a thin strip of time, the speed hardly changes, so the distance covered in that strip is almost speed × (width of strip): the area of a thin rectangle. Add up all the thin strips and you get the total distance, which is the total area.

Areas under curves: add up thin rectangles

For a curved graph, cut the region into thin rectangles and add their areas. Here is the area under \(y = x^2\) from \(x = 0\) to \(x = 1\), cut into 4 rectangles. Each one is \(\frac14\) wide, and its height is the curve's height at its right edge.

0¼½¾1 y = x²

The total area of the rectangles is

\[\frac14\left(\frac{1}{16} + \frac{4}{16} + \frac{9}{16} + \frac{16}{16}\right) = \frac{30}{64} = 0.46875.\]

The rectangles poke out above the curve, so this is a bit too big. Use thinner rectangles and the overshoot gets smaller:

Number of rectangles \(n\) 4 10 100 1000
Total area 0.46875 0.385 0.33835 0.3338335

The totals close in on \(\frac13\). A sum of thin rectangles like this is called a Riemann sum, after the 19th-century mathematician Bernhard Riemann, and the number it closes in on is the definite integral, written

\[\int_0^1 x^2 \, dx = \frac13.\]

The tall S is a stretched letter S, for "sum". The \(dx\) reminds you of the thin widths.

Why the answer is exactly one third

With \(n\) rectangles, each is \(\frac1n\) wide and the heights are \(\left(\frac1n\right)^2, \left(\frac2n\right)^2, \ldots, \left(\frac nn\right)^2\). The total is

\[\frac{1}{n} \cdot \frac{1^2 + 2^2 + \cdots + n^2}{n^2} = \frac{1^2 + 2^2 + \cdots + n^2}{n^3}.\]

There is a formula for the sum of the first \(n\) squares (it can be proved by induction, as on the Patterns Beyond step): \(1^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}\). So the total is

\[\frac{(n+1)(2n+1)}{6n^2} = \frac13 + \frac{1}{2n} + \frac{1}{6n^2}.\]

As \(n\) grows, the last two parts shrink towards 0, and the total heads to exactly \(\frac13\).

The fundamental theorem of calculus

Adding up rectangles every time would be slow. Here is the shortcut. Let \(A(x)\) be the area under a curve \(y = f(x)\) from a fixed starting point up to \(x\). Move \(x\) a tiny bit to the right, by \(h\). The area grows by a thin strip, about \(f(x)\) tall and \(h\) wide, so

\[\frac{A(x + h) - A(x)}{h} \approx f(x), \quad \text{and as } h \to 0, \quad A'(x) = f(x).\]

The derivative of the area function is the original function. Finding areas is the reverse of finding derivatives. A function \(F\) with \(F' = f\) is called an antiderivative of \(f\), and the theorem says

\[\int_a^b f(x)\,dx = F(b) - F(a).\]

For \(f(x) = x^2\), the power rule run backwards gives \(F(x) = \dfrac{x^3}{3}\) (check: its derivative is \(\dfrac{3x^2}{3} = x^2\)). So \(\displaystyle\int_0^1 x^2\,dx = \frac{1}{3} - 0 = \frac13\), with no rectangles at all. This link between slopes and areas was found by Isaac Newton and Gottfried Leibniz, working separately, in the second half of the 1600s.

Worked examples

1. Distance from a changing speed. A cart's speed is \(v(t) = 3t^2\) m/s. How far does it go from \(t = 0\) to \(t = 2\)?

  1. Distance is the area under the speed graph: \(\displaystyle\int_0^2 3t^2\,dt\).
  2. An antiderivative of \(3t^2\) is \(t^3\) (its derivative is \(3t^2\)).
  3. Distance \(= 2^3 - 0^3 = 8\) m.

2. An area between two points. Find \(\displaystyle\int_1^2 x^2\,dx\).

\[F(2) - F(1) = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \approx 2.33.\]

Sense check: on \([1, 2]\) the curve goes from height 1 to height 4, and the interval is 1 wide, so the area must be between 1 and 4. It is.

3. A tank filling up. Water flows into a tank at a rate of \(r(t) = 6t\) litres per minute. How much flows in during the first 4 minutes? The total is the area under the rate graph: an antiderivative of \(6t\) is \(3t^2\), so the total is \(3 \times 16 - 0 = 48\) litres. (The graph is a triangle, base 4, height 24: area \(\frac12 \times 4 \times 24 = 48\). Same answer.)

Try it

  1. A car drives at a steady 25 m/s for 40 seconds. Sketch the speed-time graph. How far does it go?
  2. A sled's speed is \(v(t) = 2t\) m/s. Use a triangle to find how far it goes from \(t = 0\) to \(t = 5\).
  3. Estimate \(\displaystyle\int_0^1 x^2\,dx\) with 4 rectangles whose heights come from their left edges. Is this too big or too small? Average it with the right-edge answer, 0.46875.
  4. Use the fundamental theorem to find \(\displaystyle\int_0^3 x^2\,dx\).
  5. Find \(\displaystyle\int_0^2 (3x^2 + 1)\,dx\).
  6. Check that \(F(x) = \dfrac{x^4}{4}\) is an antiderivative of \(x^3\), and use it to find \(\displaystyle\int_0^2 x^3\,dx\).
Answers
  1. A flat line at 25 for 40 seconds: a rectangle of area \(25 \times 40 = 1000\). It goes 1000 m.
  2. A triangle with base 5 and height \(v(5) = 10\): area \(\frac12 \times 5 \times 10 = 25\) m.
  3. Left heights are \(0, \frac{1}{16}, \frac{4}{16}, \frac{9}{16}\), so the area is \(\frac14 \times \frac{14}{16} = \frac{14}{64} = 0.21875\). Too small: these rectangles sit under the curve. The average is \(\frac{0.21875 + 0.46875}{2} = 0.34375\), already close to \(\frac13\).
  4. \(\dfrac{3^3}{3} - 0 = 9\).
  5. An antiderivative is \(x^3 + x\). So \((8 + 2) - 0 = 10\).
  6. The derivative of \(\dfrac{x^4}{4}\) is \(\dfrac{4x^3}{4} = x^3\). So the integral is \(\dfrac{16}{4} - 0 = 4\).

Common mistake

Forgetting to subtract \(F(a)\). \(\displaystyle\int_1^2 x^2\,dx\) is not just \(F(2) = \frac83\). It is the area from 1 to 2, so the area from 0 to 1 must be taken off: \(F(2) - F(1) = \frac73\). Only when the lower limit gives \(F(a) = 0\) can you leave it out, and it is good practice to write it every time anyway.

Where it leads