# Starting Integral Calculus by the Summation of Riemann Integrals

Home / Starting Integral Calculus by the Summation of Riemann Integrals

## How Much Area is Hiding Under that Curve?

Last week’s introduction to differential calculus explained how to find the slope of the tangent at any point along a simple curved function.

This week we begin working at finding the area underneath a curve. “Why bother?” is the first question that people usually ask. “It’s the important concept to introduce integral calculus” is not a very satisfying answer.

## The Importance of the Area Under a Curve

Integral calculus is all about knowing the rate at which something happens, and then calculating the total amount of “something” that has happened.

A simple example comes when travelling. If you can bicycle at a steady 20Km per hour, how far will you get in 3 hours? The trivially simple answer is “60Km” = 20Km/hour * 3 hours.

A similar question is, “How many products can the factory make in a day if it makes 2.7 units per minute”? These are simple because the rate is a constant per unit of time.

The problem is more complex if the rate changes. The classic science question is “How far does an object fall in a given amount of time in Earth’s gravity, assuming no wind resistance”? The acceleration due to gravity is 9.8m/sec^2 (9.8 metres per second squared). The speed is constantly increasing, so we cannot simply multiply the rate (speed) times time to get the total distance.

## Geometry can Approximate the Area Under a Curve

This image shows the function y = 3*x^2 + 2*x + 1. In this image, the approximate area under the parabola’s curve is shown by the four green rectangles and five yellow triangular areas.

The six points shown are at (0, 1), although it looks like (0, 0), (1, 6), (2, 17), (3, 34), (4, 57) and (5, 86).

The first green rectangle’s coordinates are (1, 0), (1, 6), (2, 6) and (2, 0). Its area is (6-0)*(2-1) = 6. Each of the other three rectangles has a base length of ‘1’; their heights are 17, 34 and 57. The total area of the rectangles is 6+17+34+57 = 114.

Although the graph shows straight line segments for the parabola, the mathematical function truly is a smooth continuous curve. So the yellow triangular areas simply approximate the actual curve. The brown curve is a rough guess at the actual smooth parabolic function.

The area of a right-angle triangle is “one-half of the base times the height”. Each of these triangles is indeed right-angled. The points for the largest triangle, at the right end of the graph, comes from the points (4, 57), (5, 86) and (5, 57). The height is (86-57)=29; the base is 1; so the area is 29/2 = 14.5. The total area of all the triangles is (5 + 11 + 17 + 23 + 29)/2 = 42.5.

Therefore an approximation of the complete area under the curve is the area of the rectangles plus triangles: 114 + 42.5 = 156.5.

Let’s say that a “slice” of area is the triangle plus rectangle, if any, above the base shown by the change in the ‘x’ value.

## Accuracy, or a Lack of It

We could have made a first approximation by thinking of the whole area as one triangle, with the points (0, 1), (5, 86) and (5, 1). The base would be 5, the height 85, and the area 5*85/2 = 212.5. The faint purple line shows how much extra area is included by this approximation.

The error is caused by the difference between the function’s curve versus the straight line drawn between two neighbouring points in the graph.

We can make better approximations by calculating more points along the parabola. Readers should feel free to build a spreadsheet and populate it with ‘x’ changing by 0.5, 0.1, or 0.001. Each time the change in ‘x’ becomes smaller, the straight-line “interpolation” is closer to the actual path of the mathematical function it approximates. Remember that we need to add all the vertical slices of the area: the more slices, the more terms to add together.

Categories Uncategorized