# What is the relationship between the integral of a function and the area under the graph of that function? Not a homework question.

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What is the relationship between the integral of a function and the area under the graph of that function? Not a homework question.

In: 1 The integral of a function between two values (from *a* to *b*) will precisely equal the area between the x axis and the graphed line, between the x values of *a* and *b* (where under the curve is “negative” area).

If you’re curious *why* that is, I may have a reasonable explanation. They are they’re almost the same thing. For simple functions that are positive for all values in the range, they are equal. It gets a bit weird when you have something like a sine function or cosine, where the function can be negative.

The Integral of sin(x) from 0 to pi is:
[-cos(x) – – > – cos(pi) – (- cos(0)) – – > – (-1) – (-1) = 2]

but the Integral from zero to 2pi (one complete period) is:
[ – cos(2pi) – (-cos(0)) – – > -(1) – (-1) = 0]

Intuitively, if you look at the graph of sin(x) from 0 to 2pi you think “well it’s symmetrical so if the first half area is 2 then the total area is 4“ and you’d be correct, but the Integral is 0 because the second half of the function has a ‘negative area’ (not a real thing) according to integration rules. But, you already knew that. So let’s get a little more specific. We would like to know the area of a function traced from x = 0 to x = 137. As a rectangle, you recognize that your height would be y(0) and your width would be 137. A bit of multiplication and presto! The area you were looking for. Almost. Not very accurate for any mildly interesting function. What to do?

Let’s make TWO rectangles, the widths of both rectangles are now half of 137 and the height of the first rectangle is the same, but the height of the second rectangle is the value for y at at x = 137 divided by 2. Now there are two rectangles, you add their areas and presto! The area you were looking for, but now more accurate. Not accurate enough? Then make THREE rectangles using the same procedure, add their areas and presto! (Again).

Keep adding rectangles and watch the accuracy get better and better as you go from 3 rectangles to 10, to 100, to… 9,999,999,999,999,999 and off to infinity. And, at infinity, finally, the precise answer for area under the function from x = 0 to x = 137. Presto. Let’s say you are in a car, and you plot your distance traveled using the value on the odometer on the y axis and time on the x axis. Starting at 0 miles and 0 hours, let’s say you ride for an hour and go a mile. The distance traveled can be plotted by the function y=x, which shows a line the goes up with a slope of 1.

Since you went 1 mile in one hour, the speedometer would have shown 1 mile/hour the whole time. The slope of the distance graph would be the derivative of the distance graph, which would match what the speedometer read at those same times. The speed can be plotted by the function y=1, which is simply a horizontal line crossing the y axis at 1.

Now, just like how y=1 is the derivative of y=x, y=x is the integral of y=1, so for the next part we will consider what the area under the speed graph means for the distance traveled graph.

If I’m traveling 1 mile in an hour, the distance traveled would increase on the first graph by 1 mile. in the second graph, from time 0 to time 1, the area under the curve is a simple square with height 1 and width 1. the area is 1*1, and so that is the distance traveled from time 0 to time 1.

To further solidify this, if you want to know how far you traveled in a half hour using the speed graph, you’d just have to look at the area under the portion of the graph from 0 hours to .5 hours, which would be 1*.5, or .5 miles, which should make sense. The area is a rectangle with height 1 and width .5. in the same time period (0-.5 hours), the distance traveled graph increased by .5 miles.

That is how i intuit that the integral is the area under the curve. Thank you all for your help! I think I’m well on my way to understanding the answers you have so generously provided.