Why does dividing the final(chosen) height in a line by 2 give you the average height(y) in that line?

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I was taught, of course, that the average of something is the sum of all the parts divided by the number of those parts.

This is easy when it is a set of values(2, 6, 8) because its simply (2+6+8)/3 = 16/3 ≈ 5.33.

But when you have a line, you have an **infinite amount of values** divided by **infinity**, right?

So how is it even possible to take the average? And why can you simply divide it by 2 when its a line?

And then of course there is the question about the average of curves and whatnot…

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I feel so dumb, plz help me understand.

**Also this question comes from the idea that average velocity is half the final velocity**

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13 Answers

Anonymous 0 Comments

Let’s say I have the velocity of a car going from 0-10 MPH increasing 1 MPH every second. What is its average speed over the 10 seconds? I could do (0+1+2+3+4+5+6+7+8+9+10)/11 and get 5. I could also take 0(the initial velocity) + 10(the final velocity)/2 and it would be 5. Both are correct ways to do it, one is much easier. this works because the car is speeding up at a consistent rate. it works because it all cancels each other out. (0+10)/2 = 5. (1+9)/2=5 (2+8)/2 = 5. (3+7)/2 = 5 (4+6)/2 = 5. Leaving 5/1 = 5. Any number of points equally spread out along the line will always give 5 as the average speed. For example, (1.5+8.5)/2 = 5. We could do this for the whole .5 spectrum and determine that (0+.5+1+1.5+2+2.5+3+3.5+4+4.5+5+5.5+6+6.5+7+7.5+8+8.5+9+9.5+10)/21 = 5.

A bit less ELI5: If the car wasn’t speeding up consistently or decreasing sometimes and we wanted to know what the average speed was travelling throughout the entire trip, the longer method is better. In fact the more data points evenly spread out through the trips we use the better. To find a more exact answer we can use calculus as with an equation we could for example turn velocity into distance, determine the total distance travelled and divide that by the time it took the car to travel that distance.

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