Question

Fundamental theorem calculus problem

Answer 1

The roof and the walls of the new math building in the university are shaped by the curve\[H(x)=30-\frac{ x ^{8} }{ 2,500,000 }\] if H(x) includes the height in feet of the building what is the average height of the building?

Answer 2

Average will be \[
\frac{1}{b-a}\int\limits_a^bH(x)\;dx
\]

Answer 3

ok so I got +/- 9.6467 as the bounds since there cant be a negative value since it describes a length, but can you explain to me why is is the integral times \[\frac{ 1 }{ b-a }\]

Answer 4

Ah, I actually have to explain it now?

Answer 5

yes I just want to know because I was working in a group and no one else knew how to approach this either, so i dont think we exactly learned the concept yet

Answer 6

Let me give you the less simplified form then: \[
\frac{\int\limits_a^bH(x)\;dx}{\int\limits_a^bdx}
\]

Answer 7

Does this make a bit more sense?

Answer 8

not really

Answer 9

is it because the area under the curve would be the total height and then if you divided that by the bounds it is the same as an average?

Answer 10

Consider this: \[
\frac{\sum\limits_{k=1}^n H_k}{\sum\limits_{k=1}^n1}
\]

Answer 11

This is the discrete average:\[
\frac{\sum\limits_{k=1}^n H_k}{\sum\limits_{k=1}^n1} = \frac{1}{n}\sum\limits_{k=1}^n H_k
\]

Answer 12

When dealing with continuous functions, we use integrals instead of sums.

Answer 13

ok

Answer 14

In general, the concept of the average is going to be the sum of all data points divided by the number of data points. So \(b-a\) represents the number of data points.

Answer 15

ok so that's why it is the integral times one over the bounds? Because the area under the curve is the sum of all the data points and b-a is the number of points?