Why average value of a function is not calculated by using the formula $\frac{f(a)+f(b)}{2}$

functions

I know the formula for calculating the average value of a function as $$ \frac{1}{b-a}\int_a^b f(x)\,dx$$

But in elementry level maths and physics problems we generally use a very simple approach to find the average of a value,by taking sum of two values and then dividing them from 2.

(for example,if the initial velocity of a car is ${10ms^{-1}}$ and after 10 seconds the final velocity is ${20ms^{-1}}$ then the average velocity will be ${ \frac {10 + 20}{2}=15ms^{-1}}$ )

Let us took a very simple identity function ${f(x)=x}$,we want to find the average value between ${2}$ and ${4}$ then by integration;

$$ \frac{1}{b-a}\int_a^b f(x)\,dx$$
$$= \frac{1}{4-2}\int_2^4 f(x)\,dx$$
$$= \frac{1}{2}\left(\frac{4^2-2^2}{2}\right)$$
$$= \left(\frac{16-4}{4}\right)=3$$

But I can also calculate this by using a general formula for average of two values
$$\frac{f(a)+f(b)}{2}$$
$$\frac{4+2}{2}=3$$

By both methods we get the same value,so my question is why we don't use $\frac{f(a)+f(b)}{2}$ for calculating the average value of a function (which seems to be very simple) ?

Best Answer

The formula

$$\frac{1}{b-a}\int_a^b f(x)\,dx=\frac{f(a)+f(b)}{2}$$

doesn't hold in general.

It always holds for linear functions

$$f(x)=mx+n$$

indeed in that case we have

$$\frac{1}{b-a}\int_a^b f(x)\,dx=\frac{1}{b-a}\left[\frac12mx^2+nx\right]_a^b=\frac{1}{b-a}\left(\frac12mb^2+nb-\frac12 ma^2-na\right)=$$ $$=\frac{m(b+a)(b-a)+2n(b-a)}{2(b-a)}=\frac{m(b+a)+2n}{2}=\frac{(ma+n)+(mb+n)}{2}=\frac{f(a)+f(b)}{2}$$

Related Solutions

Why does the average of a function comes out to be in this pattern for the purpose of integration

First, consider sums instead of integrals. Let's look at what happens in the discrete case, with successive differences. Suppose you have a polynomial like $f(x)=x^2$. Look at what happens with the differences between neighboring terms, and the differences between those differences: $$\begin{array}{cccccccc}f(n)=0&&1&&4&&9&&16&&25&&36\\&1&&3&&5&&7&&9&&11\\&&2&&2&&2&&2&&2\end{array}$$ The second differences (differences between the differences) are all constant.
What happens when $f(x)=x^3$? $$\begin{array}{cccccccc}f(n)=0&&1&&8&&27&&64&&125&&216\\&1&&7&&19&&37&&61&&91\\&&6&&12&&18&&24&&30\\&&&6&&6&&6&&6\end{array}$$ Now the third differences are all constant.
Degree $n$ has constant $n$th differences. The general insight is that when $f(x)$ is a degree-$n$ polynomial, the $n$th differences are all constant (and conversely). And you can find other facts from this; for example, when $f(x)=x^n$, the difference between neighboring terms must be a polynomial of degree $n-1$ because that row's $(n-1)$th differences vanish!
Compute using patterns of successive differences. Let's return to the example $f(x)=x^2$ and our first table. Using the successive differences, you can actually write each term $f(n)$ in terms of the accumulated differences. Look at the pattern it makes: $$\begin{align*}f(0) &= 0 \\ f(1) &= 0 + 1 \\ f(2) &= 0 + 1 + (1+2) \\ f(3) &= 0 + 1 + (1+2) + (1+2+2) \\ f(4) &= 0 + 1 + (1+2) + (1+2+2) + (1+2+2+2) \\ &\vdots\\f(k) &= 0 + \; 1(k)\; + 2\left[ \frac{k(k+1)}{2}\right] \end{align*}$$
In a similar way, you could compute the cumulative value $(0, 0+1, 0+1+4, 0+1+4+9, \ldots)$ or average value of the entries in this table in terms of these successive differences. (Add a row above the row for *f(n)$!) You could find a pattern for the number of 0s and 1s and 2s in the accumulated sum, and express this pattern as a mathematical formula.

But, to save time, we will use a trick to find the mathematical formula for every case $f(x)=x^n$ instead of just our specific polynomial $f(x)=x^2$.
A trick to find the sum of the first few $x^d$. If our polynomial is $f(x) = x^{d}$, we want to find a formula for $0^d + 1 ^d + 2^d + 3^d + \ldots + b^d = \sum_{k=1}^b k^d$. This is the formula for the cumulative sum of the first $b$ entries in the table. Alternatively, we could divide by $(b+1)$ to find the average of the first $b$ entries in the table $\frac{1}{(b+1)}\sum_{k=1}^b k^d$.

The general formula is called Faulhaber's Formula. For our purposes, it says that $$0^d + 1^d + \ldots + b^d = \frac{b^{d+1}}{d+1} + \ldots$$ where the term "$\ldots$" contains lower-order terms like $b^{d-1},b^{d-2},\ldots,b^2,b^1,b^0$ which we will not need to care about for our applicatoin.

For example, when $d=1$, the formula says that $$0^1 + 1^1 + 2^1 + 3^1 + \ldots + b^1 = \frac{b^2}{2} + \ldots$$

and when $d=2$, the formula says that $$0^2 + 1^2 + 2^2 + 3^2 + \ldots + b^2 = \frac{b^3}{3} + \ldots$$

If you know about binomial coefficients, I can show you why this formula is true. You can also think of it in terms of geometry. I'll put a proof after the end of this answer.
Relating sums to integrals. Faulhaber's formula for tables of numbers looks pretty similar to the rule you want to find for averages of continuous curves. This is actually an example of how integrals are often a convenient approximation for sums.
We can use this formula to estimate the average of a continuous function $f(x)=x^d$ over the range $[0,b]$. Start with the summation formula: $$0^d + 1^d + 2^d + \ldots + m^d = \frac{m^{d+1}}{d+1} + \ldots$$ Assume $m$ is a very high number.ow do we transform this expression to make it look more like the average of $f(x)$ over the range $[0,b]$? First of all, our range of x-values is incorrect. Our sum goes from 0 to $m$ instead of from $0$ to $b$. Let's rescale it by multiplying both sides by $b/m$: $$(b/m)(0^d + 1^d + 2^d + \ldots + m^d) = (b/m)\frac{m^{d+1}}{d+1} + \ldots$$ Our range of y-values is also incorrect. Our sum involves terms from $0^d$ all the way up to $m^d$, instead of up to $b^d$. Let's rescale it by multiplying both sides by $(b/m)^d$:

$$(b/m)^d(b/m)(0^d + 1^d + 2^d + \ldots + m^d) = (b/m)^d(b/m)\frac{m^{d+1}}{d+1} + \ldots$$

This formula now gives a good approximation to the value of $\int_{0}^b x^d \, dx$. What we have done is taken the discrete points $\{(k, k^d) : k=1,\ldots,m\}$ and rescaled them so that they fit on top of the graph of $f(x)=x^d$. The rescaled sum of the $k^d$ is therefore approximately equal to the integral, and the average of the $k^d$ is approximately equal to the function's average height.

$$\operatorname{ave}(f(x)) \approx \frac{1}{b}(b/m)^d(b/m)(0^d + 1^d + 2^d + \ldots + m^d) = \frac{1}{b}(b/m)^d(b/m)\frac{m^{d+1}}{d+1} + \ldots\\=\frac{b^d}{d+1}$$

Here's an easy way to find the formula. If you want to know $\sum_{k=0}^b k^n$ for any power $n$, start by writing down the expansion for $(k-1)^{(n+1)}$. For example, when $n=2$: $$(k-1)^{3} = k^{3} - {3 \choose 1} k^{2} + {3 \choose 2 }k - {3 \choose 3}1$$ And for a different exponent $n$: $$(k-1)^{(n+1)} = k^{(n+1)} - {n+1 \choose 1} k^{n} + {n+1 \choose 2 }k^{n-1} + \ldots$$ Note that all the terms in the "$\ldots$" involve $k$ raised to lower powers like $n-2$, $n-3$, etc.

Next, rearrange terms to get:

$$k^{(d+1)} - (k-1)^{(d+1)} = {(d+1) \choose 1} k^{d} + \ldots$$

If the expressions on the left and right are equal, their sums are equ

al: $$\sum_{k=1}^b \left[k^{(d+1)} - (k-1)^{(d+1)}\right] = \sum_{k=1}^b \left[{(d+1) \choose 1} k^{d} + \ldots\right]$$

The sum on the left has a nice telescoping property, e.g. $(1^3 - 0^3) + (2^3-1^3) + (3^3-2^3) + (4^3-3^3) + \ldots$. Because of all the cancellations, the sum on the left is just $b^{(d+1)}$— the peak value (!).
The sum on the right is something like: $$(d+1) \sum_{k=1}^b k^{d} + \sum_{k=1}^b (\ldots)$$ where we still don't need to care too much about the "$\ldots$" because those terms involve small exponents of $k$. We could compute those terms; it just won't matter when we use the formula later.
Putting the left and right sides together, we find $$(d+1)\sum_{k=1}^b k^d = b^{(d+1)} - \sum_{k=1}^b (\ldots)$$ or just $$\sum_{k=1}^b k^d = \frac{1}{(d+1)}b^{(d+1)} - \frac{1}{d+1}\sum_{k=1}^b (\ldots)$$

This is exactly the summation formula we were looking for! If we wanted an average formula, we could divide both sides by $b$: $$\frac{1}{b}\sum_{k=0}^b k^d = \frac{b^{d}}{d+1} - \frac{1}{b(d+1)}\sum_{k=0}^b (\ldots)$$

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Why does the average of a function comes out to be in this pattern for the purpose of integration

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