if $\int_0^4 f(x)dx = 5$ then $\int_0^2 f(x)dx=2.5$? If $\int_0^4 f(x)dx = 5$, then is it true that $\int_0^2 f(x)dx=2.5$?
I think this is only true if $f(x)$ is constant but not for polynomial function. However I don't know how to explain it in words.
 A: This is true if $f$ is a constant function, but it is also true for some other functions. For example, it is true for the function $f(x)=1.25 + \sin(\pi x)$.
In fact, you can find infinitely many polynomials for which this is true.
Of course, this is not true for all polynomials, and it is not true for all functions, for example it is not true for $f(x)=\frac{5}{8}x$.
A: Consider
$$F(x)=\int_0^xf(t)\,dt$$
Then $$\begin{cases}F(4)=5\\F(2)=2.5\end{cases}$$
There are infinitely many polynomials that satisfy these conditions and since $f(x)=\frac{d}{dx}F(x)$, there are infinitely many polynomial solutions for $f(x)$.
A: Consider a right triangle with its vertices at $(0,0),$ $(4,0)$ and $(4,2.5).$ Its area is $5,$ but the area of the triangle with vertices $(0,0),$ $(2,0),$ and $(2,1.25)$ is not $2.5.$ Do you see how this relates?
A: This is indeed not true in general. For some functions it is true, for others it is not. You cannot classify it in any simple terms.
You are however correct that it is true for constant functions, but there will also exist polynomial functions for which it is true (and a lot of polynomials for which it is false).
For instance, $f(x) = -\frac{15}{32}(x-0)(x-4)$, which is a $2$nd degree polynomial, has the property you want.
A: $$ \underbrace{\int_0^2f(x) dx}_{ = a } + \underbrace{\int_2^4 f(x)d x}_{ = 5 - a} = \underbrace{\int_0^4f(x) dx}_{ = 5} $$
Integration is basically summation. The above equation says the sum of $f(x)dx$ from $x=0$ to $x=2$ plus the sum from $x=2$ to $x=4$ equals the sum from $x=0$ to $x=4$.
If you know the first integral is $a$, and the last integral is 5, then you can deduce the middle integral is $5-a$. Any combination that adds up to 5 is possible. The first integral could be 1 and the second integral could be 4, or the first integral could be -10 and the second integral 15.
If both $\int_0^2 f(x)dx = \int_2^4 f(x)dx = 2.5$ you still can't say much about the value of $f(x)$ at individual points $x$. All you know is how $f(x)dx$ adds up!. As other answers have mentioned, all kinds of functions $f(x)$ can integrate up to 2.5 over the regions 0 to 2 and 2 to 4.
A: Let $f(x)=a_0+a_1x+\cdots+a_nx^n$ be a polynomial function.
Notice that
$$\begin{aligned}
\int_0^4f(x)\;dx=5\quad&\Leftrightarrow\quad 4a_0+\frac{4^2a_1}{2}+\cdots+ \frac{4^{n+1}a_n}{n+1}=5\\\\
&\Leftrightarrow\quad \frac{4a_0}{2}+\frac{4^2a_1}{4}+\cdots+ \frac{4^{n+1}a_n}{2(n+1)}=2.5\\\\
&\Leftrightarrow\quad 2a_0+2\cdot \frac{2^2a_1}{2}+\cdots+ 2^n\cdot\frac{2^{n+1}a_n}{n+1}=2.5
\end{aligned}$$
and
$$\int_0^2f(x)\;dx=2a_0+\frac{2^2a_1}{2}+\cdots+ \frac{2^{n+1}a_n}{n+1}$$
Therefore,
$$\int_0^4f(x)\;dx=5\quad\Rightarrow\quad\int_0^2f(x)\;dx=2.5\tag{1}$$
if and only if
$$(2-1)\frac{2^2a_1}{2}+\cdots+ (2^n-1)\frac{2^{n+1}a_n}{n+1}=0\tag{$2$}$$
Since $(2)$ is true if $a_1=\cdots=a_n=0$ but it is false in general, we conclude that $(1)$ is true if the polynomial is constant but it is not true for all polynomials.
Remark: You can use $(2)$ to construct polynomials that satisfies the condition and polynomials that doesn't satisfy the condition.
