함수의 평균값

The question tells you $\frac{1}{1-0}\int_0^1 f(x)\,\mathrm{d}x = \frac{f(0)+f(1)}{2}$. Note $f(0)=c$, and $f(1) = a+b+c$.

$$\begin{aligned} \frac{1}{1-0}\int_0^1 f(x)\,\mathrm{d}x &= \frac{f(0)+f(1)}{2} = \frac{c+(a+b+c)}{2}\\ \int_0^1 \big(ax^2+bx+c \big)\,\mathrm{d}x&=\frac{a+b+2c}{2}\\ \left[\frac{a}{3}x^3+\frac{b}{2}x^2+cx \right]_0^1&=\frac{a}{2}+\frac{b}{2}+c \\\frac{a}{3}+\frac{b}{2}+c&=\frac{a}{2}+\frac{b}{2}+c\\ \frac{a}{3}&=a\\ a&=0 \end{aligned}$$

That is, $f(x)$ is linear.

The function $g(x)=f(a+b-x)$, on the interval $[a,b]$, is a mirror of the function $f(x)$, with $g(a)=f(b)$ and $g(b)=f(a)$. So, $\int_a^b f(a+b-x)\,\mathrm{d}x = \int_a^bf(x)\,\mathrm{d}x$, and hence $\frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x = \frac{1}{b-a}\int_a^bf(x)\,\mathrm{d}x$, so the average value of $f(a+b-x)$ on $[a,b]$ is $A$.

Alternately, we can evaluate $\frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x$ directly, using the substitution $u=a+b-x$, $\mathrm{d}x=-\mathrm{d}x$:

$$\begin{aligned} \frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x&=\frac{-1}{b-a}\int_{u(a)}^{u(b)} f(u)\,\mathrm{d}u\\ &=\frac{-1}{b-a}\int_{b}^{a} f(u)\,\mathrm{d}u \\&=\frac{1}{b-a}\int_{a}^{b} f(u)\,\mathrm{d}u \\&=A \end{aligned}$$

(a) The function $A(x)$ only gives us information about an integral when one limit of integration is zero. We can get around this by using properties of definite integrals from Section 1.2 to break our integral into two integrals, each of which has 0 as one limit of integration. So, we find the average of $f(t)$ on $[a,b]$ as follows:

$$\begin{aligned} \frac{1}{b-a}\int_a^b f(t)\,\mathrm{d}t&= \frac{1}{b-a}\left(\int_a^0 f(t)\,\mathrm{d}t + \int_0^b f(t)\,\mathrm{d}t\right)\\ &=\frac{1}{b-a}\left(-\int_0^a f(t)\,\mathrm{d}t + \int_0^b f(t)\,\mathrm{d}t \right)\\ &=\frac{1}{b-a}\left(-a\cdot\underbrace{\frac{1}{a}\int_0^a f(t)\,\mathrm{d}t}_{A(a)} + b\cdot\underbrace{\frac{1}{b}\int_0^b f(t)\,\mathrm{d}t}_{A(b)} \right)\\ &=\frac{1}{b-a}\left(-aA(a)+bA(b) \right) = \frac{bA(b) - aA(a)}{b-a} \end{aligned}$$ (b) From the definition of $A(x)$, we know $$A(x) = \frac{1}{x}\int_0^x f(t)\,\mathrm{d}t$$ That is, $$xA(x) =\int_0^x f(t)\,\mathrm{d}t$$ To find $f(x)$, we differentiate both sides. For the left side, we use the product rule; for the right side, we use the Fundamental Theorem of Calculus part 1. $$A(x)+xA'(x) = f(x)$$ So, $f(t)=A(t)+tA'(t)$.

This seems like it might be true: if $f$ is getting closer and closer to zero, as $x$ grows towards infinity, then over time the later values will become a larger and larger portion of the total interval we're looking at, and so the average should look more and more like $f(x)$ when $x$ is large–that is, like 0. That's some intuition to start us out, but it isn't a rigorous argument. To be sure we haven't overlooked something, let's use the definition of an average to express $A(x)$.

$$\begin{aligned} A(x)&=\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t\\ \lim_{x \to \infty}A(x)&=\lim_{x \to \infty}\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \\ \text{If $\int_0^\infty f(t) \,\mathrm{d}t$ converges, then this limit is 0, and the statement is true. So, suppose it does not converge. Since $f(x)$ is positive, that means $\lim\limits_{x \to \infty} \int_0^x f(t)\,\mathrm{d}(t) = \infty$, so we can use l'H\^{o}pital's rule. To differentiate the numerator, we use the Fundamental Theorem of Calculus part 1.} \\ \lim_{x \to \infty}A(x)&=\lim_{x \to \infty}\frac{\int_0^x f(t)\,\mathrm{d}t}{x} = \lim_{x \to \infty}\frac{f(t)}{1}=0 \end{aligned}$$

So, the statement is true whether $\int_0^xf(t)\,\mathrm{d}t$ converges or not.