If $f:[a,b]\to\Bbb{R}$ is a continuous function and $g:\Bbb{R}\to\Bbb{R}$ is defined by: $$g(z)=\int_a^b{(f(x)-z})dx$$

$a)$ Show that exists $m\in\Bbb{R}$ that $g(m)=0$

$b)$ What is the geometric interpretation of this result?

Do I have to use the mean value theorem? I tried but I got stuck and I am not entirely sure when this is positive or negative.

  • $\begingroup$ $g(z) = (a-b)z + K$ The geometric interpretation of the result IS the mean value theorem. $\endgroup$ – Doug M Jun 20 '17 at 23:41

Let $F: [a, b] \to \mathbb{R}$ be an antiderivative of $f$, then we can use fundamental theorem of calculus to get $$g(z) = \int_a^b(f(x) - z)dx = \int_a^b f(x) dx - \int_a^b z dx = F(b) - F(a) - (b - a) z.$$

Thus $g(m) = 0$ when $m = \frac{F(b) - F(a)}{b - a}.$

Geometrically, $g(z)$ the integral of all the differences between f(x) and z at all points between $a$ and $b.$

Consider a similar finite case. Let $f(n): \{1, 2, \dots, k\} \to \mathbb{R}$, and $g(z) = \sum_{i = 1}^k (f(i) - z)$.

Then $$g(z) = \sum_{i = 1}^k f(i) - \sum_{i = 1}^k z = k\left(\frac{\sum_{i = 1}^k f(i)}{k} - z\right).$$

Thus $g(z) = 0$ when $z = \frac{\sum_{i = 1}^k f(i)}{k} = \text{mean}(\{f(i) : i = 1, 2, \dots k\}).$

  • $\begingroup$ Thanks. Is it not $-(b-a)zx$ in the first line? $\endgroup$ – Daniel Cintra Jun 21 '17 at 0:08
  • $\begingroup$ No. $x$ is the variable we are integrating over. Once we evaluate the integral, we should not have any $x$'s in the result. $\endgroup$ – zrbecker Jun 21 '17 at 0:25
  • $\begingroup$ Of course... thanks $\endgroup$ – Daniel Cintra Jun 21 '17 at 0:27

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