-1
$\begingroup$

Determine if the Mean Value Theorem for Integrals applies to the function $f(x) = 2 − x^2$ on the interval the closed interval from $0$ to $\sqrt{2}$. If so, find the $x$-coordinates of the point(s) guaranteed by the theorem.

Help please!

  • 1
    What are your thoughts on this problem? You might want to start by looking up what the theorem says. Perhaps there is an example in the book that you might study to help. – 2017-02-23
  • 0
    Urgent need of help? – 2017-02-23
  • 0
    the mean value theorem says that the slope of the average rate of change matches at least once with the slope of the instantaneous rate of change. But I dont understand how to apply that to this problem. I really need help im terrible at this stuff – 2017-02-23
  • 0
    It applies to every continuous function on a closed interval. Finding the value of $x$ that makes it work is usually a simple computation. – 2017-02-23
  • 0
    so what would i plug in to find the point that it applies to? is there a formula? – 2017-02-23
  • 0
    i got 4/3 is that correct? – 2017-02-23
  • 0
    @DougM your formula misses a factor. – 2017-02-23
  • 0
    what is the factor? – 2017-02-23
  • 0
    @R.Mikaleson interval length on the left. – 2017-02-23
  • 0
    Do you know how to evaluate that integral? If so, $ \int f(x) dx = \sqrt 2 (2-x^2) \implies x= \sqrt {2-\frac {\sqrt 2}{2}\int f(x) dx}$ – 2017-02-23
  • 0
    but what about the intervals 0 to sqrt2 – 2017-02-23
  • 0
    but how can i use that to find the x coordinate – 2017-02-23

1 Answers 1

0

$\int_0^\sqrt2 2 - x^2 dx = 2\sqrt 2 - \frac 13 2\sqrt 2 = \frac {4\sqrt 2}3$

there exists an $x\in(0,\sqrt 2)$ such that:

$f(x) = (2-x^2) =\frac {\int_0^\sqrt2 2 - x^2 dx}{\sqrt 2}\\ (2-x^2) = \frac 43\\ x^2 = 2 - \frac 43\\ x = \sqrt {\frac23}$