More differentiation qns (1 Viewer)

leehuan · Apr 17, 2016

$$Let $f$ be continuous on $[a,b]$. Prove that there exists $c\in\(a,b)$ such that$\\ \int_{a}^{b}{f(t)dt}=f(c)(b-a)$

When I tried experimenting with the MVT I found something like this:

$f(b)-f(a)=f^{\prime}(c)(b-a)$

However they gave a hint and I have no idea how to apply it.

$$Hint: Let $m=\min\{f(t):a\le t \le b\}$ and $M=\max\{f(t):a \le t \le b\}$ (existence holds by EVT I believe), and show that $\\m \le \frac{1}{b-a}\int_{a}^{b}{f(t)dt}\le M$

InteGrand · Apr 17, 2016

I'll denote the integral of f over the interval by I.

By the hint, we have m <= f(t) <= M for all t in [a,b].

Integrating the inequality from a to b,

m(b - a) <= I <= M(b-a).

Therefore, m <= (1/(b-a))*I <= M (dividing through by (b-a)).

So (1/(b-a))*I is between m and M, so by the intermediate value theorem there'll be a c in (a,b) such that f(c) = (1/(b-a))*I, i.e. f(c)*(b-a) = I, as required.

(Since we are told f is continuous on [a,b], we can indeed use the EVT and IVT.)

leehuan · Apr 17, 2016

Oh my goodness. leehuan stop overcomplicating crap.

leehuan · Apr 17, 2016

Just going to put all the remainder questions here as well... sorry

I am most certainly tired as hell. But all I want to do is scream. Can't get anything out.

InteGrand · Apr 17, 2016

leehuan said:
Just going to put all the remainder questions here as well... sorry

I am most certainly tired as hell. But all I want to do is scream. Can't get anything out.

$\noindent Some of these are pretty tedious computationally. I'll provide some hints/starters on the process for now.$

$\noindent 106. (Btw, for this theorem, which is known as the mean value theorem for integrals, $g$ actually needs to be a function of fixed sign, rather than just non-zero.) The hint given in the question is a good one. Start off with $m\leq f(x) \leq M$ for all $x\in [a,b]$ as before. Now, in terms of inequalities, what is the benefit of knowing $g(x)$ is of fixed sign for all $x\in [a,b]$? If you want the proof, it's written out with a statement of the theorem in Wikipedia here:$

https://en.wikipedia.org/wiki/Mean_value_theorem#First_Mean_Value_Theorem_for_Definite_Integrals .

$\noindent 107. A point on the ellipse $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ can be parameterised in a number of ways, e.g. $x=A\cos t$, $y=B\sin t$ for $t\in \left[0,2\pi\right)$, or $x=x$, $y=...$ (solve ellipse equation for $y$), for $-A \leq x \leq A$. You can use your favourite parameterisation to turn this into a one-variable problem. Then focus on optimising the squared-distance from a (parameterised) point on the ellipse to the point $(a,0)$.$

$\noindent Beware though that it turns out that for some values of $a$, the \emph{minimal}-distance point on the ellipse obtained by just setting derivatives to 0 etc. will give us a point that has non-real coordinates, and thus we need a different approach for these values of $a$. For these values of $a$, you can use your geometric intuition to find what the closest point on the ellipse would be, and hence the distance. You should then prove this is the correct answer (this can probably be done in a variety of ways).$

$\noindent 108. Since we want points that are horizontal inflexion points, we want places where the first and second derivatives are 0. To make it a horizontal point of inflexion, we should also have that the first non-zero derivative is an odd one (e.g. the third derivative or the fifth derivative etc.). This is due to the \emph{higher-order derivative test}, which you can read up on here:$

https://en.wikipedia.org/wiki/Derivative_test#Higher-order_derivative_test .

seanieg89 · Apr 18, 2016

InteGrand said:
(Btw, for this theorem, which is known as the mean value theorem for integrals, g actually needs to be a function of fixed sign, rather than just non-zero.)

Well the fact it is continuous on an interval and does not vanish on this interval implies that it is of fixed sign.

InteGrand · Apr 18, 2016

Oh yeah, missed that, sorry.

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