# A Sprint to FTC Our sprint starts with Lagrange’s Mean-Value theorem which states:

A function $f(x)$ is

(1) continous on closed interval $[a, b]$

(2) differentiable on open interval $(a, b)$ $\implies\exists c \in (a, b) \ni {{f(b)-f(a)} = f'(c) (b-a) }$.

Let’s prove it.

The area of a triangle with vortices $(x, f(x)), (a, f(a))$ and $(b, f(b))$ is the absolute value of ${1 \over 2} \left|\begin{array}{ccc} x & f(x) & 1 \\ a & f(a) & 1 \\ b & f(b) & 1 \end{array}\right|$

where $x \in [a, b]$ (See “Had Heron Known Analytic Geometry“)

Let $g(x) = \left|\begin{array}{ccc} x & f(x) & 1 \\ a & f(a) & 1 \\ b & f(b) & 1 \end{array}\right|$,

Since $f(x)$ is differentiable, we have $g'(x) = (b-a)f'(x)-f(b)+f(a)$.

Clearly, $g(a) = g(b) = 0$

Therefore, by Rolle’s Theorem (See “Rolle’s theorem”) $\exists c \in (a, b) \ni g'(c) = 0$, i.e., $\exists c \in (a, b) \ni {{f(b)-f(a)}} = f'(c) (b-a)$

or $\exists c \in (a, b) \ni f'(c) = {{f(b)-f(a)} \over {b-a}}$.

The geometric meaning of Lagrange’s Mean-Value theorem is illustrated for several functions in Fig. 1. It shows that the graph of a differentiable function has at least one tangent line parallel to the cord connecting $(a, f(a))$ and $(b, f(b))$. Fig. 1

The bottom curve falsifies the theorem due to its missing differentiability at one point.

Following Lagrange’s Mean-Value theorem are two corollaries. We have encountered and accepted the first one without proper proof in the past (See “Inching towards Definite Integral“)

Let’s state and prove them now.

Corollary 1. A function $f(x)$ on an open interval $I$ has derivative zero at each point $\implies \forall x \in I, f(x) = c$, a constant.

It is true due to the fact that $\forall x_1, x_2 \in I, x_1 \neq x_2, f(x)$ is both continous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$. By Lagrange’s Mean-Value theorem, $\exists c \in (x_1, x_2) \ni f(x_2)-f(x_1) = f'(c)(x_2-x_1)$. Since $\forall x \in I, f'(x)=0 \implies f'(c)=0$, We have $f(x_2)-f(x_1)=0 \cdot (x_2-x_1) = 0$. i.e., $f(x_2) = f(x_1)$. Hence $f(x)=c$, a constant on $I$.

Corollary 2. Two functions, $f_1(x)$ and $f_2(x)$ have the same derivative at each point on an open interval $I \implies \forall x \in I, f_1(x) - f_2(x) = c$, a constant.

For $\forall x \in I, f_1'(x)=f_2'(x) \implies (f_1(x) - f_2(x))' = f'_1(x) - f'_2(x) = 0$. By corollary 1, $f_1(x) - f_2(x) = c$, a constant.

Next, we define a set $G$ as follows: $G \triangleq \{g(x) | g'(x) = f(x)\}$

i.e., $G$ is a set of function such that $\forall g \in G, g'(x) = f(x)$ for all $x$‘s. $G$ is certainly not empty for we have proved that $\int\limits_{a}^{x}f(x)\;dx \in G$ by showing $(\int\limits_{a}^{x}f(x)\;dx)'= f(x)$ (See “Inching towards Definite Integral“). It follows that $\forall g \in G$, $(g(x)-\int\limits_{a}^{x}f(x)\;dx)' = g'(x)-(\int\limits_{a}^{x}f(x)\;dx)'=f(x)-f(x)=0$

By Corollary 2, $g(x)-\int\limits_{a}^{x}f(x)\;dx = c$.

Let $x=a$, we have $g(a)-\int\limits_{a}^{a}f(x)\;dx = g(a)-0=g(a)=c$.

Hence $g(x)-\int\limits_{a}^{x}f(x)\;dx = g(a)$

or $\int\limits_{a}^{x}f(x)\;dx = g(x)-g(a)$.

Let’s summarize the result of our exploration in a theorem:

On an open interval $I$ containing $a$, a function $g(x)$ is differentiable and, $g'(x) = f(x)\implies \forall x \in I$, $\int\limits_{a}^{x} f(x)\;dx = g(x)-g(a)\quad\quad\quad\quad(1)$

This is FTC, The Fundamental Theorem of Calculus.

Let $x=b$, (1) becomes the Newton-Leibniz formula $\int\limits_{a}^{b}f(x)\;dx = g(b)-g(a)\quad\quad\quad\quad$,

our old friend for evaluating $\int\limits_{a}^{b} f(x) \;dx$ (See “Inching towards Definite Integral“)

# Knock Euler’s Line

my imagination
is a piece of board
my sole instrument
is a wooden stick

I strike the board
yes—yes
no—no

“A Knocker” by Zbigniew Herbert

Euler’s line theorem states
In every triangle
the intersection of the medians
the intersection of the heights
and the center of the circumscribed circle
are on a straight line

Let’s prove it
with the aid of Omega CAS Explorer

We know $x^2+y^2+d \cdot x + e \cdot y +f =0\quad\quad\quad(1)$

is a circle centered at $(-{d \over 2}, -{e \over 2})$ with radius $r^2={{d^2+e^2-4f} \over 4}\quad\quad\quad(2)$

provide (2) is positive

Find d, e, f from triangle’s vertices $(-x_1, 0), (x_1, 0), (x_2, y_2)$:

ceq:x^2+y^2+d*x+e*y+f=0;
eq1:ev(ceq, x=x1, y=0);
eq2:ev(ceq, x=-x1, y=0);
eq3:ev(ceq, x=x2, y=y2);
sol:linsolve([eq1, eq2, eq3], [d,e,f]); $\implies d=0, e=-{{y_2^2-x_2^2+x_1^2} \over y_2}, f=-x_1^2$

Evaluate (2):

ev(d^2 + e^2-4*f, sol); $\implies {(-y_2^2-x_2^2+x1^2)^2\over y_2^2} + 4x_1^2$

always positive for $x_1>0, y2 \neq 0$

Find the center of the circumscribed circle $x_c, y_c$:

xc:ev(-d/2, sol)$ $\implies 0$ yc:ev(-e/2, sol)$ $\implies {{-y_2^2-x_2^2+x_1^2} \over {2y_2}}$

Find the intersection of the medians $x_m, y_m$:

eq1:y*((x1+x2)/2+x1)=y2/2*(x+x1)$eq2:y*((x2-x1)/2-x1)=y2/2*(x-x1)$
sol:solve([eq1, eq2], [x,y])$xm:ev(x,sol); $\implies {x_2 \over 3}$ ym:ev(y, sol); $\implies {y_2 \over 3}$ is(ev(x2*y=y2*x, x=xm, y=ym)); $\implies true$ Find the intersection of the heights $x_h, y_h$: eq1:y2*y=-(x2+x1)*(x-x1)$
eq2:y2*y=-(x2-x1)*(x+x1)$sol:solve([eq1, eq2], [x,y])$
xh:ev(x, sol); $\implies x_2$

yh:ev(y, sol); $\implies -{{x_2^2-x_1^2} \over y_2}$

Compute the area of triangle with vertices $(x_c, y_c), (x_m, y_m), (x_h, y_h)$:

m:matrix([xc, yc,1], [xm, ym, 1], [xh, yh,1]);
determinant(m)\$
ratsimp(%); $\implies 0$

Indeed $(x_c, y_c), (x_m, y_m)$ and $(x_h, y_h)$ are on a straight line.