# Proof of Definite Integral

Let’s assume that $f$ is continuous and positive on the interval $[a, b]$. Then the definite integral $\int^b_a f(x) dx$ represents the area of the region bounded by the graph of $f$ and the x-axis, from x = a to x = b. First, we partition the interval $[a, b]$ into n subintervals, each of width $\Delta x = (b - a)/n$ such that $a = x_0 < x_1 < x_2 < . . . < x_n = b$ Then we can form a trapezoid for each subinterval and the area of the ith trapezoid = $[\frac{f(x_{i-1}) + f(x_i)}{2}](\frac{b-a}{n})$. This implies that the sum of the areas of the n trapezoids is Area = $\frac{b - a}{2n}[f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)] = \frac{b - a}{2n}[f(x_0) + f(x_n) + 2\displaystyle\sum\limits_{i=1}^{n-1} f(x_i)] = \frac{b - a}{2n}(f(x_0) + f(x_n)) + \displaystyle\sum\limits_{i=1}^{n-1} f(x_i)(\frac{b - a}{n}) = \frac{b - a}{2n}(f(x_0) + f(x_n) - 2f(x_n)) + \displaystyle\sum\limits_{i=1}^{n} f(x_i)(\frac{b - a}{n}) - 2f(x_n)(\frac{b - a}{2n}) = \frac{b - a}{2n}(f(x_0) - f(x_n)) + \displaystyle\sum\limits_{i=1}^{n} f(x_i)\Delta x = \lim_{n\to\infty}\frac{b - a}{2n}(f(x_0) - f(x_n)) + \lim_{n\to\infty}\displaystyle\sum\limits_{i=1}^{n} f(x_i)\Delta x = 0 + \lim_{n\to\infty}\displaystyle\sum\limits_{i=1}^{n}f(x_i)\Delta x = \int^b_a f(x) dx$

# Differentiability Implies Continuity

A theorem in Calculus states that if $f$ is differentiable at $x = c$, then $f$ is continuous at $x = c$. With this theorem in mind it can be proven that $f$ is continuous at $x = c$ by showing that $f(x)$ approaches $f(c)$ as $x \rightarrow c$. To do this, we can use the differentiability of $f$ at $x = c$ and consider the following limit.

$\displaystyle\lim_{x\to\ c} [f(x) - f(c)] = \displaystyle\lim_{x\to\ c} \bigg[(x - c)\bigg( \frac{f(x) - f(c)}{x - c} \bigg) \bigg]$

$= \bigg[\displaystyle\lim_{x\to\ c} (x - c) \bigg] \bigg[\displaystyle\lim_{x\to\ c} \frac{f(x) - f(c)}{x - c}\bigg]$

$= (0)[f\prime(c)]$

In this post I am going to introduce the formal definition of a limit. I am still trying to figure out the symbol formatting so bear with me. If $f(x)$ becomes arbitrarily close to a single number $L$ as $x$ approaches $c$ from either side, then the limit of $f(x)$ as $x$ approaches $c$ is $L$, which can be written as $\displaystyle\lim_{x\to\ c} f(x) = L$. This definition may seem formal but isn’t because exact meanings have not yet been given to the two phrases  “$f(x)$ becomes arbitrarily close to $L$” and “$x$ approaches $c$“. In the figure below we can let $\epsilon$ represent a (small) positive number. Then the phrase $"f(x)$ becomes arbitrarily close to $L"$ means that $f(x)$ lies in the interval $(L - \epsilon, L + \epsilon)$. Using absolute value, we can write this as $|f(x) - L| < \epsilon$. Similarly, the phrase “$x$ approaches $c$” means that there exists a positive number $\delta$ such that $x$ lies in either the interval $(c - \delta, c)$ or the interval $(c, c + \delta)$. This fact can be concisely expressed by the double inequality $0 < |x -c| < \delta$. The first inequality $0 < |x -c|$ states that the distance between $x$ and $c$ is more than 0, and expresses the fact that $x \neq c$. The second inequality $|x - c| < \delta$ says that $x$ is within $\delta$ units of $c$.

This brings us to a formal definition which can be stated as following: Let $f$ be a function defined on an open interval containing $c$ (except possibly at $c$) and let $L$ be a real number. The statement $\displaystyle\lim_{x\to\ c} f(x) = L$ means that for each $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < |x-c| < \delta$, then $|f(x) - L| < \epsilon$. This can be expressed symbolically as $\forall \epsilon > 0 \exists \delta > 0 : \forall x (0 < |x - c| < \delta \Rightarrow |f(x) - L| < \epsilon)$  .

# Heron’s Area Formula

Here is the definition of Heron’s Formula which is an another way of finding the area of a triangle:

Given any triangle with sides of lengths a, b, and c, the area of the triangle is Area = √[s(s – a)(s – b)(s – c)] where s = (a + b + c)/2

And here is the proof:

The area of an oblique triangle is 1/2bc sin A where the height of the triangle is b sin A. If we square each side of the equation for area we get (Area)*2 = 1/4b*2c*2sin*2A. Now if we take the square root of each side of the equation we get Area = √[1/4b*2c*2sin*2A]. If we replace  sin*2A with the corresponding pythagorean identity, (1 – cos*2A), we get the equation √[1/4b*2c*2(1 – cos*2 A)]. Now we can factor the right side of the equation to get √[(1/2bc(1 + cos A))(1/2bc(1 – cos A)].

Using the Law of Cosines, we can show that 1/2bc(1 + cos A) = [(a + b + c)/2] · [(-a + b + c)/2] and 1/2bc(1 – cos A) = [(a – b + c)/2] · [(a + b – c)/2]. The middle dot signifies a multiplicative operation and not the dot product. I have uploaded a proof  below showing how to arrive at the first equivalence

.

Now letting s = (a + b + c)/2, these two equations can be written as 1/2bc(1 + cos A) = s(s – a) and 1/2bc(1 – cos A) = (s – b)(s – c). By substituting into the last formula for area, we can conclude that Area = √[s(s – a)(s – b)(s – c)]

# Proof of Sum and Difference Formulas

The sum and difference formulas can be used to find exact values of trigonometric functions involving sums or differences of special angles. Here are the sum and difference formulas that I will be providing the proofs for:

cos(u + v) = cos u cos v – sin u sin v

cos(u – v) = cos u cos v + sin u sin v

tan(u ± v) = (tan u – tan v)/(1 + tan u tan v)

Proof: You can use the figures in the pictures below to better understand the proof of the formulas for cos(u ± v). In the left figure, let A be the point (1,0) and then use and to locate the points B=(x¹, y¹), C=(x², y²), and D=(x³, y³) on the unit circle. So, (x¹)*2 + (y¹)*2=1, (x²)*2 + (y²)*2=1, and (x³)*2 + (y³)*2=1. Now this notation might seem strange but allow me to explain the signification of particular characters and state that the choice of these strange characters can be attributed to the fact that I am limited to the characters in the Word Press Kitchen Sink. Let *n signify the operation of exponentiation where n is an integer, and let ¹, ², and ³ signify the coordinates corresponding to B, C, and D respectively. It is slightly counterintuitive but stays consistent with the logic.

Now assume that 0 < v < u < 2pi and note that in the figure to the right, arcs AC and BD have the same length. So, line segments AC and BD are also equal in length, which implies that √[(x² – 1)*2 + (y² – 0)*2] = √[(x³ – x¹)*2 + (y³ – y¹)*2].

Now to do some algebra so as to isolate x².

After applying the proper foiling operations we obtain: (x²)*2 – 2x² + 1 + (y²)*2 = (x³)*2 – 2x¹x³ + (x¹)*2 + (y³)*2 – 2y¹y³ + (y¹)*2. Now we can group the squared coordinates of on the left side of the equation and group the squared coordinates of D and on the right side of the equation. After this operation the equation will look like this: [(x²)*2 + (y²)*2] +1 – 2x² = [(x³)*2 + (y³)*2 + [(x¹)*2 + (y¹)*2] – 2x¹x³ – 2y¹y³. Now since we know that x*2 + y*2 =1, we know that by replacing the or coordinates with the respective coordinates of either A, B, C, or the equation will still equal 1, so we can replace [(x²)*2 + (y²)*2], [(x³)*2 + (y³)*2], and [(x¹)*2 + (y¹)*2] with 1. After doing this we obtain 1 + 1 – 2x² = 1 + 1 – 2x¹x³ – 2y¹y³ which simplifies to x² = x³x¹ + y³y¹.

Now we can substitute the values x² = cox(u – v), x³ = cos u, x¹ = cos v, y³ = sin u, and y¹ = sin v which changes the above equation to cos(u – v) = cos u cos v + sin u sin v. The formula for cos(u + v) can be obtained by considering u + v = u – (-v) and using the formula just derived above to obtain: cos(u + v) = cos[u – (-v)] = cos u cos(-v) + sin u sin(-v) which equals cos u cos v – sin u sin v.

Now to get the formula for sin(u + v), we are going to take cos(u + v) = cos u cos v – sin u sin v and replace  by pi/2 – u. By doing this we the equation cos(pi/2 – u + v) = cos(pi/2 – u) cos v – sin(pi/2 – u) sin v. Since we know that sin(pi/2 – u) = cos ucos(pi/2 – u) = sin u, and cos[pi/2 -(u – v)] = sin (u – v) we can conclude that sin(u – v) = sin u cos v – cos u sin v. If we want to obtain sin(u + v) all we have to do is replace by -v.

Finally we can use the sum and difference formulas for sine and cosine to prove the formulas for tan(u ± v). The proof is as follows:

tan(u ± v) = sin(u ± v)/cos(u ± v). Then if we plug in the appropriate sum and difference formulas for sin and cos then we obtain (sin u cos v ± cos u sin v)/(cos u cos v ± sin u sin v). Then if we divide the numerator and the denominator by cos u cos v we obtain the equation [(sin u cos v ± cos u sin v)/(cos u cos v)]/[(cos u cos v ± sin u sin v)/(cos u cos v)] Now if we write the numerator of the fraction in the numerator of the expression and the numerator of the fraction in the denominator of the expression we obtain the equation: [(sin u cos v)/(cos u cos v) ± (cos u sin v)/(cos u cos v)]/[(cos u cos v)/(cos u cos v) ± (sin u sin v)/(cos u cos v). Now in the numerator of the expression cos u /cos u occurs twice and therefore cancels twice; once in the fraction containing sin u in its numerator and once in the fraction containing sin v in its numerator. In the denominator of the expression the fraction containing cos u cos v in its numerator cancels out completely because their is symmetry between the numerator and denominator leaving the value 1 remaining. After the operation of cancellation we are left with the expression: [(sin u/cos u) ± (sin v/ cos v)]/[(1±sin u/cos u)·(sin v/cos v)]. Using the tangent quotient identity we obtain the expression (tan u ± tan v)/(1 ± tan u tan v) which is the sum and difference formula for tan(u ± v).