Let $a,b \in \mathbb{R}, a < b$. By a partition of the closed interval $[a,b]$ is meant a finite sequence of numbers $x_0,x_1,...,x_N$ such that $a = x_0 < x_1 < ... < x_N = b$. The width of this partition is defined to be $$\text{max}\{x_i - x_{i-1} : i = 1,2,...,N\}$$ If $f$ is a real-valued function on $[a,b]$, by a Riemann sum for $f$ corresponding to the given partition is meant a sum $$\sum_{i = 1}^{N} f(x_i')(x_i - x_{i-1})$$ where $x_{i-1} \leq x_i' \leq x_i$ for each $1,2,...N$.
Let $a,b \in \mathbb{R}, a < b$, and let $f$ be a real-valued function on $[a,b]$. We say that $f$ is Riemann Integrable on $[a,b]$ if there exists a number $A \in \mathbb{R}$ such that, for any $\epsilon > 0$, there exists a $\delta > 0$ such that $|S-A| < \epsilon$ whenever $S$ is a Riemann sum for $f$ corresponding to any partition of $[a,b]$ of width less than $\delta$. In this case $A$ is called the Riemann integral of $f$ between $a$ and $b$ and it is denoted $\int_a^b f(x) dx$.

Proposition: Linearity and Order Properties of the Integral

Riemann Integration has the following properties:
(1) If $f,g$ are integrable real-valued functions on the interval $[a,b]$ then $f+g$ is integrable on $[a,b]$ and $$\int_a^b (f(x) + g(x))dx = \int_a^b f(x) dx + \int_a^b g(x) dx$$ (2) If $f$ is an integrable real-valued function on the interval $[a,b]$ and $c \in \mathbb{R}$ then $cf$ is integrable on $[a,b]$ and $$\int_a^b cf(x) dx = c \int_a^b f(x) dx$$
If $f$ is an integrable real-valued function on the interval $[a,b]$ and $f(x) \geq 0$ for all $x \in [a,b]$, then $$\int_a^b f(x) dx \geq 0$$

Proofs:

Given any $\epsilon$, there exists $\delta_1, \delta_2$ such that if $S_1, S_2$ are Riemann sums for $f,g$ with partitions of widths less than $\delta_1, \delta_2$, then we have: $$|S_1 - \int_a^bf(x)dx| < \frac{\epsilon}{2} , |S_2 - \int_a^bg(x)dx| < \frac{\epsilon}{2}$$ We can easily verify that the error between $S$ and $\int_a^bf(x)dx +\int_a^bg(x)dx$ is bounded by $\epsilon$ if we use a partition of width $\text{min}\{\delta_1,\delta_2\}$.

For part (2), notice that given any $\epsilon$ there is a $\delta$ such that any Riemann Sum $S$ for $f$ corresponding to a partition of $[a,b]$ with width less than $\delta$ satisfies $|S - \int_a^b f(x)dx| < \epsilon/|c|$. Then a partition of $cf(x)$ has error $\epsilon$ when compared to the value $c\int_a^bf(x)dx$.

A real-valued function $f$ on the interval $[a,b]$ is called a step function if there exists a partition $x_0,x_1,...x_N$ of $[a,b]$ such that $f$ is constant on each open subinterval $(x_0,x_1),(x_1,x_2),...(x_{N-1},x_N)$.

Proposition: Existence of the Integral

A real-valued function $f$ on the interval $[a,b]$ is integrable on $[a,b]$ if and only if, given any $\epsilon > 0$, there exists a number $\delta > 0$ such that $|S_1 - S_2| < \epsilon$ whenever $S_1$ and $S_2$ are Riemann sums for $f$ corresponding to partitions of $[a,b]$ of width less than $\delta$.
A step function is integrable. In particular, if $x_0,x_1,...x_N$ is a partition of the interval $[a,b]$, if $c_1,...,c_N \in \mathbb{R}$ and if $f: [a,b] \to \mathbb{R}$ is such that $f(x) = c_i$ if $x_{i-1} < x < x_i$ for $i = 1,...,N$, then $$\int_a^bf(x)dx = \sum_{i = 1}^N c_i (x_i - x_{i-1})$$
The real-valued function $f$ on the interval $[a,b]$ is integrable on $[a,b]$ if and only if, for each $\epsilon$, there exist step functions $f_1,f_2$ on $[a,b]$ such that $$f_1(x) \leq f(x) \leq f_2(x) \qquad \text{ for each } x \in [a,b]$$ and $$\int_a^b(f_2(x) - f_1(x))dx < \epsilon$$
If $f$ is a continuous real-valued function on the interval $[a,b]$ then $\int_a^bf(x)dx$ exists.

Proofs:

First let's do the forward proof. Assume that $f$ is integrable on $[a,b]$. Then for any $\epsilon$, we have that there exists some $\delta$ such that $|S - \int_a^bf(x)dx| < \epsilon/2$ whenever $S$ is a partition of width less than $\delta$. If $S_1,S_2$ are two such Riemann Sums then: $$|S_1 - S_2| = |(S_1 - \int_a^bf(x)dx) - (S_2 - \int_a^bf(x)dx)|$$ $$ \leq |S_1 - \int_a^bf(x)dx| + |S_2 - \int_a^bf(x)dx| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ proving the forward proof. Now for the reverse proof, assume the right hand side of the proposition. For $n = 1,2,3,...$ choose any partition of $[a,b]$ of width less than $1/n$ with a corresponding Riemann sum $S^n$. Then $S^1, S^2,S^3,...$ is a Cauchy sequence, and has a limit $A$. Now, for any $\epsilon$, pick a corresponding $\delta$ such that $|S_1 - S_2| < \epsilon/2$ whenever $S_1,S_2$ have widths less than $\delta$. By the convergence of $S^n$, pick an $N$ such that $|S^N - A| < \epsilon/2$ and $1/N < \delta$. We then have that for $S$ a partition with width less than $\delta$: $$|S - A| = |(S - S^N) + (S^N - A)| \leq |S - S^N| + |S^N - A|$$ $$ < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ Thus $f$ is integrable on $[a,b]$.
For $i = 1,2,...N$ define $\phi_i : [a,b] \to \mathbb{R}$ by: $$\phi_i(x) = \begin{cases} 1 \text{ if } x \in (x_{i-1},x_i) \\ 0 \text{ if } x \in [a,b], x \notin (x_{i-1},x_i) \end{cases}$$ Then $f - \sum_{i = 1}^N c_i \phi_i$ is a function on $[a,b]$ that takes on the value zero at all points except possibly the points $x_0,x_1,...x_N$, hence it is a sum of finitely many functions $g$ which take are zero everywhere except a single point. But for any Riemann Sum $S$ of width $\delta$ of such $g$, we have that $|S| < 2 |c| \delta$ where $c$ is the value of the non-zero point. This vanishes as $\delta$ vanishes, hence the Riemann integral of all such $g$ is zero. So, by the linearity of integrals, we have that: $$lim_a^b (f(x) - \sum_{i = 1}^{N} c_i \phi_i(x))dx = 0$$ But it is easy to prove that each $phi_i(x)$ is integrable too (and has value $x_i - x_{i-1}$). So by linearity we have: $$\int_a^bf(x)dx = \int_a^b(f(x) - \sum_{i = 1}^{N} c_i \phi_i(x)) + \sum_{i = 1}^{N} c_i \int_a^b \phi_i(x)dx$$ $$= \sum_{i = 1}^{N} c_i (x_i - x_{i-1})$$ Proving the proposition.
First, we prove the reverse proof, i.e. assume that there exist such step functions $f_1,f_2$ such that $f_1(x) \leq f(x) \leq f_2(x)$ with $\int_a^b (f_2(x) - f_1(x))dx < \epsilon/3$ for given $\epsilon$. Since $f_1,f_2$ are integrable, we can find a $\delta$ such that a Riemann sum of $f_1,f_2$ of width less than $\delta$ has error bounded by $\epsilon/3$. Now let $S$ be a Riemann sum of $f$ with width less than $\delta$. Then we have that: $$f_1(x) \leq f(x) \leq f_2(x)$$ $$S^1 \leq S \leq S^2$$ where $S^1,S^2$ are Riemann sums for $f_1,f_2$ using the same 'testing-variables' that $S$ is using. (The testing variable being the variable that you pick to represent each rectangle's height). Because all of these Riemann sums have width less than $\delta$, $$S^1 - \frac{\epsilon}{3} < S < S^2 + \frac{\epsilon}{3}$$ $$S \in (S^1 - \frac{\epsilon}{3}, S^2 + \frac{\epsilon}{3})$$ But by assumption, $S^2 - S^1$ has size less than $\epsilon/3$. So the open interval has length less than $\epsilon$. But this means that any $S_a,S_b$ partitions of width less than $\delta$ of $f$ are within $\epsilon$ of each other, fulfilling the Cauchy-like criterion of the first proposition, implying that $f$ is integrable.

Now assume that $f$ is integrable. According to proposition 1, the Cauchy-like proposition, given $\epsilon$, there exists a $\delta$ such that for a given partition of $f$ with width less than $\delta$, we can find two Riemann sums that differ by some $\epsilon'$, where $\epsilon'$ is less than $\epsilon$. For each Riemann sum, the integral is split into $N$ different rectangles, and we much pick a "testing point" for each rectangle within some closed set. Suppose that all but one testing point, is picked equal. For one Riemann sum, we use the testing point $x_j$ (this is the right-most end of the rectangle). The other Riemann sum uses the testing point $x_j'$ Then: $$(f(x_j') - f(x_j))(x_j - x_{j-1}) < \epsilon'$$ implying $$|f(x_j')| < \frac{\epsilon'}{x_j - x_{j+1}} + |f(x_j)|$$ This last inequality holds for all $x_j' \in [x_{j-1},x_j]$. Thus $f$ is bounded on this closed interval, and because $j$ is arbitrary, on the entire closed set. Thus we can define: $$m_i = \text{g.l.b}\{f(x_i') : x_i' \in [x_{i-1},x_i]\}$$ $$M_i = \text{l.u.b}\{f(x_i') : x_i' \in [x_{i-1},x_i]\}$$ and we can define step functions $f_1,f_2$ on $[a,b]$ by $$f_1(x) = \begin{cases} m_i \qquad \qquad \qquad \ \ \text{ if } x_{i-1} < x < x_i, i = 1,...,N \\ \text{min}\{m_1,...,m_N\} \text{ if } x = x_i, i = 0,1,...N \end{cases}$$ $$f_2(x) = \begin{cases} M_i \qquad \qquad \qquad \ \ \ \text{ if } x_{i-1} < x < x_i, i = 1,...,N \\ \text{max}\{M_1,...,M_N\} \text{ if } x = x_i, i = 0,1,...N \end{cases}$$ Clearly $f_1 \leq f \leq f_2$. Now pick an arbitrary number $\gamma > 0$. We have that for $x_i',x_i'' \in [x_{i-1},x_i], i = 1,...N$ $$f(x_i') < m_i + \gamma, f(x_i'') > M_i - \gamma$$ Then $$\sum_{i = 1}^N (f(x_i'') - f(x_i'))(x_i - x_{i-1}) > \sum_{i = 1}^N (M_i - m_i - 2\gamma)(x_i-x_{i-1})$$ $$ = \int_a^b (f_2(x) - f_1(x))dx - 2\gamma(b-a)$$ But we know that the left hand side (the difference between the two Riemann sums) is bounded by $\epsilon'$ (by our assumption that $f$ is integrable, using the Cauchy-like criterion). So we know the right hand side is bounded: $$\int_a^b (f_2(x) - f_1(x))dx - 2\gamma(b-a) < \epsilon'$$ $$\int_a^b (f_2(x) - f_1(x))dx < \epsilon' + 2\gamma(b-a)$$ But since $\gamma$ can be arbitrarily small: $$\int_a^b (f_2(x) - f_1(x))dx \leq \epsilon' < \epsilon$$ and the proof is complete.
Since the interval $[a,b]$ is compact, $f$'s continuity implies it's uniform continuity. Therefore, given any $\epsilon$ there exists a $\delta$ such that whenever $x',x'' \in [a,b]$ and $|x' - x''| < \delta$ then $|f(x') - f(x'')| < \epsilon/(b-a)$. Choose any partition $x_0,x_1,...x_N$ of $[a,b]$ of width less than $\delta$. For each $i = 1,...N$ choose $x_i',x_i'' \in [x_{i-1},x_i]$ such that $x_i'$ attains the minimum on the closed set and $x_i''$ attains the maximum. Define step functions $f_1,f_2$ on $[a,b]$ by $$f_1(x) = \begin{cases} f(x_i') \text{ if } x_{i-1} < x < x_i, i = 1,...,N \\ f(x) \ \text{ if } x = x_i, i = 0,1,...N \end{cases}$$ $$f_2(x) = \begin{cases} f(x_i'') \text{ if } x_{i-1} < x < x_i, i = 1,...,N \\ f(x) \ \text{ if } x = x_i, i = 0,1,...N \end{cases}$$ Then $f_1 \leq f \leq f_2$. Furthermore for every $i$ we have $|x_i' - x_i''| \leq x_i - x_{i-1} < \delta$ (because it is a partition of width less than $\delta$), so that $|f(x_i') - f(x_i'')| < \epsilon/(b-a)$ (by uniform continuity) and therefore $f_2(x) - f_1(x) < \epsilon/(b-a)$ for all $x \in [a,b]$. Therefore $$\int_a^b (f_2(x) - f_1(x))dx \leq \text{max} \{f_2(x) - f_1(x) : x \in [a,b]\}\cdot (b-a)$$ $$ < \frac{\epsilon}{b-a}\cdot (b-a) = \epsilon$$ thus the criterion of the last proposition is proved (that the function can be squeezed inbetween two step functions that differ by some $\epsilon$), and hence $f$ is integrable.

Proposition: Fundamental Theorem of Calculus

Let $a,b,c \in \mathbb{R}, a< b < c$, and let $f$be a real-valued function on $[a,c]$. Then $f$ is integrable on $[a,c]$ if and only if it is integrable on both $[a,b]$ and $[b,c]$, in which case $$\int_a^bf(x)dx + \int_b^cf(x)dx = \int_a^cf(x)dx$$
Let $f$ be a continuous real-valued function on an open interval $U$ in $\mathbb{R}$ and let $a \in U$. Let the function $F$ on $U$ be defined by $F(x) = \int_a^xf(t)dt$ for all $x \in U$. Then $F$ is differentiable and $F' = f$.
If the real-valued function $F$ on an open interval $U$ in $\mathbb{R}$ has the continuous derivative $f$ and $a,b \in U$, then $$\int_a^bf(t)dt = F(b) - F(a)$$

Proofs:

Let's first prove that integrability of the whole function implies integrability of its parts, and vice versa. So suppose that $f$ is integrable on both the segments. Then, by the previous section, we know that there exist step functions that bound each half of $f$ that have an arbitrarily small difference. We can then 'compose' these four step functions together to create two step functions that bound all of $f$ with arbitrarily small difference. Hence, by the proposition, $f$ must be integrable.

Now suppose that $f$ is integrable as a whole. Then it is bounded by step functions that have small difference. But we can split these step functions into four step functions, two bounding each half of $f$, and each pair having small difference. But by the proposition, this implies that each half of $f$ is integrable. Hence we have proved that the integrability of the whole function iff integrability of its parts.

To complete the proof, we now know that $f$ is integrable as a whole, and its parts are also both integrable. Then we know that given $\epsilon$, we can find $\delta$ such that any Riemann sum with width less than $\delta$ for $f$ corresponding to any of the three parts generates less than $\epsilon/3$ amount of error. Let $S_1,S_2$ be the Riemann sums for the left and right half respectively. Then we have: $$|S_1 - \int_a^bf(x)dx| < \frac{\epsilon}{3}, |S_2 - \int_b^cf(x)dx| < \frac{\epsilon}{3}$$ $$|S_1 + S_2 - \int_a^cf(x)dx| < \frac{\epsilon}{3}$$ Therefore $$|\int_a^bf(x) dx + \int_b^xf(x)dx - \int_a^cf(x)dx|$$ $$\leq |\int_a^bf(x)dx - S_1| + |\int_b^cf(x)dx - S_2|$$ $$+ |S_1 + S_2 - \int_a^cf(x)dx| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$$ Since $\epsilon$ arbitrarily small, we have the desired proof.
Since $f$ is continuous, $F(x) = \int_a^xf(t)dt$ is defined for all $x \in U$. We have to show that for any fixed $x_0 \in U$ $$\lim_{x \to x_0}\frac{F(x) - F(x_0)}{x - x_0} = f(x_0)$$ For any $x \in U, x \neq x_0$ we have $$|\frac{F(x) - F(x_0)}{x - x_0} - f(x_0)| = |\frac{\int_a^x f(t)dt - \int_a^{x_0}f(t)dt}{x - x_0} - f(x_0)|$$ $$= |\frac{\int_{x_0}^xf(t)dt}{x-x_0} - \frac{\int_{x_0}^xf(x_0)dt}{x-x_0}| = |\frac{\int_{x_0}^x(f(t)-f(x_0))dt}{x-x_0}|$$ where we use the last proposition. Since $f$ is continuous at $x_0$, given any $\epsilon$ we can find a $\delta$ such that $|f(x) - f(x_0)| < \epsilon$ if $x \in U$ and $|x - x_0| < \delta$. Thus: $$|\int_{x_0}^x (f(t) - f(x_0))dt| \leq \epsilon|x - x_0|$$ Substituting this back into the previous equation, we have our required result.
Since $\frac{d}{dx}(\int_a^xf(t)dt - F(x)) = f(x) - f(x) = 0, \int_a^xf(t)dt - F(x)$ is constant. Thus $\int_a^xf(t)dt = F(x) + c$, for some $c \in \mathbb{R}$. In particular $0 = \int_a^af(t)dt = F(a)+c$, so that $c = -F(a)$. Therefore $\int_a^xf(t)dt = F(x) - F(a)$. Hence $\int_a^bf(t)dt = F(b)-F(a)$

If $x \in \mathbb{R}, x > 0,$ then $\text{log}(x) = \int_1^x\frac{dt}{t}$
$\text{exp}$ is the inverse function of $\text{log}$, that is
$\text{exp }(x) = y$ means $x = \text{log }(y)$
If $x,n \in \mathbb{R}, x > 0$, then $x^n = \text{exp }(n \text{ log }(x))$
$e = \text{exp }(1)$

Proposition: The Logarithmic and Exponential Functions

The function $\text{log} : \{x \in \mathbb{R} : x > 0\} \to \mathbb{R}$ is differentiable with $d \text{ log }(x)/dx = 1/x$, it is strictly increasing, assumes all values in $\mathbb{R}$, and satisfies the rules: $$\text{log }xy = \text{log }x + \text{log }y \text{ if }x,y > 0$$ $$\text{log }\frac{x}{y} = \text{log }x - \text{log }y \text{ if } x,y > 0$$ $$\text{log }x^n = n\text{ log }x \ \text{ if } x > 0, n \text{ an integer}$$
The function $\text{exp}: \mathbb{R} \to \{x \in \mathbb{R} : x > 0\}$ is differentiable, with $d \text{ exp } (x)/dx = \text{exp }(x)$. It is strictly increasing, assumes all positive values, and satisfies the rules: $$\text{exp }(x) \cdot \text{exp }(y) = \text{exp } (x+y) \text{ if } x,y \in \mathbb{R}$$ $$\frac{\text{exp }(x)}{\text{exp }(y)} = \text{exp }(x-y) \text{ if } x,y \in \mathbb{R}$$ $$\text{exp } (nx) = (\text{exp }(x))^n \text{ if } x \in \mathbb{R}, n \text{ an integer}$$
For $x,y,n,m \in \mathbb{R}, x,y > 0$, we have: $$x^n x^m = x^{n+m}$$ $$\frac{x^n}{x^m} = x^{n-m}$$ $$(x^n)^m = x^{nm}$$ $$(xy)^n = x^ny^n$$ $$\frac{d}{dx}x^n = nx^{n-1}$$

Proofs:

The differentiability of log, along with its formula, comes straight from the Fundamental theorem of calculus. Since $1/x > 0$ if $x > 0$, the derivative of log is always positive, so it's strictly increasing. If we set $y = ax$, then by chain rule we have $$\frac{d}{dx} \text{ log }(y) = \frac{1}{y} \frac{dy}{dx} = \frac{1}{x}$$ so that $d\text{ log }ax/dx = d\text{ log }x/dx$ and hence $\text{ log }ax = \text{ log }x + c$. Setting $x = 1$ we find that $c = \text{ log }a$, hence we have $\text{ log }(ax) = \text{ log }a + \text{ log }x$. Changing notation we get $$\text{ log }xy = \text{ log }x + \text{ log }y$$ In the special case that $x = 1/y$ we get $\text{ log }1/y = -\text{ log }y$ and so $$\text{ log }\frac{x}{y} = \text{ log }x + \text{ log }\frac{1}{y} = \text{ log }x - \text{ log }y \text{ if } x,y > 0$$ Clearly, if $x,n > 0$ with $n$ an integer then $$\text{ log }x^n = n\text{ log }x$$ by simply iterating the firsy log law over and over again. For negative values, we use the fact that $\text{ log }x^{-n} = \text{ log }1/x^n = -\text{ log }x^n = (-n)\text{ log }x$. Hence it is true for all $n$. Now, since $2 > 1$, we have $\text{ log }2 > \text{ log }1 = 0$. Since $\text{ log }2^n = n\text{ log }2$, it's clear that log attains all values (we can see this by using the intermediate value theorem).
First, let us prove that $\text{ exp }$ is continuous. Pick an $x_0$. Pick a small $\epsilon$. Pick: $$\delta = \text{min}\{x_0 - \text{ log }(\text{ exp }(x_0)-\epsilon),\text{ log }(\text{ exp }(x_0) + \epsilon) - x_0\}$$ Then we have that at a point $x_0$, if $x$ is contained within a $\delta$-ball around $x_0$, then $x \in (\text{ log }(\text{ exp }(x_0)-\epsilon),\text{ log }(\text{ exp }(x_0) + \epsilon))$. Now $\text{ exp }(x) \in (\text{ exp }(x_0) - \epsilon, \text{ exp }(x_0) + \epsilon)$. This implies that exp is continuous.

To prove its differentiability, let $x_0$ be fixed, and write $\text{exp }(x_0) = y_0, \text{exp }(x) = y$. Then $\lim_{x \to x_0} y = y_0$ and $$lim_{x \to x_0} \frac{\text{exp }(x) - \text{exp }(x_0)}{x-x_0} = \lim_{y \to y_0} \frac{y - y_0}{\text{log }(y) - \text{log }(y_0)} = \frac{1}{\lim_{y \to y_0} \frac{\text{log }y - \text{log }y_0}{y - y_0}}$$ $$= \frac{1}{\frac{d\text{log }y}{dy}(y_0)} = \frac{1}{\frac{1}{y}} = y_0 = \text{exp }(x_0)$$ This ends the proof. The other properties can be easily verified by applying log to them, and seeing that they are equivalent to the log laws.
These mostly just follow from previously proven things and the definitions.