We might think of Calculus as that part of mathematics that is dependent on the notion of limit.

In Calculus I, the main objects of study are derivatives and integrals - though, you probably only just started integration.

I've got a little table that outlines the parallels between the development of differential calculus and that of integral calculus:

- Techiques of integration
- Applications
- Sequences and series

- The geometric interpretation
- The definition
- The basics of anti-differentiation
- The fundamental theorem of calculus

You don't need to be an expert on those topics! They should sound familiar, though.

You should know that the definite integral can be interpreted as signed area. Roughly, $$\begin{align} \int_a^b f(x) \, dx & = \text{the area under the graph and over the } x \text{ axis} \\ & - \text{the area over the graph and under the } x \text{ axis.}\end{align}$$

The complete graph of a function $f$ is shown below; it consists of a quarter circle and a straight line segment. Compute $$\int_{-2}^3 f(x) \, dx.$$

The green portion above the $x$-axis consists of a quarter of a circle of radius 2 (so that it's area is $\frac{1}{4}\pi \times 2^2 = \pi$) and a triangle whose area is $2$. The red portion below the $x$-axis consists of a single triangle whose area is $1/2$. Thus, $$\int_{-2}^3 f(x) \, dx = \pi + 2 - \frac{1}{2}.$$

$$ \begin{aligned} \int_a^b f(x) \, dx &= \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \, \Delta x \: \text{ (or)}\\ & = \lim_{n\to\infty} \sum_{i=1}^n f(x_{i-1}) \, \Delta x, \end{aligned} $$ where $\Delta x = \frac{b-a}{n}$ and $x_i = a+i\Delta x$.

Use the second version (with the $x_{i-1}$) when you want a left sum and use the first version (with the $x_i$) when you want a right sum.

These sums literally mean - please add up the areas of the rectangles. $$ \text{Left: }\sum_{i=1}^n f(x_{i-1}) \, \Delta x \: \: \: \: \: \: \: \: \text{ Right: } \sum_{i=1}^n f(x_{i}) \, \Delta x $$

Estimate $$ \int_0^2 (2 + 4 x - x^2) \, dx$$ using a left Riemann sum with $n=4$ terms.

We break the interval $[0,2]$ into $n=4$ pieces so $$\Delta x = \frac{b-a}{n} = \frac{2-0}{4} = \frac{1}{2}.$$ Thus, $x_i = a + i\,\Delta x = 0 + i\times\frac{1}{2} = i/2$ and $$ \int_0^2 (2 + 4 x - x^2) \, dx \approx \sum_{i=1}^4 \left(2 + 4 \left(\frac{i-1}{2}\right) - \left(\frac{i-1}{2}\right)^2\right)\,\frac{1}{2} = 8.25.$$ You can compute this type sum using Sage very easily.

If you can visualize the sum, it's pretty easy to see that the points you need to plug in are $0$, $0.5$, $1$, and $1.5$.

Thus, $$ \begin{align}\int_0^2 (2 + 4 x - x^2) \, dx & \approx (2+0+0)\times0.5 \\ &+ (2 + 4\times0.5 - 0.5^2)\times0.5 \\ &+ (2 + 4\times1 - 1^2)\times0.5 \\ &+ (2 + 4\times1.5 - 1.5^2)\times0.5 \\ &= 8.25.\end{align}$$

The fundamental theorem of calculus states that $$\int_a^b f(x) \, dx = F(b) - F(a),$$ provided that $F'=f$. Thus, we can evaluate definite integrals by finding anti-derivatives. Because of this close, inverse connection between integration and differentiation, we define the so-called indefinite integral as a convenient notation to represent and anti-deriviate, i.e. $$ \int f(x) \, dx = F(x) + c, \text{ where } F' = f. $$

$$\int (x^2 + x + 1) \, dx = ?$$

Using the power and sum rules for integrals, we find that $$\int (x^2 + x + 1) \, dx = \frac{1}{3}x^3 + \frac{1}{2} x^2 + x + c.$$

$$\int_0^2 (2x - x^2) \, dx = ?$$

$$\int_0^2 (2x - x^2) \, dx = \left.\left(x^2 - \frac{1}{3}x^3\right)\right|_0^2 = 2^2 - \frac{1}{3}2^3 = \frac{4}{3}.$$

Don't let the simplicity of the previous examples lull you into complacency. Integration is quantifiably more difficult than differentiation, as this demonstration concretely illustrates.

Over the next few weeks, we'll carefully discuss

- $u$-substitution,
- Numerical integration,
- Integration by parts,
- Trigonometric integrals, and
- Trig substitution.