Assume that you need to calculate the rate of heat flow through a window. The flow depends on many factors. The window area is one of them. How do you find this area?

For a rectangular widow, you just multiply the widths of the window by the
height of the window.

Let us write this in a slightly awkward form.

Let the x-axis be the horizontal axis and the y-axis the vertical axis.

Let the left corner of the window be at x_{1} and the right corner at x_{2}.

The width of the window is ∆x = x_{2} - x_{1}.

Let the height of the window be y = f(x) = h.

The area of the window then is A = f(x)∆x = h(x_{2} - x_{1}).

Now assume that the window is not rectangular, but has a parabolic shape.
The base of the window still extends from x_{1} to x_{2}. But
for the height y we have

y = f(x) = (4h/∆x^{2})[-x^{2} + (2x_{1} + ∆x)x - x_{1}(x_{1}
+ ∆x)].

This is the equation of a parabola which crosses the x-axis at x_{1} and
x_{1} + ∆x and has a height h.

How do we find the area of this window?

Let us find the approximate area by dividing the region between x_{1}
and x_{2} into N equal intervals ∆x_{i}, i = 1 to N.

If N is large enough and therefore ∆x_{i} is small enough, the area of
the window above the jth interval ∆x_{j} is very close to the area of
the rectangle f(x_{j})∆x_{j}.

We can then find the area of the widow by summing the areas of all the small rectangles.

A = ∑_{1}^{N} f(x_{i})∆x_{i}.

The sum is over all i, from 1 to N.

We can do this, for example, using a spreadsheet. The smaller we make ∆x_{i}, the closer our calculated area approaches
the true area of the window.
Letting ∆x_{i} --> 0 ,we convert the sum to a **definite integral**.

The notation is lim_{∆xi-->0} ∑_{x1}^{x2}f(x_{i})∆x_{i}
= ∫_{x1}^{x2}f(x)dx.
Here dx stands for the infinitesimally small interval.

A definite integral represents the area under a curve
f(x) from some starting position x_{i} to some ending position x_{f}.

Areas above the x-axis are positive, and areas below the x-axis are negative.

A spreadsheet (or other computer programs) can be used to evaluate a definite
integral **numerically** by converting it into a sum over a large number of very
small intervals. Evaluating a definite integral **analytically** (if possible) is a quicker way of
finding the area under a curve.

For many common function f(x) you can look up a formula ∫f(x)dx in a table of
integrals or online.

For example if f(x) = c*x^{n}, with c a constant and n any number not
equal to -1, then

∫f(x)dx = ∫c*x^{n}dx = F(x) = c*x^{n+1}/(n+1).

The definite integral ∫_{x1}^{x2}f(x)dx is the found by
evaluating F(x) at the integration limits.

∫_{x1}^{x2}f(x)dx = F(x)|_{x1}^{x2} = F(x_{2})
- F(x_{1}).

For example ∫_{x1}^{x2} c*x^{n}dx = c*x_{2}^{n+1}/(n+1)
- c*x_{1}^{n+1}/(n+1).

Finding the area of our window with

f(x) = (4h/∆x^{2})[-x^{2}
+ (2x_{1} + ∆x)x - x_{1}(x_{1} + ∆x)],

we integrate f(x) from x_{1} to x_{2} = x_{1} + ∆x.

∫f(x)dx = (4h/∆x^{2})[-∫x^{2}dx + (2x_{1} + ∆x)∫x^{1}dx
- x_{1}(x_{1} + ∆x) ∫x^{0}dx]

=-(4h/∆x^{2})[-x^{2}/3 + (2x_{1} + ∆x)x^{2}/2 -
x_{1}(x_{1} + ∆x) x^{1}/1] = F(x)

∫_{ x1}^{x2}f(x)dx = F(x_{2}) - F(x_{1}).

Note that you can factor out constants to simplify the notation.

Let us simplify and chose x_{1} = 0, x_{2} = 1 and h = 1. The ∆x = 1.

Then f(x) = 4[-x^{2}
+ x], ∫f(x)dx = 4[-∫x^{2}dx + ∫xdx] = 4[-x^{3}/3
+ x^{2}/2] = F(x)

F(x_{2}) - F(x_{1})
= 4[-1/3 + 1/2] = 2/3.

Compare this with the result of the numerical integration using the
spreadsheet.

This semester we will only encounter a few integrals.

The once you may encounter are

- ∫x
^{n}dx = x^{n+}1/(n+1) n ≠ -1, - ∫x
^{-1}dx = ln(x), - ∫e
^{-ax}dx = e^{-ax}/a, - ∫sin(x)dx = -cos(x),
- ∫cos(x)dx = sin(x).

Resources:

Definite Integral Calculator

Wolfram online integral calculator