# Definite integral

Definite integration deals with calculating the area under the curve of a function.

Here is a function, and the area under its curve from a to b is:

This is assuming, of course, that f(x) is integrable on the closed interval [a, b], and it has a primitive function on this interval.

This primitive function is called F(x): the indefinite integral.

If such primitive function does not exist, then calculating the area under the curve turns into a nightmare.

Nightmares will be discussed in a separate slideshow.

Instead, let's try how this theorem works, and find the area under the curve of x2 from 0 to 1.

According to Newton and Leibniz, this area is:

Here comes the primitive function:

And this is where we have to substitute first 1, and then 0.

Let's see how we could make all this a bit more exciting.

For example, let’s find the area of the region that is bounded by functions f and g.

Here is the plan:

First, we calculate the area under the red function from a to b,

then we do the same for the yellow function,

and finally, we subtract them.

At the same time, it would not hurt to know what a and b are.

Well, the special thing about a and b is that the two functions are equal there.

So, we have to solve this equation.

These type of regions of which we calculated the area are called normal regions.

A normal region is bounded both above and below by a function,

and on the sides by lines x=a and x=b.

Sometimes the two functions meet on one side,

but they can meet on both sides, too.

The area of a normal region is:

or, if function g happens to be on top, like here in our drawing,

then the other way around.

The advantage of this method is that we only have to integrate once. Let’s see it.

For example, let’s find the area of the region that is bounded by functions f and g.

First we calculate the intersections,

Here is a function,

to which we draw a tangent at x=3.

This way we get two regions.

One is bounded by the function, the tangent and the y axis,

the other one is bounded by the function, the tangent and the x axis.

We need to find the area of these regions.

Well, we will likely need the equation of the tangent.

Well, aren't we lucky? Here it comes:

Now let's get down to business.

It is much easier to calculate the area that is bounded by the y axis.

Since this is a normal region, it is sufficient to integrate the difference of the two functions:

Calculating the other area will be a lot more unpleasant.

First, we will need these intersections.

And now let's see the areas.

The area in question:

And now we will do a very funny thing.

We will integrate to infinity.

Let’s calculate, for instance, this area:

We will integrate to by first integrating to ,

and then we say “Dear , please go to “.

Let’s see what this limit would be.

These may be handy for the calculation:

But it is easier to remember them like this.

Here comes another one:

Well, it could happen that both limits are infinite:

In that case, we split the integration, let’s say, at zero.

Actually, this area is 1/3+1/3, so 2/3, but definite integration works in a way that areas under the x axis will get a negative sign.

So that’s why we got zero.

These integrals that stretch into infinity are called improper integrals.

Those that we saw so far were stretching into infinity in the x axis direction, but there are others that do the same along the y axis.

Here it is:

But the solution process is the same.

If we integrate this function over the positive number line:

then we get an improper integral from 0 to 1,

and also from 1 to infinity.

Let’s see first what happens if we integrate from 0 to 1.

Well, we will discuss the case of separately.

Now let's see what this limit is.

First we substitute 1,

and then see what happens if .

That means if the exponent is a positive number,

then we will get zero here.

If the exponent is negative...

then the limit is infinite.

If happens to be 1:

Now let’s see what is going on from 1 to infinity.

If , then the exponent is positive,

and in that case the integral is divergent.

And if happens to be 1:

Summarizing all this: if , then from 0 to 1 the integral is divergent, and from 1 to infinity it is convergent.

If , then from 0 to 1 the integral is convergent, and from 1 to infinity it is divergent.

And if , well, then it is divergent everywhere.

And now we will do a very funny thing.

We will integrate to infinity.

Let’s calculate, for instance, this area:

We will integrate to by first integrating to ,

and then we say “Dear , please go to “.

Let’s see what this limit would be.

These may be handy for the calculation:

But it is easier to remember them like this.

Here comes another one:

Well, it could happen that both limits are infinite:

In that case, we split the integration, let’s say, at zero.

Actually, this area is 1/3+1/3, so 2/3, but definite integration works in a way that areas under the x axis will get a negative sign.

So that’s why we got zero.

These integrals that stretch into infinity are called improper integrals.

Those that we saw so far were stretching into infinity in the x axis direction, but there are others that do the same along the y axis.

Here it is:

But the solution process is the same.

If we integrate this function over the positive number line:

then we get an improper integral from 0 to 1,

and also from 1 to infinity.

Let’s see first what happens if we integrate from 0 to 1.

Well, we will discuss the case of separately.

Now let's see what this limit is.

First we substitute 1,

and then see what happens if .

That means if the exponent is a positive number,

then we will get zero here.

If the exponent is negative...

then the limit is infinite.

If happens to be 1:

Now let’s see what is going on from 1 to infinity.

If , then the exponent is positive,

and in that case the integral is divergent.

And if happens to be 1:

Summarizing all this: if , then from 0 to 1 the integral is divergent, and from 1 to infinity it is convergent.

If , then from 0 to 1 the integral is convergent, and from 1 to infinity it is divergent.

And if , well, then it is divergent everywhere.