The improper integral | mathXplain
Összes egyetemi tantárgy
Legnépszerűbb tantárgyak:

Contents of this Calculus 1 episode:

Definite integration, Improper integrals, Area under the curve of a function, Newton-Leibniz formula, Fundamental Theorem of Calculus, Area of region that bounded two functions.

Text of slideshow

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.

# The improper integral

03
Let's see this
Calculus 1 episode
Enter the world of simple math.
• Much better than any of my university lectures.

Daniel, 20
• It makes sense, it's fun, it's worth all the money.

Thomas, 23
• It was recommended by senior students with the title 'mandatory'.

Richard, 19
• It is the best clear, interpretable, usable learning opportunity for the lowest price.

Ellen, 23