# 7e. Difference Between Indefinite and Definite Integrals

## Integration Mini Video Lecture

This video explains the differences and similarities between indefinitie and definite integration.

## Video transcript

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What's the difference between indefinite and definite integrals?

### Indefinite integral

With an indefinite integral there are no upper and lower limits on the integral here, and what we'll get is an answer that still has x's in it and will also have a K, plus K, in it.

A definite integral has upper and lower limits on the integrals, and it's called definite because, at the end of the problem, we have a number - it is a definite answer.

OK. Let's do both of them and see the difference.

The integral x to the five dx is equal to ... (int x^5 dx =?)

... Now, for integration, I have to add one to the index.

So, it's going to be x to the 6

...and then I divide by the new number ... x to the sixth over 6 (int x^5 dx = x^6/6...)

...and then I must remember the plus K. (int x^5 dx = x^6/6+K)

K is just some constant. We might have some information elsewhere in the problem that will help us to find the constant. In this case we don't have any extra information, so it's just plus K.

### Definite integral

Now, let's see what it looks like as a definite integral, this time with upper and lower limits, and we'll see what happens. (int_1^2 x^5 dx = ?)

Step 1 is to do what we just did. We integrate, and I'm going to have once again x to the six over 6, but this time I do not have plus K - I don't need it, so I don't have it.

And I write it like this,

... and then I have a lower limit of 1

... and an upper limit of 2. (int_1^2 x^5 dx = [x^6/6]_1^2)

These are the numbers that come from here in the question - 1 comes from here, 2 comes from here.

So, what I've done is I've integrated, and ... written in the upper and lower limits correctly.

The next step is to actually substitute these upper and lower limits.

So, what I can do is like this ... and like this ... and I'm going to substitute 2 into this position, and 1 into this position.

So, what have I done? (int_1^2 x^5 dx = [x^6/6]_1^2=2^6/6-1^6/6)

I've found the integral, and I've substituted 2 into this position ... and 1 into this position ... and I'm going to subtract the difference ... and I'm just going to use Scientific Notebook now to give me that, as a decimal.

And it turns out to be 10.5.

int_1^2 x^5 dx = [x^6/6]_1^2 =2^6/6-1^6/6 =10.5

And this is a definite answer.

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