# Question Video: Evaluating a Double Integral over a Rectangular Region

Evaluate the double integral ∫_(−1) ^(1) ∫_(−1) ^(2) 1 d𝑥 d𝑦.

02:45

### Video Transcript

Evaluate the double integral shown here.

Now when we’re looking to evaluate this double integral, what it is is actually an integral inside another integral. And I can rewrite it like this. So, we’ve got the definite integral of one. And that’s with the limits two and negative one, inside the definite integral with limits one and negative one. But this time we’ll be taking a look at why.

So, the first thing we’ll do is we’re gonna integrate. And we’ve got a one here. If there was a 𝑦 term involved, then the 𝑦 term would just be a constant value. And we wouldn’t deal with it with this integral. However, as we’ve just got one, we’re gonna integrate this and see what the result is.

So, when we’ve integrated our one, we get 𝑥. But then how do we work out a definite integral? Well, to remind ourselves what to do, what we do if we want the definite integral of 𝑓 of 𝑥 between the limits 𝑏 and 𝑎 is we integrate. And then we substitute in 𝑏 instead of 𝑥. And then we subtract from this the integral with 𝑎 substituted in instead of 𝑥.

So, when we do that, inside the parentheses, we’re gonna get two minus negative one. And that’s because we’ve substituted in our bounds for 𝑥. So, now we’re left with an integral. And this is another definite integral. And this time, it’s between the limits of one and negative one. And this time, it’s the integral of three. And we’re looking at d𝑦 here. So, we’re looking in terms of 𝑦.

So, now what we’ve done is we’ve completed the second integral of our double integral. And when we integrate three, we get three 𝑦. Cause remember, this time we’re dealing with 𝑦. So, we’ve now got three 𝑦. And what we need to do is we substitute in one for 𝑦 and negative one for 𝑦. So, when we do this, we’re gonna get three multiplied by one minus three multiplied by negative one. So therefore, this is gonna give us an answer of six.

So, we can say that when we have evaluated the double integral, as shown, the value is going to be six. And we achieved that by dealing with it as an integral inside an integral. So, we first integrated and we were dealing in 𝑥. Then we next integrated dealing in 𝑦, each time using the bounds shown.

And the question here is, why would we use a double integral? Well, the double integral gives us the volume below the surface of 𝑓 of 𝑥 comma 𝑦 above a region R. And the region R will be a rectangular region that we’d look out. So, that’s what we get if you found the double integral.