00:01
All right, hello, in this question we're going to evaluate this integral x times the square root of 2x squared plus 7 dx.
00:06
And in order to do this, we're first going to have to ask ourselves, what approach do we think we need to take to do this? and because we have this function here inside of our larger function, the first thought that we should have is can we do a u substitution? and that is, can we say that this thing inside the square root, this 2x squared plus 7, can we set that equal to u? and if we do that, u substitution idea, are we going to run into a problem? and so to do u sub, we define our u, which is the thing inside of our inner function.
00:39
And then we go ahead and differentiate.
00:41
So if we take the derivative of this, we're going to get 4x times dx.
00:45
Another way to think of this is du dx is going to be 4x.
00:50
And so if we want just du, we just multiply both sides by dx.
00:54
That's not technically what's happening.
00:56
But for the sake of use of concepts, that's the way that we can think about it.
01:01
So we have our du there.
01:04
That means this integral, well, we know that this term here is going to be equal to u.
01:09
So we're going to have the square root of u inside there.
01:12
And then we have an x dx here.
01:15
But we need a 4 in order for that to be equal to du.
01:19
So we have a x dx.
01:23
So if we rearrange this equation here, that is equal to du over 4.
01:30
And so since we have an x dx, this is going to be du divided by 4.
01:36
The important part here is now that we've done our u sub, we can take this integral and it's all in terms of one variable...