00:01
Okay, so we've been asked to integrate from 0 to 3 a function of x with respect to x.
00:17
Okay, so it seems pretty straightforward.
00:21
We need to integrate f of x and then plug in the limited integration and subtract.
00:27
What is the function that we were trying to integrate? now, this is the hard part.
00:34
We have a numerator and a denominator.
00:42
Awesome.
00:45
So there's a bunch of integration rules and a big table in your textbook.
00:51
When i first see this, i see square root 9 plus x squared.
00:56
That's tipping me off that i should go look in my table of integration rules and see if any of them match this one.
01:02
Unfortunately, none of the rules in our table of integration rules matches this one.
01:08
One.
01:10
Oh, well, we can still integrate this.
01:14
And typically in these scenarios where it doesn't match one of the integration rules, we have to make some sort of substitution.
01:23
What substitution do we make? you're probably not going to get the right substitution the first time, unless you've practiced a bunch of these.
01:34
It's a little bit of intuition, a little bit of gut feeling, and a little bit of if you didn't get it right the first time, try try again.
01:43
In this case, the correct substitution is root 9 plus x squared.
01:56
So notice how i took that little bit of the original function we want to integrate, and i'm going to now call it u.
02:03
And whenever we make this type of substitution to integrate a function, we're going to differentiate our substitution.
02:14
Okay, so 9 plus x squared to the one -half power.
02:18
It's going to be a power rule combined with a chain rule.
02:24
Take that one half, drop it down.
02:26
The chain rule, the derivative of the inside is 2x, and now it's the original inside, so the one -half minus one, or negative one -half.
02:44
Okay, let's simplify that.
02:47
So, the derivative of you, derivative of the left side, the two is going to cancel out, and we have x on top.
02:56
And now instead of negative one half, it's going to be a radical in the denominator.
03:09
And i forgot to do it here, but when we differentiate the right side, we need to put in that differential.
03:19
Okay, look how similar that right side is to our original function.
03:26
In fact, i'm going to show how similar it is right here...