00:02
Of the function from bound 2x to bound 3x of u squared minus 1 over u squared plus 1 so the first step that you should automatically know is you should know that you're essentially going to plug in the upper bound and then subtract the lower bound and then you're going to multiply each individual term by the derivative of the specific bound so this is a really simple question that's when you're finding the derivative of an integral fundamental theorem of calculus step 1 f to c 1 okay, so here's what we're going to do.
00:37
So first we know that the derivative of 3x is 3.
00:40
So we're going to plug that in.
00:41
So it's 3 times.
00:43
When we plug in 3x to this function, it becomes 9x squared minus 1 over 9x squared plus 1.
00:56
And then we're going to subtract the derivative of 2x, which is 2 times what happens when you plug in 2x into this function.
01:06
It becomes 4x squared minus 1 over 4x squared plus 1.
01:18
Okay, so now all we have to do is simplify this question.
01:21
So what is 3 times 9x squared minus 1? that's going to be 27x squared minus 3 over 9x squared plus 1 minus, then you multiply everything by 2.
01:45
So it's going to be 8x squared minus 1, i'm sorry, minus 2 over, excuse me, i miswrote that, 8x squared minus 2 over 4x squared plus 1.
02:16
So let's go back through this question just to make sure that everybody understands what's going on.
02:22
So the first thing we did is we know that we're using the fundamental theorem of calculus 1...