00:01
In this problem, we have to compute the definite integral from 0 to 8, 2x, divided by square root of x squared plus 36 dx.
00:14
Now, the first thing we notice here is if i take the derivative of x square plus 36, we end up getting 2x.
00:24
And i see that there is the expression 2x in our problem, right? so let's substitute.
00:34
So let's take y is equal to x squared plus 36.
00:38
This implies, dy is equal to 2x followed by dx.
00:44
Now, let's find our new lower limit and new upper limit.
00:49
So let's make a chart.
00:51
So this is my y and i have my x as my old variable.
00:56
So if i take x as equal to 0, then i see that y is equal to x square plus 36 which is basically 0 plus 36 so 36 and then if i take x is equal to 8 then 8 times 8 is 64 so 64 plus 36 is equal to 100 so that's going to be my new lower limit and new upper limit so let's write down the problem then so this becomes integration.
01:32
Now the lower limit is 36 and the upper limit is 100.
01:36
And then 2x, tx becomes d .y divided by square root of y.
01:44
Now, this can be integrated as follows.
01:48
So this is nothing but y to the power negative half.
01:52
So this becomes y to the point negative half plus 1 by the standard power rule divided by negative half plus 1...