00:01
We are given v is the velocity of a space vehicle after it's been launched particularly from ground.
00:07
And we want to find its altitude after 10 seconds.
00:11
Let us be the altitude.
00:13
And so we have to integrate v with respect to t.
00:19
So we'll be integrating lawn t -q plus 1.
00:24
Let me put the square outside of the lawn and t -square over t -q plus 1 d -t.
00:34
There are a couple of ways to integrate this.
00:38
For this topic, we are using the substitution method and the integration power rule.
00:44
So let you be this whole part over here, which is long t -cube plus 1.
00:54
D -u -d -t will be 1 over t -cube plus 1.
01:01
And differentiating t -cube, we get 3t -square plus, plus.
01:07
Differentiating 1 we get 0 so du is 3 t square over t cubed plus 1 d t so this part here is our almost our du except for a tree so let's put the tree in there and because i put a tree here i got to put one third in front so now i'm ready to substitute s equals to 1 3rd.
01:44
Now, this part here is my u, and so here is the square.
01:50
Now, this whole part here is my du...