00:01
Okay, let's evaluate this integral by making an appropriate substitution and then integrating by parts.
00:09
So let's do that.
00:11
First, i'm going to rewrite this expression that is 15x cube as like that.
00:17
That is, i'm going to rewrite this as 15x squared times x because x squared times x is x cube.
00:26
That is basically i'm relating x cube as x squared times x squared.
00:30
And then writing this as 15x squared times x and we then have e raised to the power of 15 x squared dx.
00:44
So let's make the substitution now.
00:46
So we do this substitution that is let 15x squared equals p.
00:59
Now let me take differentials on both sides.
01:02
So i have 15 times of derivative of x squared is 2x.
01:07
And when we take the differential, we put this dx.
01:10
And this equals derivative of t is one.
01:14
And then we put the differential dt.
01:16
So we can solve for x dx from this one.
01:20
15 times 2 is 30x.
01:24
Dx and this equals 1 times d t is d t.
01:28
I'm going to solve for x dx from this one by dividing both size by 30.
01:33
So i get x dx equals d t over 30.
01:39
Now let's use this substitution that is 15 x squared equals t.
01:45
So therefore when we apply this substitution into this integral, we replace all instances of 15x squared by t.
01:53
So therefore this integral becomes this 15x squared will be replaced as t and then we have this x times of e raised to the power of 15x squared, which is replaced as t, we then have x dx.
02:10
Now this can be written as t times of e raised to the power of t.
02:15
We then have this x dx.
02:19
Now notice that this x dx can be replaced as d t over 30.
02:24
So let's do that.
02:26
So therefore this becomes t times of e raised to the power of t multiplied with d t over 30.
02:34
And we can now see that we have completely replaced all x in the original integral in terms of the t variable.
02:44
So this is the integral.
02:45
In the next step, i'm going to factor this 1 over 30 out of the integral since it is a constant.
02:52
So it becomes 1 over 30 t times e raised to the power of t d t.
03:01
So now we see that we have to integrate this expression, that is t times e raise to the power of t.
03:09
So this looks like a product of two different functions.
03:13
That is t is algebra and e raised to the power of t.
03:19
This is exponential function.
03:21
So when we have a product of two different types of function, we use integration by parts method.
03:28
Integration by parts and in this method we have a formula that is an integral of u dv and this equals uv minus integral of v d u so basically we have to determine the u as well as the dv from this integral and there is a rule to choose you rule to choose u which is called as i late rule that is i l a t e and each letter in this acronym stands for type of function i replace represents inverse function l represents logarithmic function e represents algebra function t represents trigonometry function and e represents exponential function and we should look this rule in this order that is from left to right and when you see this rule from left to right inverse function comes first so we have to look for any inverse function in this integral do we have any enforce function if then we have to choose that as you since we don't have any inverse function basically we have only two types of function that is algebraic and exponential.
04:58
So we just ignore this, then move on to the next letter that is l.
05:03
And then we have to look for any logarithmic function in this integral.
05:08
We don't have logarithmic function.
05:09
So we then move on to the next letter as like this.
05:13
So the next letter is a, which stands for algebra...