00:01
Right? so the first thing i need to do here when i'm looking for the area is i need to figure out the bounds for my integral.
00:09
And the bounds will be determined by the fact that the area is enclosed by my curve and my x -axis.
00:17
That is when y is equal to zero.
00:23
So when y is equal to zero, then t minus t squared is equal to zero.
00:32
So if i factored that, that be t times 1 minus t.
00:41
So t would be equal to 0 or 1.
00:51
These would be my bounds.
00:59
So since those are my bounds, i will have the integral from 0 to 1.
01:06
And using the substitution rule, i will have y, which is t squared, or t, minus t squared times the derivative of x we recall x is our one plus e to the t so the derivative of that will just be e to the t so i'm taking the integral of all of that so first thing i'll do is i'll distribute my e to the t so that will be from zero to one of t of t e to to the t minus t squared.
02:25
Actually, i don't think i will distribute this.
02:35
Okay, i think i'm going to use integration by parts.
02:47
So recall integration by parts was if you had the integral of u dv, then this would be u times v minus integral of v d u, right? so this is going to be my u and this will be my dv okay so if i'm using integration by parts well i will have u times v so t minus t squared times v so the antiderivative of e to the t is simply e to the t and i will have to evaluate all of that using the fundamental theorem of calculus and i'll do minus the integral from zero to one v's et times du.
04:04
So i just take the derivative of my u, which will be 1 minus 2t, dt.
04:20
So now i'm actually going to have to use integration by parts again.
04:25
This will be, et will still be my dv, and this will be my u.
04:38
So this is t minus t squared times et 01.
04:50
Minus be careful here because i'm going to be subtracting a sum so first i have uv or let me do a different color here so first i have uv so my u is now 1 minus 2 t my v is the antiderivative of e to the t which will still be e to t and i'll have to apply fundamental theorem minus the integral from 0 to 1 well v is still e to the t, and i take the derivative of u, which will be negative 2.
05:37
Now, this integral is something i can easily find the anti -derivative of, so i do not have to do integration by parts again.
05:50
Okay, so let's evaluate some things...