https://www.youtube.com/watch?v=T5Y2dx0JmH8
this isnt mine just uploaded it to listen to it. there is also instrumental included.
go watch the original video.

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Algebra but It's the Final Part (FNF Dave and Bambi: Golden Apple Edition)

Evaluate the line integral where C is the given curve. We're integrating over the curve C, y to the third ds, and C is the curve with parametric equations x = t cubed, y = t. We're going from t = 0 to t = 2. So we're going to integrate over that curve C of y to the third ds. We're going to convert everything into our parameter t in terms of our parameter t. So I'm going to be integrating from t = 0 to t = 2. Those will be my limits of integration. Now y is equal to t, so I'm going to replace y with what it's equal to in terms of t. So I'm going to be integrating the function t to the third. Now ds we're going to write as a square root of dx dt squared + dy dt squared, squared of all that as we said dt. So we're integrating now everything with respect to t. So this is going to be equal to the integral from 0 to 2 of t to the third times the square root of -- see the derivative of x with respect to t is 3 t squared. So we have 3 t squared squared + dy dt; well, that's just 1 squared dt. So we have the integral from 0 to 2 of t to the third times the square root of 9 t to the fourth + 1 dt. So this is a pretty straightforward integration here. We're going to let u be equal to 9 t to the fourth + 1 then du is equal to 36 t to the third dt and so that tells me I can replace a t to the third dt with a du over 36. And so we're going to have the integral then from -- well, new limits of integration. I'm just going to put some squiggly marks there to remind myself that we switched variables. So I'm not going from t = 0 to t = 2. I'm doing things in terms of you right now. But I have a 1 over 36. I'll put that out front, and we're going to have the square root of u. So u to the 1/2, t to the third dt was replaced by du over 36. We got the 36 out front. And so now this is a pretty easy antiderivative in terms of u. It's u to the 3/2 times 2/3. And again, different limits of integration. We could figure out what they are in terms of u, but I'm going to convert back into t. So we're going to have 1 over 36 times 2/3 times u to the 3/2. Now, u is 9 t to the fourth + 1, that to the 3/2 power. And now we can go ahead and go from original limits of integration 0 to 2. So let's see, when I put a 2 in here, we're going to have -- 1 over 36 times 2/3. That's going to be 1 over 54, isn't it? So we'll have 1 over 54 times -- putting a 2 in, we have 9 times 2 to the fourth. That's 9 times 16, which is 144 + 1, is 145. So we put the 2 in there, we get 145 to the 3/2 minus, putting the 0 in, we get 9 x 0 to the fourth. That's 0. 0 + 1 is 1. So we just get 1 to the 3/2 or 1. So let's see, what's the best way to write this. How about 1 over 54 -- I guess we could leave it like that. We could also write 145 to the 3/2 as 145 times the square root of 145 and then minus 1. And that is that line integral of y to the third ds over the given curve C.
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SoundCloud falsely terminated Moldy's account, i will delete this if in the future he gets it back or if Moldy asks me to delete it. Anyways, Algebra Trouble from the game Dave's Fun Algebra Class Remastered fellas.
The background image is fanmade, i just took a screenshot in the game and edited a few things here and there.

(Game): https://gamejolt.com/games/DFAC_remastered/772531

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