^I did a(10)-a(0)/10…would that still be right???
@sjwon3789 average velocity is just f(b)-f(a)/b-a
@beavergod Interesting approach for the distance one; I think I took the integral from 0 to 1 of the speed formula from the previous part.
On the bike one average value is 1/10integral of velocity as mvt states for integrals @sjwon3789 and for the vel its just the initial 3+integral of 1-2 of the vel function @JuicyMango
@synack that is arc length for parametric Lol same hthing
@sjwon3789 I did 1/10 of (f(10)-f(0))/10-0. Because you would have to solve for the constant if you took the integral.
F(b)-f(a) is the average rate of change of vel
@beavergod oh yeah xD
@synack Same formula essentially.
@accidentalprone I thought it was similar to “average rate of change”…
EDIT: Oh I see, okay. I messed up on that problem :/// I thought you HAD to apply that formula
@Frigidcold it was a definite integral
For the distance one I took the individual integrals and used Pythagorean theorem.
Did anyone get just two terms for the last FRQ dealing with the third degree taylor polynomial???
@beavergod I honestly forget the difference. Too much calc. I got the question right, so whatever. 
@agupte That’s what the formula is doing, in a more direct manner.
@JuicyMango 3 terms, sorry.
Haha make sure u guys put +c on the partial fractions integral or that’s like minus 2 points
@beavergod Always lol, but I doubt it’s 2, maybe 1.
@Frigidcold Yeah 3 terms, because x, x^2, and x^3. I added things wrong and I noticed it at the last second so I don’t know if I got the correct answer lol (hopefully I foiled it correctly)
Did anyone get 0.75 for the sinx/x rate of change problem? I completely guessed that…
@beavergod yeah I think I forgot that lol. Oh well…