Question

MVT with integrals!

Answer 1

\[\int\limits_{0}^{x^2} sint dt\]
At how many points in the closed interval [0, sqrt pi] does the instantaneous rate of change of f equal the average rate of change of f on that interval?

Answer 2

oh and f(x)= that integral

Answer 3

I'd find average rate of change for that interval first.

Answer 4

so f(x) = ∫ sint dt from x^2 to zero
Using FTC
f(x) = -cos(x^2) - -cos(0)
slope formula:
(-cos(0^2) - -cos(0)) - (-cos((sqrt(pi)^2) - -cos(0))/ (0-sqrt(pi)) = average rate of change

Answer 5

Oh yeah i got that to be \[\frac{ 2 }{ \sqrt{\pi} }\]

Answer 6

Then to find the instantaneous rate of change we would find f'(x).
From the previous post: f(x) = -cos(x^2) - -cos(0)
f'(x) = sin(x^2)
where IRC = ARC
sin(x^2) = 2/sqrt(pi)

Answer 7

looks like a quadratic to find the values of x

Answer 8

so then how would you do that??

Answer 9

I mean you could set it equal to zero and graph to find zeros

Answer 10

fastest way for sure

Answer 11

Yeah there's no unit circle values where sqrt(theta) = 2/sqrt(pi) im pretty sure. Just graph it.

Answer 12

set sint-(2/sqrt pi) = 0 ?

Answer 13

yeah

Answer 14

Okay so the only way to finish it would be to graph it then?

Answer 15

no but thats the only way i'd be willing to do it.

Answer 16

there's a more complicated way that isnt worth your time.

Answer 17

lol okay thx :P What did you mean by the quadratic tho? Just for future reference on no calc tests

Answer 18

Quadratics are the ax^2 + bx +c

Answer 19

You could rewrite Sin(x^2) = 2/sqrt(pi)
x^2 = arctan(2/sqrt(pi))
x^2 - arctan(2/sqrt(pi)) = 0
then use the quadratic formula.

Answer 20

ahh okay. Thank you for your help!

Answer 21

Yeah no problem