Integral

[MeENCaNTaMiDiTa]

Let F(x) = integral from 0 to x of sin(t^2)dt for 0<x<3 (equal to also)

a) use the trapezoidal rule with 4 equal subdivisions of the closed interval (0,1) to approximated F(1)
- i did this part and got F(1) = 3.16

b) Where does F have a relative maximum on 0<x<3? Justify your answer
-i know that the derivative of F is sin(x^2) but i don't know how to get the zeros from 0 to 3 and then find the maximum

c) If the average rate of change of F on the closed interval [1,3] is k, then solve for the integral for 1 to 3 sin(t^2)dt in terms of k
-on this part i am very confused and don't know where to begin!

Thanks so much for any help provided!

 

[tkhunny]

a) Since ALL the area under y = 1 on [0,1] is only 1, it seems a little unlikely that only part of the area is as much as 3. You seem to have a magnitude problem.

b) sin(x²) = 0 ==> x² = 0 ==> \(\displaystyle x\,=\,0,\,\) \(\displaystyle \pi\), \(\displaystyle 2*\pi\), \(\displaystyle 3*\pi\)...
We're interested in [0,3], so \(\displaystyle x\,=\,0,\,\sqrt{\pi},\,\sqrt{2*\pi}\)
Of course, derivatives don't know about endpoints, so you'll have to check
out x = 3 by yourself.