Approximating f(x) with Midpoint, Trapezoidal & Simpson's Rule

[413]

a) With f(x)= e^-x^2 , compute approximations using midpoint, trapezoidal and simpson's rule with n=2.
I found that from midpoint rule gives 0.88420, trapezoidal rule gives 0.87704, and simpson's rule gives 0.82994.

Then it asks me to
Compute the error esimates for midpoint, trapezoidal, simpson's. Carefully examine the extreme values of f ''(x). You may use that |f ''''(x)| < 12 for 0<x< 2. (the < signs all mean less than or equal to).
I found that EM=0.14875, ET=0.07438, ES=0.13333

do my answers seem correct?

My question is how big should we take n to guarntee absolute value of ET <0.0001? and for EM<0.0001 and ES<0.0001?
How would you figure this question out?

 

[jkh4]

for how big should take n to guarntee absolute value

for example with ET <0.0001 plug in all the value for K, a, b

so in your case, K is 2, a is 0, b is 2

so it would be (2*2^3)/(12n^2) < 0.0001 and just do it as algebra question

(2*2^3)/(12*0.0001) < n^2 and find the square root of n

just curious, are you also taking MATH 152 in SFU? coz i did this question too =)