Question

Prove that both trapezoid rule and Simpson's rule give an estimate within 0.005 of true value of integral?

Answer 1

Using Maple I've found:
trapezoid rule gives 2.432066146
Simpson's rule gives 2.430797145

Answer 2

\[\int\limits_{a}^{b}\frac{ x^2 }{ \sqrt{(1-x^2)(k^2-x^2)} }\]
a = 3
b = 5
k = 2

Answer 3

@SolomonZelman

@pooja195

Answer 4

Even if someone just has an idea that might be helpful.

Answer 5

error bound on simpsons rule

Answer 6

you want to confirm this numerically or you want to prove it using the error bound theorem

Answer 7

i think to find error bounds for trapezoidal one

Answer 8

take all right end points and all left end points

Answer 9

the true value is between those 2 answers

Answer 10

if you show that the difference between trap right end points trap left end points method is less than 0.005 then your answer is within 0.005 of the true value

Answer 11

Okay thanks I'll try and see if I can figure out how to do that.

Answer 12

I am not completely sure about this argument though

Answer 13

it could be possible that both the right end points and the left end points are both greater than the true value, or both less than the true value

Answer 14

yeah throw that method out of the window, there is a more mathematical way

Answer 15

https://www.youtube.com/watch?v=qVXIU6mKank
This tells you about simpsons error bound

Answer 16

Hah that's funny I just found that. I don't see one for the Trapezoid rule though :-/

Answer 17

isnt simpsons just a higher order trap

Answer 18

I don't really know what that means.

Answer 19

simpsons rule is just like beight a bit smarter about the trapezoidal rule

Answer 20

it will take the average of the areas of the trapezoids in a smarter way

Answer 21

Okay. So it will be found in a similar way?

Answer 22

yeah u should understand the pictures of these operations that will make it a lot clearer

Answer 23

https://www.youtube.com/watch?v=uc4xJsi99bk

Answer 24

ok i see now

Answer 25

basically for trapezoidal