1. ## Midpoint

a)With f(x)= e^-x^2 , compute approximations using midpoint, trapezoidal and simpson's rule for

∫ [0 to 2] f(x) dx

with n=2 (5 decimal places)

2. Simpson's rule won't work unless $\displaystyle n \ge 3$. Is your n correct?

3. that's what it says.

4. Well for midpoint, divide each subinterval into width $\displaystyle \frac{b-a}{n}$. And the height is of course the midpoint values. Thus we can use these rectangles to approximate the area. So...

$\displaystyle A \approx (\frac{2-0}{2}) \left(f(.5)+f(1.5) \right)$

5. Originally Posted by 413
a)With f(x)= e^-x^2 , compute approximations using ..., trapezoidal ... rule for
∫ [0 to 2] f(x) dx
with n=2 (5 decimal places)
Hi,

I presume that you know this rule already.

I presume also, that you mean the function $\displaystyle f(x)=e^{-x^2}$

So you have to calculate:

$\displaystyle \frac{f(0)+f(1)}{2}+\frac{f(1)+f(2)}{2}\approx 1.75407...$

I've attached a diagram showing the 2 trapezoids.

EB

6. I found that from midpoint rule gives 0.88420, trapezoidal rule gives 0.87704, and simpson's rule gives 0.82994. The next question 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).

How would you do this?

7. Originally Posted by 413
I found that ... trapezoidal rule gives 0.87704, ...
Hi,

you are right. Certainly you have noticed that I've forgotten to divide by 2. So your result is correct.

So sorry.

EB

8. its okay.

How do you compute the error estimates, and why would you use f ''(x) and f ''''(x)?

9. Originally Posted by 413

How do you compute the error estimates, and why would you use f ''(x) and f ''''(x)?
It is shown in numerical analysis that given a function,
$\displaystyle f$ continous on $\displaystyle [a,b]$ with a countinous $\displaystyle f''$ then, let $\displaystyle T_n$ be the result of using the trapezoidal rule $\displaystyle n$ subdivisons. Then, the error term
If $\displaystyle E_n=\left| \int_a^b f(x)dx -T_n \right|$
satisfies,
$\displaystyle E_n\leq \frac{(b-a)^3}{12n^2}\cdot \max_{a\leq x\leq b}\{f''(x)\}$ (which exists by EVT)

10. is f '' (x)= -2xe^-x^2 and f ''''(x)=-2e^-x^2 + 2xe^x^2 ?

11. Originally Posted by 413
is f '' (x)= -2xe^-x^2 and f ''''(x)=-2e^-x^2 + 2xe^x^2 ?
We have,
$\displaystyle f=e^{-x^2}$
Thus,
$\displaystyle f'=-2xe^{-x^2}$
Thus, (product rule you forgot to use it )
$\displaystyle f''=-2e^{-x^2}+4x^2e^{-x^2}$

12. does En< 0.14875 seem right?

13. Originally Posted by 413
does En< 0.14875 seem right?
For what? Trapezoids? Does not seem corret to me.

The following table are the error bounds for using the trapezoidal rule $\displaystyle n$ times:

14. not correct?..hmm...then how do you actually compute it with the formula's you gave me?

15. and another question,
How big should we take n to guarntee absolute value of ET <0.0001? and also for EM and ES.