Approximations rules of a function

[Ocasta]

Homework Statement

Given the following integral and value of n, approximate the integral using the methods indicated (round your answers to six decimal places):

<br /> \int_0^1 e^{-3x^2} dx<br />

<br /> n = 4<br />


(a) Trapezoidal Rule

(b) Midpoint Rule
This is the only one I'm having trouble with.

(c) Simpson's Rule

Homework Equations

<br /> \Delta x = \frac{b-a}{n}<br />

Trapezoid Rule
<br /> \Delta X \cdot \frac{1}{2} \cdot [ f(0) + 2f(\frac{1}{4}) + 2f(\frac{1}{2}) + 2f(\frac{3}{4}) + f(1) ]<br />

Midpoint Rule <- incorrect
<br /> \Delta X \cdot [ f(\frac{0 + \frac{1}{4}}{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{3} + \frac{3}{4}}{2}) +f(\frac{\frac{3}{4} + 1}{2})]<br />

Simpson's Rule
<br /> \Delta X \cdot \frac{1}{3} \cdot [ f(0) + 4f(\frac{1}{4}) + 2f(\frac{1}{2}) + 4f(\frac{3}{4}) + f(1) ]<br />

The Attempt at a Solution

I set variable for my f-values.
<br /> a = f(0) = 1<br />

<br /> b = f(\frac{1}{4}) \approx 0.82903<br />

<br /> c = f(\frac{1}{2}) \approx 0.47237<br />

<br /> d = f(\frac{3}{4}) \approx 0.18498<br />

<br /> e = f(1) \approx 0.04979<br />

So I plugged everything in and I get

Midpoint Rule
<br /> \Delta X \cdot [ f(\frac{0 + \frac{1}{4}}{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{3} + \frac{3}{4}}{2}) +f(\frac{\frac{3}{4} + 1}{2})]<br />

(a) Trapezoidal Rule
0.5028176513

(b) Midpoint Rule
0.5028176513

(c) Simpson's Rule
0.5042135205

However, my midpoint rule answer is wrong. I do think it's strange that I'm getting exactly the same answer for B as I am for A, but I can't seem to see what I'm doing wrong.

 

[Mark44]

Homework Statement

Given the following integral and value of n, approximate the integral using the methods indicated (round your answers to six decimal places):

<br /> \int_0^1 x^{-3x^2} dx<br />

<br /> n = 4<br />


(a) Trapezoidal Rule

(b) Midpoint Rule
This is the only one I'm having trouble with.

(c) Simpson's Rule

Homework Equations

<br /> \Delta x = \frac{b-a}{n}<br />

Trapezoid Rule
<br /> \Delta X \cdot \frac{1}{2} \cdot [ f(0) + f(\frac{1}{4}) + f(\frac{1}{2}) + f(\frac{3}{4}) + f(1) ]<br />

Midpoint Rule
<br /> \Delta X \cdot [ \frac{f(0) + f(\frac{1}{4})}{2} + \frac{ f(\frac{1}{4}) + f(\frac{1}{2}) }{2} + \frac{f(\frac{1}{3}) + f(\frac{3}{4})}{2} +\frac{f(\frac{3}{4}) + f(1)}{2}]<br />

Simpson's Rule
<br /> \Delta X \cdot \frac{1}{3} \cdot [ f(0) + 4f(\frac{1}{4}) + 2f(\frac{1}{2}) + 4f(\frac{3}{4}) + f(1) ]<br />

The Attempt at a Solution

I set variable for my f-values.
<br /> a = f(0) = 1<br />

<br /> b = f(\frac{1}{4}) \approx 0.82903<br />

<br /> c = f(\frac{1}{2}) \approx 0.47237<br />

<br /> d = f(\frac{3}{4}) \approx 0.18498<br />

<br /> e = f(1) \approx 0.04979<br />

So I plugged everything in and I get

Midpoint Rule
<br /> \Delta X \cdot [ \frac{f(0) + f(\frac{1}{4})}{2} + \frac{ f(\frac{1}{4}) + f(\frac{1}{2}) }{2} + \frac{f(\frac{1}{3}) + f(\frac{3}{4})}{2} +\frac{f(\frac{3}{4}) + f(1)}{2}]<br />

(a) Trapezoidal Rule
0.5028176513

(b) Midpoint Rule
0.5028176513

(c) Simpson's Rule
0.5042135205

However, my midpoint rule answer is wrong. I do think it's strange that I'm getting exactly the same answer for B as I am for A, but I can't seem to see what I'm doing wrong.

I think you should be getting different values for the trapezoid and midpoint approximations. Your formula for the midpoint rule is wrong, I'm pretty sure. Instead of averaging the function values at the left and right ends of the subinterval (which is what the trapezoid rule does), it should evaluate the function at the midpoint of the subinterval. For your work, you want to evaluate [f(1/8) + f(3/8) + f(5/8) + f(7/8)]\cdot \Delta x.

 

[Mark44]

Also, your formula for the Trapezoid Rule is wrong. This is what it should be:

\Delta X \cdot \frac{1}{2} \cdot [ f(0) + 2f(\frac{1}{4}) + 2f(\frac{1}{2}) + 2f(\frac{3}{4}) + f(1) ]

 

[Ocasta]

Thanks, Mark44. I must have transcribed my trapezoid rule incorrectly.

Thanks for the clarification on my midpoint rule. I'm going to give that a try!

 

[Ocasta]

I think you should be getting different values for the trapezoid and midpoint approximations. Your formula for the midpoint rule is wrong, I'm pretty sure. Instead of averaging the function values at the left and right ends of the subinterval (which is what the trapezoid rule does), it should evaluate the function at the midpoint of the subinterval. For your work, you want to evaluate [f(1/8) + f(3/8) + f(5/8) + f(7/8)]\cdot \Delta x.

I think you were absolutely right, Mark but I plugged this in but I'm getting something just above 0.3, which is way too low.

 

[Mark44]

I think you made an error. I'm getting something right around .5.

 

[Ocasta]

It took some mathematical gymnastics, but I eventually got the right number. I'm running into a strange problem with these rules. I don't seem to need to multiply by dx for some reason I can't fathom.

 

[Mark44]

Then I think you're making another mistake. You have to multiply by \Delta x, the width of each subinterval.