Calculus II Simpson's, Trapezoidal, and Midpoint Rule

In summary, the conversation was about finding approximations for a mathematical problem using different methods - trapezoidal, midpoint, and Simpson's rule. The person asking for help used various equations and values to compute their answers, but had some discrepancies in their results. They received tips and hints from others in the conversation, ultimately leading to a successful approximation.
  • #1
ardentmed
158
0
Hey guys, I'd appreciate some tips for this question.

1391f1257f53a17199f9_3.jpg


For question 7a, I used
1/2 * 4/8 [1f(x) + 2 + 2 + 2 + 1.. etc]
and computed 1.732865 as the answer.

For part b, I used 1/4, 3/4.. (etc.).. up to 15/4 for the midpoints, added the f(x) values of those points together, and divided by four. Oddly enough, I only got 0.893713 this time.

Finally, for question 7c, I used
1/3 * 4/8 [1f(x) + 4 + 2+ 4+ 2 + 4.. etc]
and computed 1.719693 as the answer, which is close to the trapezoidal approximation, but the midpoint approximation is what's throwing me off.

Thanks a ton, I really appreciate the help guys.
 
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  • #2
For the trapezoidal one:
$a=0$
$b=4$
$n=8$
$\Delta x=\frac{4-0}{8}=\frac{1}{2}$

$T_{8}=\frac{0.5}{2}[f(0)+2f(0.5)+2f(1)+2f(1.5)+2f(2)+2f(2.5)+2f(3)+2f(3.5)+f(4)]=0.25[(\sqrt{0} \sin\left({0}\right))+2(\sqrt{0.5} \sin\left({0.5}\right))+2(\sqrt{1} \sin\left({1}\right))+2(\sqrt{1.5} \sin\left({1.5}\right))+2(\sqrt{2} \sin\left({2}\right))+2(\sqrt{2.5} \sin\left({2.5}\right))+2(\sqrt{3} \sin\left({3}\right))+2(\sqrt{3.5} \sin\left({3.5}\right))+(\sqrt{4} \sin\left({4}\right))]=1.73287$
For the Midpoint one:
$\Delta x$ is the same.
$M_{8}=0.5[f(\frac{1}{4})+f(\frac{3}{4})+f(\frac{5}{4})+f(\frac{7}{4})+f(\frac{9}{4})+f(\frac{11}{4})+f(\frac{13}{4})+f(\frac{15}{4})]=0.5[(\sqrt{1/4}*\sin\left({1/4}\right))+(\sqrt{3/4}*\sin\left({3/4}\right))+(\sqrt{5/4}*\sin\left({5/4}\right))+(\sqrt{7/4}*\sin\left({7/4}\right))+(\sqrt{9/4}*\sin\left({9/4}\right))+(\sqrt{11/4}*\sin\left({11/4}\right))+(\sqrt{13/4}*\sin\left({13/4}\right))+(\sqrt{15/4}*\sin\left({15/4}\right))]=1.78743$
For the Simpsons one:
$\Delta x$ is the same.
$S_{8}=\frac{0.5}{3}[f(0)+4f(0.5)+2f(1)+4f(1.5)+2f(2)+4f(2.5)+2f(3)+4f(3.5)+f(4)]=[(\sqrt{0} \sin\left({0}\right))+4(\sqrt{0.5} \sin\left({0.5}\right))+2(\sqrt{1} \sin\left({1}\right))+4(\sqrt{1.5} \sin\left({1.5}\right))+2(\sqrt{2} \sin\left({2}\right))+4(\sqrt{2.5} \sin\left({2.5}\right))+2(\sqrt{3} \sin\left({3}\right))+4(\sqrt{3.5} \sin\left({3.5}\right))+(\sqrt{4} \sin\left({4}\right))]=1.77214$


i hope that helps
 
Last edited:
  • #3
ardentmed said:
For question 7a, I used
1/2 * 4/8 [1f(x) + 2 + 2 + 2 + 1.. etc]
and computed 1.732865 as the answer.
I agree.

ardentmed said:
For part b, I used 1/4, 3/4.. (etc.).. up to 15/4 for the midpoints, added the f(x) values of those points together, and divided by four. Oddly enough, I only got 0.893713 this time.
Hint: your answer is two times less than the correct one.

ardentmed said:
Finally, for question 7c, I used
1/3 * 4/8 [1f(x) + 4 + 2+ 4+ 2 + 4.. etc]
and computed 1.719693 as the answer, which is close to the trapezoidal approximation
I got 1.77214.
 
  • #4
You guys are brilliant. Thanks!
 

FAQ: Calculus II Simpson's, Trapezoidal, and Midpoint Rule

What is the significance of Simpson's, Trapezoidal, and Midpoint Rule in Calculus II?

Simpson's, Trapezoidal, and Midpoint Rule are numerical integration methods used in Calculus II to approximate the value of a definite integral. They are useful when the function being integrated cannot be easily integrated by hand or when the limits of integration are not simple.

How does Simpson's Rule differ from Trapezoidal and Midpoint Rule?

Simpson's Rule is more accurate than Trapezoidal and Midpoint Rule because it uses parabolic curves to approximate the function, whereas Trapezoidal and Midpoint Rule use straight lines. This allows for a more precise approximation of the definite integral.

Can Simpson's, Trapezoidal, and Midpoint Rule be used for any type of function?

Yes, these numerical integration methods can be used for any type of function, as long as the function is continuous over the interval of integration. They can also be used for both single and multiple integrals.

How do you calculate the error in Simpson's, Trapezoidal, and Midpoint Rule?

The error in Simpson's, Trapezoidal, and Midpoint Rule can be calculated using the error formula, which takes into account the fourth derivative of the function being integrated and the chosen number of intervals. The error decreases as the number of intervals increases.

In which situations would Simpson's, Trapezoidal, and Midpoint Rule be preferred over other numerical integration methods?

Simpson's, Trapezoidal, and Midpoint Rule are preferred when the function being integrated is smooth and has a continuous second derivative. They are also useful when the limits of integration are not simple or when the function cannot be easily integrated by hand.

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