MHB Calculus II Simpson's, Trapezoidal, and Midpoint Rule

[ardentmed]

Hey guys, I'd appreciate some tips for this question.

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.

 

[ineedhelpnow]

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

 

[Evgeny.Makarov]

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.

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.

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.

 

[ardentmed]

You guys are brilliant. Thanks!