Riemann sum for definite integral -- solved

W

WillGabriel

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Jan 12, 2006
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Hey, I've been stuck on one particular problem all day. I'd hugely appreciate it if anyone were able to help:

problem.jpg


I don't even know if I'm going about the problem right... Thanks in advance!
 
How are you "going about the problem"? Please show your steps, and we'll be glad to check your work.

Thank you.

Eliz.
 
Alright, sure thing. Thanks for the reply, by the way.

work.jpg


Sorry about the low quality/messy writing--I'm lacking a scanner and I've never been known for my good penmanship. Also, programs like LaTeX have just appeared to be more trouble than they're worth... I hope this will do; if not, let me know and I'll try to make a copy that's a bit more clear.

I figured that I could be at least heading in an acceptable direction, but I found out that I have no idea how to solve for what I've got. :(
 
I'm short on time at the moment, so I'll just LaTeX the graphic for now. (I couldn't read the image -- I have very poor eyesight -- so I downloaded and enlarged it.) Please post corrections, as necessary.


. . . . .∫ba x2 dx = limitn→∞ n∑i=1 F(xi) Δx\displaystyle \large{\int_b^a{\,x^2\,}dx\,= \,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,F(x_i)\,\Delta x}



. . . . .= limitn→∞ n∑i=1 [a + bi − ain]2  b − an\displaystyle \large{=\,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,{\left[a\,+\,\frac{b_i\,-\,a_i}{n}\right]}^2\;\frac{b\,-\,a}{n}}



. . . . .= limitn→∞ b − an n∑i=1(a2 + 2a(bi − ain) + (bi − ain2))\displaystyle \large{=\,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\frac{b\,-\,a}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\left(a^2\,+\,2a{\left(\frac{b_i\,-\,a_i}{n}\right)}\,+\,{\left({\frac{b_i\,-\,a_i}{n}}^2\right)}\right)}



. . . . .= limitn→∞ b − an [a2n + 2an n∑i=1 bi − 2an n∑i=1 ai + 1n2 n∑i=1 bi2 − 2n n∑i=1 aibi + 1n2 n∑i=1ai2]\displaystyle \large{=\,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\frac{b\,-\,a}{n}\,\left[{a^2 n\,+\,\frac{2a}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,b_i\,-\,\frac{2a}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,a_i\,+\,\frac{1}{n^2}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,{b_i}^2\,-\,\frac{2}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,a_i b_i\,+\,\frac{1}{n^2}\,\begin{array}{c}n\\\sum\\i=1\end{array}{a_i}^2}\right]}


Eliz.
 
Yes, that's correct--thanks a ton for that. If anyone else could offer some help with the problem, I'd be very thankful.
 
Actually there should be no subscripts in the b−a.
Let \(\displaystyle \L
\Delta x = \frac{{b - a}}{n}\).

The a left hand SUM is \(\displaystyle \L
\sum\limits_{k = 0}^{n - 1} {\left( {a + k\Delta x} \right)^2 \Delta x}\).

You need to find \(\displaystyle \L
\{{\rm{lim}}}\limits_{n \to \infty }\)
 
That's my problem, though. The subscripts referred to multiplying the delta x with the index--that's just how I've been taught to do it. The issue I've got is what to do now that I'm there--I can't choose a value for k right off, and I can't isolate it within the sum because it's part of the function (x squared).
 
Hm... Sorry for the double post/bump, but I am getting somewhat desperate. It is my fault that I'm in this situation (If I had started before yesterday, there were help sessions I could have gone to earlier in the week) but I'm stuck now so I might as well do what I can. For now, I'm heading to bed--the assignment's due today so if anyone offers help before then, I'll get it in the morning. Hopefully I can get this finished for today's class, but it's still not the end of the world.

Even if I miss a post, though, your help won't go unappreciated. Outside of the grounds of my assignment, I've still got to learn how to do these proper, don't I? ;)
 
\(\displaystyle \begin{array}{c}
\sum\limits_{k = 0}^{n - 1} {(a + k\Delta x)^2 \Delta x} & = & \sum\limits_{k = 0}^{n - 1} {(a^2 \Delta x + 2ak\left[ {\Delta x} \right]^2 + k^2 \left[ {\Delta x} \right]^3 )} \\
& = & a^2 (n)\Delta x + (2a)\left( {\frac{{n(n - 1)}}{2}} \right)\left[ {\Delta x} \right]^2 + \frac{{(n - 1)(n)(2n - 1)}}{6}\left[ {\Delta x} \right]^3 \\
& = & a^2 (n)\left[ {\frac{{b - a}}{n}} \right] + (2a)\left( {\frac{{n(n - 1)}}{2}} \right)\left[ {\frac{{b - a}}{n}} \right]^2 + \frac{{(n - 1)(n)(2n - 1)}}{6}\left[ {\frac{{b - a}}{n}} \right]^3 \\
\end{array}\)
 
Ah! Thank you, pka. :)

Unfortunately, I was late with my assignment, but what's important is that I now know how to do it proper.

Thanks again, and to you as well, stapel. :D
 
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