Riemann sum for definite integral -- solved

[WillGabriel]

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

I don't even know if I'm going about the problem right... Thanks in advance!

 

[stapel]

How are you "going about the problem"? Please show your steps, and we'll be glad to check your work.

Thank you.

Eliz.

 

[WillGabriel]

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

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.

 

[stapel]

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.


. . . . .\(\displaystyle \large{\int_b^a{\,x^2\,}dx\,= \,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,F(x_i)\,\Delta x}\)



. . . . .\(\displaystyle \large{=\,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,{\left[a\,+\,\frac{b_i\,-\,a_i}{n}\right]}^2\;\frac{b\,-\,a}{n}}\)



. . . . .\(\displaystyle \large{=\,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\frac{b\,-\,a}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\left(a^2\,+\,2a{\left(\frac{b_i\,-\,a_i}{n}\right)}\,+\,{\left({\frac{b_i\,-\,a_i}{n}}^2\right)}\right)}\)



. . . . .\(\displaystyle \large{=\,\begin{array}{c}limit\\n\rightarrow\infty\end{array}\,\frac{b\,-\,a}{n}\,\left[{a^2 n\,+\,\frac{2a}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,b_i\,-\,\frac{2a}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,a_i\,+\,\frac{1}{n^2}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,{b_i}^2\,-\,\frac{2}{n}\,\begin{array}{c}n\\\sum\\i=1\end{array}\,a_i b_i\,+\,\frac{1}{n^2}\,\begin{array}{c}n\\\sum\\i=1\end{array}{a_i}^2}\right]}\)


Eliz.

 

[WillGabriel]

Yes, that's correct--thanks a ton for that. If anyone else could offer some help with the problem, I'd be very thankful.

 

[pka]

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 \(\displaystyle \L
\{{\rm{lim}}}\limits_{n \to \infty }\)

 

[WillGabriel]

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).

 

[WillGabriel]

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?

 

[pka]

\(\displaystyle \begin{array}{c}
\sum\limits_{k = 0}^{n - 1} {(a + k\Delta x)^2 \Delta x} & = & \sum\limits_{k = 0}^{n - 1} {(a^2 \Delta x + 2ak\left[ {\Delta x} \right]^2 + k^2 \left[ {\Delta x} \right]^3 )} \\
& = & a^2 (n)\Delta x + (2a)\left( {\frac{{n(n - 1)}}{2}} \right)\left[ {\Delta x} \right]^2 + \frac{{(n - 1)(n)(2n - 1)}}{6}\left[ {\Delta x} \right]^3 \\
& = & a^2 (n)\left[ {\frac{{b - a}}{n}} \right] + (2a)\left( {\frac{{n(n - 1)}}{2}} \right)\left[ {\frac{{b - a}}{n}} \right]^2 + \frac{{(n - 1)(n)(2n - 1)}}{6}\left[ {\frac{{b - a}}{n}} \right]^3 \\
\end{array}\)

 

[WillGabriel]

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.