Calculating area using sigma notation

[Zack K]

Homework Statement

The question involves using sigma notation of Riemann sums to find the area under the graph of $x^2+\sqrt {1+2x}$. I managed to calculate most of the values and I have $16+\frac 8 3 + \Sigma {\frac 2 n \sqrt {9 + \frac {4i} n}}$

Homework Equations

[/B]
$\Sigma i= \frac {n(n+1)} 2$
$\Sigma i^2= \frac {n(n+1)(2n+1)} 6$

The Attempt at a Solution

What I can think of doing to get rid of the root is to square the whole expression, since the area is equal to the expression and I will get area². Then I can just square root my final answer to get the actual area. But I'm not sure if that will work.

 

[BvU]

the area under the graph

Wouldn't that area be infinity ? Please post a complete problem statement

 

[Zack K]

Wouldn't that area be infinity ? Please post a complete problem statement

Sorry my bad. It's between the interval [4, 6]. and n→infinity

 

[BvU]

First term alone should already yield around 50. Whence the 16 ? the 8/3 ?
Please post your steps in more detail.

square the whole expression, since the area is equal to the expression and I will get area²

O, is that so ? So the area of $\sin^2 x$ is the area of $\sin x$ squared ?

 

[Ray Vickson]

Homework Statement

The question involves using sigma notation of Riemann sums to find the area under the graph of $x^2+\sqrt {1+2x}$. I managed to calculate most of the values and I have $16+\frac 8 3 + \Sigma {\frac 2 n \sqrt {9 + \frac {4i} n}}$

Homework Equations

[/B]
$\Sigma i= \frac {n(n+1)} 2$
$\Sigma i^2= \frac {n(n+1)(2n+1)} 6$

The Attempt at a Solution

What I can think of doing to get rid of the root is to square the whole expression, since the area is equal to the expression and I will get area². Then I can just square root my final answer to get the actual area. But I'm not sure if that will work.

No: it won't work. We have $(\sqrt{a}+\sqrt{b})^2 \neq (a + b)$, so it does not even work for just two terms.

 

[Zack K]

No: it won't work. We have $(\sqrt{a}+\sqrt{b})^2 \neq (a + b)$, so it does not even work for just two terms.

So what's recommended that I do? I know that I have to isolate i, but I can't seem to figure it out.

 

[BvU]

Please post your steps in more detail

 

[Ray Vickson]

So what's recommended that I do? I know that I have to isolate i, but I can't seem to figure it out.

If you are given a value of $n$, such as $n = 30$, you can compute the 30-term sum numerically. If you use something like a spreadsheet, or Matlab, or Maple, or Mathematica, etc., it should not be hard. Even summing $n = 1000$ or more such terms should be do-able in a second or two, but doing it manually might take a few hours.

Paradoxically, in this case doing the $n = \infty$ sum (= the integral) is quite easy, but the finite sums are much harder.

 

[vela]

You could set up a Riemann sum using horizontal rectangles, i.e., integrate with respect to y, to avoid the square root.