Riemann Sums

[f1player]

Use the limit definition to evaluate the following integral:

integral{-1 to 1} x^3 dx

Now the answer is obviously 0 but by using the riemann sum i get a different answer.

Area = lim{n->infinity} sum{i=1 to infinity} (i delta x)^3 delta x

Using this and simplifying it down using summation formulas i get area = 4 and not 0.

Can anyone help?

 

[galactus]

\(\displaystyle \L\\{\Delta}x=\frac{1-(-1)}{n}=\frac{2}{n}\)

\(\displaystyle \L\\x_{k}=a+k{\Delta}x=-1+\frac{2k}{n}\)

\(\displaystyle \L\\\left(-1+\frac{2k}{n}\right)^{3}\frac{2}{n}\)

If you expand this, you get:

\(\displaystyle \L\\\frac{16k^{3}}{n^{4}}-\frac{24k^{2}}{n^{3}}+\frac{12k}{n^{2}}-\frac{2}{n}\)

Can you finish?.

Just remember that \(\displaystyle \L\\\sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4}\)
\(\displaystyle \L\\\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}\)

\(\displaystyle \L\\\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\)

Sometimes these things are algebra nightmares.

 

[f1player]

Yes it is an algebra nightmare!

But why cant x_k = k delta x

why is it x_k = a + k delta x

(Does this mean that if the lower limit a is not 0 you have to add it to k delta x?)

But even with your working I still get Area = 2 at the end and not 0.

This is a nightmare!

 

[galactus]

I am using the right endpoint method. The subinterval [-1,1] is divided by points
\(\displaystyle \L\\x_{1}, x_{2}, x_{3}, .........., x_{n-1}\) into n equal parts each of length
\(\displaystyle \L\\{\Delta}x\), and if we let \(\displaystyle \L\\x_{0}=-1 \;\ and \;\ x_{n}=1\), then
\(\displaystyle \L\\x_{k}=-1+k{\Delta}x\), for k=0, 1, 2, .........., n.

So, \(\displaystyle \L\\x_{k}=-1+k{\Delta}x\)

For your problem, \(\displaystyle \L\\{\Delta}x=\frac{1-(-1)}{n}=\frac{2}{n}\)

and \(\displaystyle \L\\x_{k}=-1+k{\Delta}x=-1+\frac{2k}{n}\)


So, \(\displaystyle \L\\\left(-1+\frac{2k}{n}\right)^{3}(\frac{2}{n})\)

=\(\displaystyle \L\\\frac{16}{n^{4}}\sum_{k=1}^{n}k^{3}-\frac{24}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{12}{n^{2}}\sum_{k=1}^{n}k-\frac{2}{n}\sum_{k=1}^{n}1\)

Using what we know about summations, we get:

\(\displaystyle \L\\\frac{4(n+1)^{2}}{n^{2}}-\frac{4(n+1)(2n+1)}{n^{2}}+\frac{6(n+1)}{n}-2\)

Believe it or not, this whittles down to \(\displaystyle \L\\\frac{2}{n}\)(you can do the algebra, it's not as bad as it looks).

\(\displaystyle \L\\\lim_{n\to\infty}\frac{2}{n}=0\)

 

[f1player]

Yes! This makes perfect sense now.

I can see my mistake from your working. There were also a couple of other similar questions that I couldn't do before but now its all fine.

Thank you so much.

It's not a nightmare after all lol