Area as limit problem: under y = x^3 on interval [0, b]

[hgaon001]

There is a homework problem that I'm messing up somewhere throughout the process.

The problem says:
Show that the area under the graph of y=x^3 and over the interval [0,b] is b^4/4... The answer i keep getting is b/4, can someone please explain to me where they get the exponent 4 from?

 

[pka]

Re: Area as limit problem

\(\displaystyle \sum\limits_{k = 1}^n {k^3 } = \left[ {\frac{{n(n + 1)}}{2}} \right]^2\)
Well you asked!
Do you understand approximating sums?

 

[hgaon001]

Re: Area as limit problem

\(\displaystyle \sum\limits_{k = 1}^n {k^3 } = \left[ {\frac{{n(n + 1)}}{2}} \right]^2\)
Well you asked!
Do you understand approximating sums?

Not quite. I'm following the steps that the teacher did on some examples but when i take out the constant in front of the sigma i dont have the right one.

 

[galactus]

Re: Area as limit problem

Sorry, I did not realize you were working with Riemann sums. Is that what you must do?. Sounds like it.

Right endpoint: \(\displaystyle a+k{\Delta}x=0+k(\frac{b-0}{n})=\frac{kb}{n}\)

\(\displaystyle (\frac{kb}{n})^{3}\cdot\frac{b}{n}\)

Now, expand out and use the identity pka showed you for k^3. Then, let n-->inf and you should get the limit as what you are looking for.