1. Aug 23, 2014

### axe34

Hi,

I'm trying to prove that the integral of x^3 (x cubed) between the limits of a (lower limit) and b (upper limit) is:

(b^4)/4 - (a^4)/4

I'm using the traditional method of dividing the area into n rectangles (where n tends to infinity). Hence the width of 1 rectangle is (b-a)/n

The x coordinate (left side of each thin rectangle) of any rectangle is: a + (k-1) * ((b-a)/n)

I can prove other integrations using this 'traditional' approach but cannot get the correct answer here.

Does anyone have a link to the proof or can provide it?

Thanks. This is driving me mad.

2. Aug 23, 2014

Maybe you could post your calculation that gave you the wrong answer.

3. Aug 23, 2014

### Simon Bridge

So you divide the area into N rectangles width $\Delta x : N\Delta x = b-a$
The area of the nth rectangle is $A_n=(x_n)^3\Delta x:x_n=a+n\Delta x$

The area between a and b is the sum:$$A=\frac{b-a}{N}\sum_{n=0}^N \left(a+n\frac{b-a}{N}\right)^3$$ ... expand the cubic and sum each term separately.
Then take the limit as $N\to\infty$

This what you tried?
Where did you get stuck?

Or just google for "riemann sum for x^3" ;)

4. Aug 23, 2014

### Ray Vickson

It is probably a lot less messy to first get $F(c) =\int_0^c x^3 \, dx$ using the traditional approach, then getting your answer as $F(b) - F(a)$ (assuming $0 < a < b$). To get $F(c)$ you just need to perform summations of the form $\sum_{n=1}^N n^3$.