Traditional integration of X^3

[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.

 

[Quesadilla]

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

 

[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" ;)

 

[Ray Vickson]

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.

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 $\sum_{n=1}^N n^3$.