Question

how would I find the integral of x^3 dx from 1 to 0?

Answer 1

anti derivative of \(x^3\) is \(\frac{x^4}{4}\) evaluate at 1 and get \(\frac{1}{4}\)

Answer 2

ah from one to zero! so it is \(-\frac{1}{4}\) because you are going backwards

Answer 3

I need to show that this works for the formula for the sum of n cubes....
just like we did for the sum of squares.....integral of x^2 from 0 to 1 = lim [(n+1)(2n+1)]/6n^2....it says to use this algorithm to evaluate the integral of x^3 from 0 to 1

Answer 4

the riemann sum?

Answer 5

think he used that..

Answer 6

oh really??? have fun!

Answer 7

we can do it if you like, it will take a while

Answer 8

first you have to divide the unit interval up in to n pieces, each of which is of length \(\frac{1}{n}\)

Answer 9

that is your \(\Delta x\)

Answer 10

he said to "evaluate" and I believe it involves just using the formula we found for the sum of the fist n cubes....

Answer 11

the we have to say what the \(x_i\) are. you have \(x_0=0, x_1=\frac{1}{n}, x_2=\frac{2}{n}, x_3=\frac{3}{n}\) and in general
\(x_i=\frac{i}{n}\)

Answer 12

then we have to compute
\[\sum_{i=0}^nf(x_i)\Delta x\]

Answer 13

we get
\[\sum_{i=1}^n(\frac{i}{n})^3\frac{1}{n}\]