Basic Integrals

[shiditso]

"Consider the function f(x) = ? [(x2)/3] ? 2.

In this problem you will calculate ?(0 to 4) (? [(x2)/3] ? 2) dx by using the definition

?b
?a f(x)dx = {lim[n:?] ?[i=1,n]f(xi)?x} = {lim[n:?] R_n}


The summation inside the brackets is Rn which is the Riemann sum where the sample points are chosen to be the right-hand endpoints (xi) of each sub-interval.
Calculate Rn for f(x) = ? [(x2)/3] ? 2 on the interval [0, 4] and write your answer as a function of n without any summation signs."

I know the following:
?x = 4/n
xi = 4i/n

I also know that
?[i=1,n] (i) = n(n+1)/2 and (i^2) = (n(n+1)(2n+1)/6)

Any help would be greatly appreciated for finding R_n.

 

[galactus]

Is this it:

\(\displaystyle \int_{0}^{4}\left[\frac{-x^{2}}{3}-2\right]dx\)

If so, use the ^ for exponent. Writing x2 looks like x multiplied by 2.

If the above is the correct integrand, then

Each subinterval has length:

\(\displaystyle {\Delta}x=\frac{4-0}{n}=\frac{4}{n}\)

The right endpoint method is \(\displaystyle x_{k}=a+k{\Delta}x=0+\frac{4k}{n}\)

Thus, rectangle k has area:

\(\displaystyle \sum_{k=1}^{n}\left[\frac{-(\frac{4k}{n})^{2}}{3}-2\right]\cdot \frac{4}{n}\)

\(\displaystyle \frac{-64}{3n^{3}}\sum_{k=1}^{n}k^{2}-\frac{8}{n}\sum_{k=1}^{n}1\)

Now, since you know the closed form for the sum of squares. The summation of the 1 is just n.

Put it altogether and take the limit \(\displaystyle n\to \infty\)

 

[shiditso]

Thank you very much! I forgot the carrot (^) but I am eternally grateful!

Shiditso