Question

For the function given​ below, find a formula for the Riemann sum obtained by dividing the interval ​[0​,3​] into n equal subintervals and using the​ right-hand endpoint for each c[Subscript]k. Then take a limit of this sum as n approaches infinity to calculate the area under the curve over ​[0​,3​].

f(x)=2x^2

Answer 1

We divide [0, 3] into $n$ subintervals,

$\left[0,\dfrac3n\right]\cup\left[\dfrac3n,\dfrac6n\right]\cup\left[\dfrac6n,\dfrac9n\right]\cup\cdots\cup\left[\dfrac{3(n-1)}n,3\right]$

so that the right endpoint of each subinterval is given according to the arithmetic sequence,

$r_k=\dfrac{3k}n$

for $1\le k\le n$.

The Riemann sum is then

$\displaystyle\sum_{k=1}^nf(r_k)\Delta x_k$

where

$\Delta x_k=r_k-r_{k-1}=\dfrac{3k}n-\dfrac{3(k-1)}n=\dfrac3n$

With $f(x)=2x^2$, we have

$\displaystyle\frac3n\sum_{k=1}^n2\left(\frac{3k}n\right)^2=\frac{54}{n^3}\sum_{k=1}^nk^2$

Recall that

$\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

The area under the curve $f(x)$ over the interval [0, 3] is then

$\displaystyle\int_0^32x^2\,\mathrm dx=\lim_{n\to\infty}\frac{54n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}9\left(2+\frac3n+\frac1{n^2}\right)=\boxed{18}$