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,\frac3n\right]\cup\left[\frac3n,\frac6n\right]\cup\left[\frac6n,\frac9n\right]\cup\cdots\cup\left[\frac{3(n-1)}n,3\right]`

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

`r_k=\frac{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)=\frac{3k}n-\frac{3(k-1)}n=\frac3n`

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

`\displaystyle\frac3n\sum_(k=1)^n2\left(\frac{3k}n\right)^2=(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)(54n(n+1)(2n+1))/(6n^3)=\lim_(n\to\infty)9\left(2+\frac3n+\frac1{n^2}\right)=\boxed{18}`