EXAMPLE $2$ For the region under $f(x)=5x^2$ on $[0,4],$ show that the sum of the areas of the upper approximating rectangle approaches $\frac{320}{3},$ that is

$\underset{n}{lim}{R}_n=\frac{320}{3}$

SOLUTION $,R_n$ is the sum of the areas of the $n$ rectangles in the figure. Each rectangle has width $\frac{4}{n}$ and the heights are the values of the function $f(x)=5x^2$ at the points $\frac{4}{n},\frac{8}{n},\frac{12}{n},$dots,$\frac{4n}{n}$; that is$,$ the heights are $5(\frac{4}{n}{)}^2,5(\frac{8}{n}{)}^2,5(\frac{12}{n}{)}^2,$dots,$5(\frac{4n}{n}{)}^2.$ Thus,

$R_n=\frac{4}{n}*5(\frac{4}{n}{)}^2+\frac{4}{n}*5(\frac{8}{n}{)}^2+\frac{4}{n}*5(\frac{12}{n}{)}^2+$dots$+\frac{4}{n}*5(\frac{4n}{n}{)}^2$

$=\frac{20}{n}*,*(1^2+2^2+3^2+$dots$+n^2)$

$$

$(1^2+2^2+3^2+$dots$+n^2)$

Video Example

Here we need the formula for the sum of the squares of the first $n$ positive integers:

$1^2+2^2+3^2+$dots$+n^2=\frac{n(n+1)(2n+1)}{6}$

Perhaps you have seen this formula before. Putting this formula into our expression for $R_n,$ we get

$R_n=$

$$

Thus we have

$\underset{n}{lim}{R}_n,=\underset{n}{lim},$

$,=\underset{n}{lim},(\frac{n+1}{n})(\frac{2n+1}{n})$

$,=\underset{n}{lim},(1+\frac{1}{n})(2+\frac{1}{n})$

$,=,1*2=$