Question

For the function​ below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subintervals. Then take a limit of these sums as n right arrow infinityn → [infinity] to calculate the area under the curve over [a comma b ][a,b]. f (x )equals x squared plus 3f(x)=x2+3 over the interval [0 comma 4 ]

Answer 1

See below

Explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

`[0,\displaystyle(4)/(n)],[\displaystyle(4)/(n),\displaystyle(2*4)/(n)],[\displaystyle(2*4)/(n),\displaystyle(3*4)/(n)],...,[\displaystyle((n-1)*4)/(n),4]`

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

`\displaystyle(4)/(n)f(\displaystyle(1*4)/(n))+\displaystyle(4)/(n)f(\displaystyle(2*4)/(n))+...+\displaystyle(4)/(n)f(\displaystyle(n*4)/(n))=\\\\=\displaystyle(4)/(n)((\displaystyle(1*4)/(n))^2+3+(\displaystyle(2*4)/(n))^2+3+...+(\displaystyle(n*4)/(n))^2+3)=\\\\\displaystyle(4)/(n)((1^2+2^2+...+n^2)\displaystyle(4^2)/(n^2)+3n)=\\\\\displaystyle(4^3)/(n^3)(1^2+2^2+...+n^2)+12`

but

`1^2+2^2+...+n^2=\displaystyle(n(n+1)(2n+1))/(6)`

so the upper sum equals

`\displaystyle(4^3)/(n^3)(1^2+2^2+...+n^2)+12=\displaystyle(4^3)/(n^3)\displaystyle(n(n+1)(2n+1))/(6)+12=\\\\\displaystyle(4^3)/(6)(2+\displaystyle(3)/(n)+\displaystyle(1)/(n^2))+12`

When
`n\rightarrow \infty` both
`\displaystyle(3)/(n)` and
`\displaystyle(1)/(n^2)` tend to zero and the upper sum tends to

`\displaystyle(4^3)/(3)+12=\displaystyle(100)/(3)`