Question

How would I write the Riemann sum needed to find the area under the curve given by the function `f(x)=5x^2+3x+2` over the interval [2,6]?

Answer 1

Divide the interval into n equal parts, each one of length `4/n`, with end points `2+4/n, 2+8/n, 2+12/n ... 6`. Consider the upper Riemann sums,

`sum 4/n[ 5{(2+4/n)^2 +(2+8/n)^2 +(2+12/9)^2 +..}+3{(2+4/n) +(2+8/n) +(2+12/n) +.......} +{ 2+2+2+...}]` up to n terms.

=#sum20/n{(2^2 +2^2+2^2...)+(16/n +32/n+48/n...)+(16/n^2 +64/n^2+144/n^2+..)} +sum12/n{(2+2+2...)+4/nsum (1+2+3+...)}+ 4/n *2n #

=`20/n *4n +20/n *16/n (n(n+1))/2+20/n sum *16/n^2( 1^2 +2^2 +3^2 +..)+ 12/n*2n +12/n*4/n *(n(n+1))/2 +8`

=`80+160(1+1/n)+320/n^3 * (n(n+1)(2n+1))/6+24+24(1+1/n)+8`
=`80+160(1+1/n)+320/6*( 2+3/n+ 1/n^2)+24+24(1+1/n)+8`
Now apply limit as `n-> oo`
= 80+160+320/3+24+24+8= 1208/3