Expert: Abe Mantell - 5/20/2006

Question
I am horrible at series. On my math final, it said to evaluate the intergral using Riemann sums
3
S  x(squared)+2x  dx
0
S standing for the antiderivative sign.......we also had to use the equations i=n(n+1) all divided by 2 and i(squared)=n(n+1)(2n+1) all divided by 6. I was hoping that you could solve this problem for me. Thank you, Hilary


Answer
Hello Hilary,

delta_x=3/n, so...
The integral = lim(n->inf) SUM[{(3i/n)^2+2(3i/n)}(3/n), i=1..n]
=lim(n->inf) SUM[{(9/n^2)*i^2+i*6/n}(3/n), i=1..n]
=lim(n->inf) [(9/n^2)SUM[i^2] + (6/n)SUM[i]]*(3/n)
=lim(n->inf) [(27/n^3)*n(n+1)(2n+1)/6 + (18/n^2)*n(n+1)/2]
=lim(n->inf) [(9/2)(2n^3+3n^2+n)/n^3 + 9*(n+1)/n]
=9 + 9
=18

I hope you can follow this!  It's tough to type this all
in with using nice math notation!  :-/

TTYL, Abe