Question

evaluate the following integral as the limit of a sum [0,3] (x^2+5x-6)dx and please show steps!

Answer 1

\[\Delta x=(b-a)/n=3/n\], xi=3i/n
\[\lim_{? \rightarrow ?}\sum_{n}^{\infty}[3/n][3i/n]^2 +5(3i/n)-6]\]

Answer 2

=\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}[3/n][9i^2/n^2+15i/n-6]\]

Answer 3

\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}[(27/n^3)(n(n+1)(2n+1)/6)]+(45/n^2)(n(n+1)/2)-18]\]

Answer 4

dividing and use the limit
=\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n}[(27/6)(2+2/n+1/n+1/n^2]+(45/2)(1+1/n)-18]\]

Answer 5

=27(2)/6 +45/2 -18=27/2=13.5 ans.................