Express the integral as a limit of Riemann sums. Do not evaluate the limit.
Answer I got for this is...
(2+5/ni) - 3ln(2+5/ni) * 5/n.
I thought it was right, but apparently it's wrong.
Express the integral as a limit of Riemann sums. Do not evaluate the limit.
Answer I got for this is...
(2+5/ni) - 3ln(2+5/ni) * 5/n.
I thought it was right, but apparently it's wrong.
As the function in the integral is continuous is $\displaystyle [2,7]$ we know the integral exists and thus we can choose the partition of the interval as we want. We choose
to subdivide $\displaystyle [2,7] $ in n subintervals of equal length $\displaystyle \frac{7-2}{n}=\frac{5}{n}$ , and we evaluate the function at right end of each subinterval:
$\displaystyle \lim_{n\to\infty}\sum^n_{i=i}\frac{5}{n}\left[2+\frac{5i}{n}-3\ln\left(2+\frac{5i}{n}\right)\right]$ ...this looks similar to what you got but I can't be sure since it isn't clear the way you wrote it.
Tonio