# Riemann's sum question

#### Jgirl689

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.

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#### tonio

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