Hey there I am having some trouble understanding a practice problems and problems of this type in general,
the question asks to show, and or give an argument for the inequality below $$ \int_1^n lnx dx \le ln(1)+ln(2)+ln(3)+ln(4)+…+ln(n) \le \int_1^{n+1}ln(x)dx $$ for $$ n \ge 1 $$
Now, I know it has something to do with Rieman sums i.e. left reiman sums and right reiman sums but I am having some trouble figuring out how to know which one is which. I am also thinking that the value in-between the inequality is the true value of the integral?
I know if you started from n+1 and made rectangles to the left you would have an overestimation, and if you started from 1 to n going to the right you would have an underestimation, but I am still not sure how I can tie it all together, and maybe make it more clear and such.
Hope all is well, thank you! Let me know if you want me to try to show more work or anything like that!
Let's evaluate the upper sum for the function $ \ln x $ on the interval $ [1, n] $. First we subdivide the given interval on subintervals of the form $ [k, k + 1], 1 \leq k \leq n - 1, k \in \Bbb{Z}$. Since the function is strictly growing on the interval, $ \sup_{[k, k + 1]} \ln x = ln (k + 1)$. Therefore, the upper sum is: $$ \ln 2 + \ln 3 + \ln 4 + \dots + \ln n $$ We know that the integral of the function over an interval can't be greater than the upper sum. Hence: $$ \int_1^n \ln x dx \leq (\ln 1 = 0) + \ln 2 + \ln 3 + \ln 4 + \dots + \ln n $$
A similar argument for the right-side integral will take you there (look at the lower sum)!