Conclude $\left|\sum_{n = 0}^{100} \frac{e^{-n^2/10000}}{100} -\int_{0}^{1}e^{-x^2}dx\right|\le 0.05$ Conclude $$\left|\sum_{n = 0}^{100} \frac{e^{-n^2/10000}}{100} -\int_{0}^{1}e^{-x^2}dx\right| \le 0.05$$
Ok so this is a question by parts for integrals according to the definion, meaning upper and lower sums and we are not allowed to use the fundemental theorem of calculus for definite integrals. This is the last part of the question and I need to showcase this.
I want to say that the left part (the sum) and the right part are equal and this their difference is $=0 < 0.05$ but I dont know how to explain it formally. I am not even sure its the right direction in which case I am definitely bit lost
I can say $P$ is a partition for $f(x) = e^{-x^2/10000}$ for which $x_k = 1/100$ for every $x$ in the range but I am not sure how to say that the sum which is also the lower sum is equal to the upper integral and reach that conclusion any help would be appreciated
 A: Since the function $e^{-x^2}$ is decreasing on the interval $[0,1]$, a right hand Riemann sum will be smaller than the integral, which will be smaller than a left hand Riemann sum. Therefore
$$
\sum_{n=1}^{100}\frac{e^{-(n/100)^2}}{100}
\le \int_0^1 e^{-x^2}\ dx
\le \sum_{n=0}^{99} \frac{e^{(n/100)^2}}{100}
\le \sum_{n=0}^{100} \frac{e^{(n/100)^2}}{100}.
$$
For $n=0$ we have $\frac{e^{-n^2/10000}}{100}=\frac{1}{100}<.05.$
So
$$
\sum_{n=0}^{100}\frac{e^{-n^2/10000}}{100}-.05<
\sum_{n=1}^{100}\frac{e^{-n^2/10000}}{100}
\le \int_0^1 e^{-x^2}\ dx
\le \sum_{n=0}^{100} \frac{e^{n^2/10000}}{100}.
$$
Then
$$
\Bigg\lvert\sum_{n=0}^{100} \frac{e^{n^2/10000}}{100} - \int_0^1 e^{-x^2}\ dx
\Bigg \rvert<.05
$$
follows.
A: Notice that
$$D:=\Big|\sum^{100}_{k=0}\frac{e^{-(k/100)^2}}{100}-\int^1_0e^{-t^2}\,dt\Big|\leq\frac{1}{100}+\Big|\sum^{100}_{k=1}\int^{k/100}_{(k-1)/100}e^{-(k/100)^2}-e^{-t^2}\,dt\Big|$$
By the mean value theorem, for $t\in[(k-1)/100,k/100]$ there is $t<\xi_t<k/100$ such that
$$|e^{-t^2}-e^{-(k/100)^2}\big|= 2\xi e^{-\xi^2}\big|t-\tfrac{k}{100}\big|\leq \frac{2k}{100}\frac{1}{100}$$
$$\int^{k/100}_{(k-1)/100}\big|e^{-t^2}-e^{-(k/100)^2}\big|\,dt\leq\int^{k/100}_{(k-1)/100}\frac{2k}{100}\frac{1}{100}dt=\frac{k}{500000}$$
$$\frac{1}{500000}\sum^n_{k=1}k=\frac{5050}{500000}=.0101$$
Putting things together $D\leq .01+.0101=.0201<0.05$.
