For a homework problem, I'm supposed to compute the integral of $f(x) = x e^{-x^2}$, which I did, and got a result of $F(x) = -\frac{1}{2} e^{-x^2} + C$.
Secondly, I'm supposed to find the smallest value of N such that the area under the graph $f(x)$ between N and N+1 is, at most, 0.01.
From that problem description, I came up with the following inequality:
$$ -\frac{1}{2}[e^{-(N+1)^2} - e^{-N^2}] \leq 0.01 $$
Which I then changed to:
$$ \frac{1}{2}[e^{-N^2} - e^{-(N+1)^2}] \leq 0.01 $$
Multiplying both sides by 2, I get:
$$ e^{-N^2} - e^{-(N+1)^2} \leq 0.02 $$
However, I'm stumped at this point. I was able to play around with the equation in Desmos and get N=2, which satisfies the inequality, but that's hardly a good solution.
How do I solve the above inequality analytically? Any pointers on the next steps I should take? Is trying to solve via the inequality even the correct approach?
Textbook: OpenStax Calculus Volume 1
Section: 5.6
Exercise: 377, Page 606 (Note that the problem erroneously states the upper bound to be N+10 instead of N+1)
Making the problem more general with $x$ instead of $n$, you are looking for the zero of function $$f(x)=e^{-x^2} - e^{-(x+1)^2}-\epsilon$$ where $\epsilon$ is a small number. If $x$ is supposed to be large, since $$e^{-x^2} - e^{-(x+1)^2}=e^{-x^2}\left(1-e^{-(2x+1) } \right)\sim e^{-x^2}$$ and an estimate of the solution is $$x_0=\sqrt{-\log (\epsilon )}$$ Now, since $f(x)$ varies very fast, consider instead that you look for the zero of $$g(x)=\log\left(e^{-x^2} - e^{-(x+1)^2} \right)-\log(\epsilon )$$ and use Newton method with $$g'(x)=\frac{2-2 \left(e^{2 x+1}-1\right) x}{e^{2 x+1}-1}$$
For illustration, let us try with $\epsilon=10^{-2}$. This would give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.145966026289347239636184 \\ 1 & 2.144787850787186005128209 \\ 2 & 2.144787523122507226762567 \\ 3 & 2.144787523122481882718073 \end{array} \right)$$