TITLE: Finding largest $r$ and smallest $q$ in $\mathbb{N}$ so that $\exp(r)\lt\int_{1}^{2}\exp(x^3+x)\mathrm dx\lt \exp(q)$ by hand
QUESTION [0 upvotes]: Initially I was given that $k_{1}\le \int_{1}^{2}\exp(x^3+x)\mathrm dx\le k_{2}$, with few options of what $k_{1},k_{2}$ could be. With the very basic technique of approximation, one can argue that the definite integral in question lies between $\exp(2)$ and $\exp(10)$. But I could not arrive at a direct method for comparing the integral with a different value inside this interval say $\exp(3)$.
So what if we were supposed to find the largest $r$ and smallest $q$ with $r,q\in\mathbb{N}$ such that $\exp(r)\lt\int_{1}^{2}\exp(x^3+x)\mathrm dx\lt\exp(q)$. Any hints are appreciated. Thanks.

REPLY [2 votes]: Let $I = \displaystyle\int_{1}^{2}e^{x^3+x}\,dx$. Using integration by parts, we have
\begin{align*}
I &= \int_{1}^{2}\dfrac{(3x^2+1)e^{x^3+x}}{3x^2+1}\,dx 
\\
&= \left[\dfrac{e^{x^3+x}}{3x^2+1}\right]_{1}^{2}+\int_{1}^{2}\dfrac{6xe^{x^3+x}}{(3x^2+1)^2}\,dx
\\
&= \left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right)+\int_{1}^{2}\dfrac{6xe^{x^3+x}}{(3x^2+1)^2}\,dx
\end{align*}
You can check that $\dfrac{12}{169} \le \dfrac{6x}{(3x^2+1)^2} \le \dfrac{3}{8}$ and $e^{x^3+x} \ge 0$ for all $1 \le x \le 2$. Hence, $$\dfrac{12}{169}I = \int_{1}^{2}\dfrac{12}{169}e^{x^3+x}\,dx \le \int_{1}^{2}\dfrac{6xe^{x^3+x}}{(3x^2+1)^2}\,dx \le \int_{1}^{2}\dfrac{3}{8}e^{x^3+x}\,dx = \dfrac{3}{8}I.$$
Using the above facts, we have $$\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right)+\dfrac{12}{169}I \le I \le \left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right)+\dfrac{3}{8}I,$$ which can be rearranged to $$\dfrac{169}{157}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right) \le I \le \dfrac{8}{5}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right).$$
Using $e^2 \le 8$, the upper bound can be relaxed to $I \le \dfrac{8}{5}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right) \le \dfrac{8}{65}e^{10} \le \dfrac{64}{65}e^8 \le e^8$.
Using $e^8 \ge e^3 \ge 20$, the lower bound can be relaxed to $I \ge \dfrac{169}{157}\left(\dfrac{e^{10}}{13}-\dfrac{e^2}{4}\right) \ge \dfrac{e^{10}}{13}-\dfrac{e^2}{4} = \dfrac{e^{10}}{13}\left(1-\dfrac{13}{4e^8}\right) \ge \dfrac{e^{10}}{13}\left(1-\dfrac{13}{80}\right) \ge \dfrac{e^{10}}{20} \ge e^7$.
Hence, $e^7 \le I \le e^8$.
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