<s>[INST] Find the area between $f(x) = x^2+3x+7 $ and $g(x) = xe^{x^3+4}$ for $x \in [3,5]$. 
Calculate the area between the two functions, $f(x)$, $g(x)$, for $x \in [3,5]$.
$$f(x)=x^2+3x+7$$
$$g(x)=xe^{x^3+4}$$

To determine the area between the functions I used the formula $A= \int_a^b|f(x)-g(x)|dx$. Therefore, I have:
\begin{align}
A&=\int_3^5|x^2+3x+7-xe^{x^3+4}|dx \\
&= \int_3^5|x^2|dx+\int_3^5|3x|dx+\int_3^5|7|dx-\int_3^5|e^{x^3+4}|dx\\
&= \left(\frac{x^3}{3}+\frac{3x^2}{2}+7x\right)_3^5 - \int_3^5|xe^{x^3+4}|dx
\end{align}
Here I can use $u$-subsitution and the Incomplete Gamma Function to find the integral of $xe^{x^3+4}$. All in all I get:
\begin{align}
A&=\left(\frac{x^3}{3}+\frac{3x^2}{2}+7x-e^4\frac{1}{2}\left(-\frac{x^2Γ\left(\frac{1}{\frac{3}{2}},\:-\left(x^2\right)^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-\left(x^2\right)^{\frac{3}{2}}}}\right)\right)_3^5\\
&=-\frac{98}{3}-24-14+e^4\frac{1}{2}\left(-\frac{25Γ\left(\frac{1}{\frac{3}{2}},\:-25^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-25^{\frac{3}{2}}}}-\left(-\frac{9Γ\left(\frac{1}{\frac{3}{2}},\:-9^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-9^{\frac{3}{2}}}}\right)\right)
\end{align}
Can this be simplified further or is the given solution enough?
 [/INST] You can do this:
$$The \ area \ between \ the \ two \ functions = \int_{3}^{5}{xe^{x^3+4}}dx \ - \ \int_3^5{x^2+3x+7}dx$$ Because $\forall x \in [3, 5] \ \ \ xe^{x^3+4} > x^2+3x+7$.
You can look here: https://www.desmos.com/calculator/csh4alwmeu
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