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

Question

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?

Answer 1

$$\int{x\,e^{x^3+4}}\,dx=-\frac{e^4 x^2 \Gamma \left(\frac{2}{3},-x^3\right)}{3 \left(-x^3\right)^{2/3}}=-\frac{1}{3} e^4 x^2 E_{\frac{1}{3}}\left(-x^3\right)$$ $$\int_3 ^5 {x\,e^{x^3+4}}\,dx=\frac{e^4}{3} \left(9 E_{\frac{1}{3}}(-27)-25 E_{\frac{1}{3}}(-125)\right)$$ Since the arguments ar quite large, for an evaluation, you could use the expansion

$$E_{\frac{1}{3}}\left(-x^3\right)=-e^{x^3} \left(\frac{1}{x^3}+\frac{1}{3 x^6}+\frac{4}{9 x^9}+O\left(\frac{1}{x^{12}}\right)\right)$$

Answer 2

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