I have an interesting intergral that seems pretty challenging. After attempting to solve it for some time, I've come up empty handed. Can you solve this integral?Indefinite Integral of:[tex] {x}^{4} { {e}^{x} }^{3} [/tex]
Accepted Solution
A:
Answer:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \sum^{\infty}_{n = 0} \frac{x^{3n + 5}}{(3n + 5)n!} + C[/tex][tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{\Gamma (\frac{5}{3}, \ -x^3)}{3} + C[/tex]General Formulas and Concepts:CalculusDifferentiationDerivativesDerivative NotationBasic Power Rule:
f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹
IntegrationIntegrals[Indefinite Integrals] Integration Constant CIntegration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]U-SubstitutionSequencesSeriesTaylor Polynomials and ApproximationsMacLaurin PolynomialsTaylor PolynomialsPower SeriesPower Series of Elementary FunctionsTaylor Series: [tex]\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(c)}{n!}(x - c)^n[/tex]Integration of Power Series:
[tex]\displaystyle f(x) = \sum^{\infty}_{n = 0} a_n(x - c)^n[/tex]
[tex]\displaystyle \int {f(x)} \, dx = \sum^{\infty}_{n = 0} \frac{a_n(x - c)^{n + 1}}{n + 1} + C_1[/tex]
Multivariable CalculusGamma Functions[tex]\displaystyle \Gamma (s, x) = \int\limits^{\infty}_x {t^{s - 1}e^{-t}} \, dt[/tex]Incomplete Gamma FunctionsStep-by-step explanation:*Note:If we are talking single-variable calculus, then we would have to write this integral as a power series.You can derive the power series for eˣ using Taylor Polynomials (not shown here)If we are talking multi-variable calculus, then we could integrate this and get an "actual" value.Single-variable CalculusWe are given the integral:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx[/tex]We know that the power series for [tex]\displaystyle e^x[/tex] is:[tex]\displaystyle e^x = \sum^{\infty}_{n = 0} \frac{x^n}{n!}[/tex]To find the power series for [tex]\displaystyle e^\big{x^3}[/tex] , substitute in x = x³:[tex]\displaystyle e^\big{x^3} = \sum^{\infty}_{n = 0} \frac{(x^3)^n}{n!}[/tex]Simplify it, we have:[tex]\displaystyle e^\big{x^3} = \sum^{\infty}_{n = 0} \frac{x^{3n}}{n!}[/tex]Rewrite the original function:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \int {x^4 \sum^{\infty}_{n = 0} \frac{x^{3n}}{n!}} \, dx[/tex]Rewrite the integrand by including the x⁴ in the power series:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \int {\sum^{\infty}_{n = 0} \frac{x^{3n + 4}}{n!}} \, dx[/tex]Integrating the power series, we have:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \sum^{\infty}_{n = 0} \frac{x^{3n + 5}}{(3n + 5)n!} + C[/tex]Multivariable CalculusLet's set our variables for u-substitution:u = x⁵ → du = 5x⁴ dxUse u-substitution on the integral to obtain:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{1}{5}\int {e^\big{u^{\frac{3}{5}}}} \, dx[/tex]We see that the integral is an incomplete gamma function:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{1}{5} \bigg[ \frac{5 \Gamma (\frac{5}{3}, \ -u^\big{\frac{3}{5}})}{3} \bigg] + C[/tex]Simplifying it, we have:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{\Gamma (\frac{5}{3}, \ -u^\big{\frac{3}{5}})}{3} + C[/tex]Back-substituting u will give us the final result:[tex]\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{\Gamma (\frac{5}{3}, \ -x^3)}{3} + C[/tex]