Respuesta :
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:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integration
- Integrals
- [Indefinite Integrals] Integration Constant C
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
U-Substitution
Sequences
Series
Taylor Polynomials and Approximations
- MacLaurin Polynomials
- Taylor Polynomials
Power Series
- Power Series of Elementary Functions
- Taylor 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 Calculus
Gamma Functions
- [tex]\displaystyle \Gamma (s, x) = \int\limits^{\infty}_x {t^{s - 1}e^{-t}} \, dt[/tex]
- Incomplete Gamma Functions
Step-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 Calculus
We 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 Calculus
Let's set our variables for u-substitution:
u = x⁵ → du = 5x⁴ dx
Use 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]
