Mathematics, 16.01.2020 21:31 dominikkovar

Ihave 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:

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Answer from: alvarezreinier2

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \sum^{\infty}_{n = 0} \frac{x^{3n + 5}}{(3n + 5)n!} + C$

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{\Gamma (\frac{5}{3}, \ -x^3)}{3} + C$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Sequences

Series

Taylor Polynomials and Approximations

MacLaurin PolynomialsTaylor Polynomials

Power Series

Power Series of Elementary FunctionsTaylor Series: $\displaystyle P(x) = \sum^{\infty}_{n = 0} \frac{f^n(c)}{n!}(x - c)^n$

Integration of Power Series:

$\displaystyle f(x) = \sum^{\infty}_{n = 0} a_n(x - c)^n$ $\displaystyle \int {f(x)} \, dx = \sum^{\infty}_{n = 0} \frac{a_n(x - c)^{n + 1}}{n + 1} + C_1$

Multivariable Calculus

Gamma Functions

$\displaystyle \Gamma (s, x) = \int\limits^{\infty}_x {t^{s - 1}e^{-t}} \, dt$ 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:

$\displaystyle \int {x^4e^\big{x^3}} \, dx$

We know that the power series for $\displaystyle e^x$ is:

$\displaystyle e^x = \sum^{\infty}_{n = 0} \frac{x^n}{n!}$

To find the power series for $\displaystyle e^\big{x^3}$ , substitute in x = x³:

$\displaystyle e^\big{x^3} = \sum^{\infty}_{n = 0} \frac{(x^3)^n}{n!}$

Simplify it, we have:

$\displaystyle e^\big{x^3} = \sum^{\infty}_{n = 0} \frac{x^{3n}}{n!}$

Rewrite the original function:

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \int {x^4 \sum^{\infty}_{n = 0} \frac{x^{3n}}{n!}} \, dx$

Rewrite the integrand by including the x⁴ in the power series:

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \int {\sum^{\infty}_{n = 0} \frac{x^{3n + 4}}{n!}} \, dx$

Integrating the power series, we have:

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \sum^{\infty}_{n = 0} \frac{x^{3n + 5}}{(3n + 5)n!} + C$

Multivariable Calculus

Let's set our variables for u-substitution:

u = x⁵ → du = 5x⁴ dx

Use u-substitution on the integral to obtain:

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{1}{5}\int {e^\big{u^{\frac{3}{5}}}} \, dx$

We see that the integral is an incomplete gamma function:

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{1}{5} \bigg[ \frac{5 \Gamma (\frac{5}{3}, \ -u^\big{\frac{3}{5}})}{3} \bigg] + C$

Simplifying it, we have:

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{\Gamma (\frac{5}{3}, \ -u^\big{\frac{3}{5}})}{3} + C$

Back-substituting u will give us the final result:

$\displaystyle \int {x^4e^\big{x^3}} \, dx = \frac{\Gamma (\frac{5}{3}, \ -x^3)}{3} + C$

Answer from: Quest

answer: the answer is c.

step-by-step explanation:

she’s dividing 1/3 of the flour into 2 jars.

hope this : )

Use the expression at the top right to answer the question (1/3 divided by 2)

Answer from: Quest

I’m a smart person so that’s why i think it was 6

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Ihave 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:

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Ihave 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: