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: $\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

$\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

which is the required equation for finding the value of x,

$\frac{x}{sin40^{\circ}}= \frac{12}{sin60^{\circ}}$

$x= 12\times \frac{sin40^{\circ}}{sin60^{\circ}}$

$x= 12\times 0.74222719896$

$x= 8.90672638762\approx 8.967$

Answer from: Quest

go hook jvc hgv cj xk jvc 6756

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