Question

(integral)x^2015*e^(-x)dx

Answer 1

Ответ:

$\gamma(2016,x)+C_1=C_2-2015!e^{-x}\sum\limits_{k=0}^{2015}\dfrac{x^k}{k!}$

Пошаговое объяснение:

$\int \underbrace{x^{2015}e^{-x}}_{f(x)}dx$

Очевидно, $f(x)$ непрерывна на  $[0;+\infty)$. Но тогда первообразной для нее будет являться интеграл с переменным верхним пределом $\int\limits_0^x f(t)dt$, откуда $\int x^{2015}e^{-x}dx=\underbrace{\int\limits_0^x t^{2015}e^{-t}dt}_{I} +C$.

А $I_{}$ - неполная нижняя Гамма-функция $\gamma(2016,x)$, которая, т.к. первый аргумент натуральный, выражается через полином относительно x:

$\gamma(2016,x)=2015!-2015!e^{-x}\sum\limits_{k=0}^{2015}\dfrac{x^k}{k!}$

Answer 2

The integral of $x^{2015}e^{-x}dx$ can be solved using integration by parts. The integration by parts formula is given by:

$$\int u \, dv = uv - \int v \, du$$

Let's choose:

$u = x^{2015}$ (polynomial part)

$dv = e^{-x}dx$ (exponential part)

Now, let's find $du$ and $v$:

$$du = 2015x^{2014}dx \quad \text{(derivative of } u \text{)}$$

$$v = -e^{-x} \quad \text{(antiderivative of } dv \text{)}$$

Now we can apply the integration by parts formula:

$$\int x^{2015}e^{-x}dx = -x^{2015}e^{-x} - \int (-e^{-x}) \cdot 2015x^{2014}dx$$

Simplify the expression:

$$= -x^{2015}e^{-x} + 2015 \int x^{2014}e^{-x}dx$$

Now, you have a similar integral on the right-hand side. You can apply integration by parts again. Repeat this process until you reach an integrable expression. After several iterations, you'll get an expression that can be easily integrated.

Keep in mind that the repeated integration by parts can become quite complex for high powers, and it may be more practical to use specialized software or tables for such cases.