What is the integral of 2xe^x?

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What is the integral of 2xe^x?

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Answer 1 · 2018-02-24T17:35:06

$\int 2 x e^{x} d x = 2 (x e^{x} - e^{x}) + C$ $int2xe^xdx=2(xe^x-e^x)+C$

Explanation:

To find:

$\int 2 x e^{x} d x$ $int2xe^xdx$

By constant rule

$\int 2 x e^{x} d x = 2 \int x e^{x} d x$ $int2xe^xdx=2intxe^xdx$

Let $I = \int x e^{x} d x$ $I= intxe^xdx$

$\int 2 x e^{x} d x = 2 I$ $int2xe^xdx=2I$

We can integrate using the parts rule

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

Here,

$u = x \to d u = d x$ $u=xtodu=dx$

$d v = e^{x} d x \to v = e^{x}$ $dv=e^xdxtov=e^x$

Substituting

$\int x e^{x} d x = x e^{x} - \int e^{x} d x$ $intxe^xdx=xe^x-inte^xdx$

$\int e^{x} d x = e^{x}$ $inte^xdx=e^x$

$\int x e^{x} d x = x e^{x} - e^{x}$ $intxe^xdx=xe^x-e^x$

$I = x e^{x} - e^{x}$ $I=xe^x-e^x$

$\int 2 x e^{x} d x = 2 (x e^{x} - e^{x}) + C$ $int2xe^xdx=2(xe^x-e^x)+C$

Answer 2 · 2018-02-24T18:44:44

2xe^(x)-2e^(x)+C

Explanation:

We have: $\int (2 x e^{x}) d x$ $int(2xe^(x))dx$

This integral can be evaluated using integration by parts.

Let $u = 2 x \Rightarrow \frac{d u}{d x} = 2$ $u=2x Rightarrow frac(du)(dx)=2$ and $\frac{d v}{d x} = e^{x} \Rightarrow v = e^{x}$ $frac(dv)(dx)=e^(x) Rightarrow v=e^(x)$ :

$\Rightarrow \int (2 x e^{x}) d x = 2 x e^{x} - \int (2 e^{x}) d x$ $Rightarrow int(2xe^(x))dx=2xe^(x)-int(2e^(x))dx$

$therefore int(2xe^(x))dx=2xe^(x)-2e^(x)+C$

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What is the integral of 2xe^x?

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2 Answers

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