What is $f (x) = \int x e^{x + 2} - x d x$ $f(x) = int xe^(x+2)-x dx$ if $f (- 1) = 2$ $f(-1) = 2$ ?

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What is $f (x) = \int x e^{x + 2} - x d x$ $f(x) = int xe^(x+2)-x dx$ if $f (- 1) = 2$ $f(-1) = 2$ ?

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Answer 1 · 2016-08-22T16:31:53

$f (x) = x e^{x + 2} - e^{x + 2} - \frac{x^{2}}{2} + \frac{5}{2} + 2 e$ $f(x)=xe^(x+2)-e^(x+2)-x^2/2+5/2+2e$ .

Explanation:

$f (x) = \int (x e^{x + 2} - x) d x = \int x e^{x + 2} d x - \int x d x = I - \frac{x^{2}}{2}$ $f(x)=int(xe^(x+2)-x)dx=intxe^(x+2)dx-intxdx=I-x^2/2$ , where,

$I = \int x e^{x + 2} d x$ $I=intxe^(x+2)dx$ .

To evaluate $I$ $I$ , we use the Rule of Integration by Parts :

$\int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$ $intuvdx=uintvdx-int[(du)/dxintvdx]dx$ .

We take, $u = x \Rightarrow \frac{d u}{d x} = 1$ $u=x rArr (du)/dx=1$ , and,

$v = e^{x + 2} \Rightarrow \int v d x = e^{x + 2}$ $v=e^(x+2) rArr intvdx=e^(x+2)$ . Hence,

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

$:. f(x)=xe^(x+2)-e^(x+2)-x^2/2+C...........(1)$ .

To determine $C$ , we use the cond. $: f(-1)=2$ in $(1)$ .

$:. -e-e-1/2+C=2 rArr C=5/2+2e$ . Therefore, $(1)$ gives,

$f(x)=xe^(x+2)-e^(x+2)-x^2/2+5/2+2e$ .

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What is $f (x) = \int x e^{x + 2} - x d x$ $f(x) = int xe^(x+2)-x dx$ if $f (- 1) = 2$ $f(-1) = 2$ ?

1 Answer

Explanation:

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What is $f (x) = \int x e^{x + 2} - x d x$ $f(x) = int xe^(x+2)-x dx$ if $f (- 1) = 2$ $f(-1) = 2$ ?