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Hi I am unsure of my attempt to solve the following:

$$ \int_{-1}^1 xde^{|x|} $$

my attempt is the following:

Using substitution where $ u = x, v = ?, dv = e^{|x|} $ Now :

$$ v = \int_{-1}^1 dv = \int_{-1}^1 e^{|x|}dx$$

$$= \int_{-1}^0 e^{-x}dx + \int_0^1 e^x dx $$

$$= -e^{-x}|_{-1}^0 + e^x|_0^1$$

$$= -1 + e + e -1 $$

$$= 2e -2 $$

$$ \int_{-1}^1 udv = uv|_{-1}^1 - \int_{-1}^1 vdv $$

$$= x(2e-2)|_{-1}^1 - \int_{-1}^1 (2e - 2)dx$$

$$=x(2e - 2)|_{-1}^1 - (2e - 2)x|_{-1}^1 $$

$$=4e -4 -4e - 4 $$

$$=0$$

Did
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user146387
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    Title: "Stohcastic Finance Stochastic integrals" ? Tag: "(stochastic-calculus)" ? – Did May 11 '14 at 19:02
  • @Jean-ClaudeArbaut Sorry but these changes do not help. At all. – Did Jun 23 '14 at 07:55
  • @Jean-ClaudeArbaut Sorry but why did you think there was some hint of a stochastic integral in there? – Did Jun 23 '14 at 11:21
  • @Jean-ClaudeArbaut I know all this. I also know that you replaced an irrelevant "stochastic" tag by another, equally irrelevant, "stochastic" tag. Sure, this is not a crime, nevertheless I see no motive to do so. – Did Jun 23 '14 at 15:03
  • @Jean-ClaudeArbaut You may calm down, I just did (edit the post). – Did Jun 23 '14 at 15:30

2 Answers2

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Let $I$ be the given integral. Then:

$I = \int_{-1}^0 x\cdot d(e^{-x}) + \int_{0}^1 x\cdot d(e^x) = (x\cdot e^{-x})|_{-1}^0 - \int_{-1}^0 e^{-x}dx + (x\cdot e^x)|_{0}^1 - \int_{0}^1 e^x dx = e - (-e^{-x})|_{-1}^0 + e - (e^x)|_{0}^1 = 2$

DeepSea
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Why not explicit out the differential, namely:

$$I=\int xd(e^{|x|}),$$

but

$$ d(f(x))=f'(x)dx \Rightarrow d(e^{|x|})=e^{|x|}\frac{d(|x|)}{dx}dx=e^{|x|}\frac{|x|}{x}dx, $$

hence:

$$ I=\int |x|e^{|x|}dx, $$

split it as you did, and the result should be straightforward.

7raiden7
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