Solving Integral: $\int_{-\infty}^{\infty} e^{-|x|} dx$

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Homework Help Overview

The discussion revolves around the integral $\int_{-\infty}^{\infty} e^{-|x|} dx$, specifically focusing on the interpretation of the absolute value function and its implications for splitting the integral into two parts. The subject area includes calculus and integral evaluation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why the integral can be split into two parts, questioning the sign of the expression in the first integral. Some participants provide definitions of absolute value to clarify the reasoning behind the split.

Discussion Status

Participants are exploring the definition of absolute value and its application in the context of the integral. There is a mix of confusion and clarification regarding the properties of absolute values, with some guidance offered on how to correctly interpret the function in different intervals.

Contextual Notes

There appears to be some confusion regarding the definition of absolute value, particularly in relation to negative inputs. This has led to differing interpretations of how to express the integral based on the definition provided.

G01
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[tex]\int_{-\infty}^{\infty} e^{-|x|} dx[/tex]

Could someone tell me why this integral, when you split it comes out to be:

[tex]\int_{-\infty}^0 e^x dx + \int_0^{\infty} e^{-x} dx[/tex]

I keep thinking it should be e^(-x) in the first integral. I don't know why its positive. I can solve this integral otherwise. Thanks again.
 
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The absolute value is defined for a real number [itex]x[/itex] by
[tex] |x|= \left \lbrace<br /> \begin{array}{l l}<br /> x & \mbox{if} \ x \geq 0 \\<br /> -x & \mbox{if} \ x\leq 0<br /> \end{array}<br /> \right.[/tex]

when [itex]x[/itex] is in the interval [itex](-\infty,0)[/itex], then [itex]|x| = -x[/itex]. Likewise, when [itex]x[/itex] is in the interval [itex](0,\infty)[/itex], then [itex]|x| = x[/itex]. Therefore,

[tex]\int_{-\infty}^0 e^{-|x|}dx = \int_{-\infty}^0 e^{-(-x)}dx<br /> = \int_{-\infty}^0 e^x dx[/tex]
 
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Im sorry that definition confuses me... I always thought the absolute value was the positive distance of that number from the origin, leading me to believe that |-x|= x. This seems contrary.
 
G01 said:
Im sorry that definition confuses me... I always thought the absolute value was the positive distance of that number from the origin
This is true.
leading me to believe that |-x|= x. This seems contrary.
This is false. For example take [itex]x=-3[/itex]. Then your saying that [itex]|-(-3)| = -3[/itex], which is the same as saying that [itex]|3| = -3[/itex], which is clearly false. I used paranthenses to delimit the value which was substituted for x, namely -3.

However, if we use the definition of absolute value, we arive at the correct result. For example, if [itex]x=-3[/itex], we have that [itex]x<0[/itex] so we use the second case of the definition to arrive that [itex]|x| = -x[/itex], which gives [itex]|-3| = -(-3) = 3[/itex].

Try plugging in different values for x into the definition to convince yourself that this definition works.

The definition for when [itex]x>0[/itex] should be clear. The absolute value of a positive number is a positive quantity and is equal to that number. However if [itex]x<0[/itex], then [itex]-x>0[/itex], so [itex]-x[/itex] is a positive number (don't let the negative sign confuse you, remember that the negative of a negative number is positive).
 
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