Integration Problem involving absolute bracket, sine,cosine

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SUMMARY

The integral of the absolute value of sine, ∫(1)to(-1) |sin(x)| dx, requires splitting the integral into regions where sin(x) maintains a consistent sign. The initial attempt incorrectly evaluates the integral without considering the sign changes of sin(x) over the interval [-1, 1]. The correct approach involves calculating the integral separately for the intervals where sin(x) is positive and negative, leading to a different final result.

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Homework Statement


Evaluate

∫(1)to(-1) l sin(x) l dx


Homework Equations





The Attempt at a Solution



If I just leave it like this, is it wrong?

1. ∫(1)to(-1) l sin(x) l dx = l -cos (x)l

2. l -cos (1) l — l -cos (-1)l = 0 - 0 = 0
 
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Way wrong. Split the integral up into regions where sin(x) has a definite sign. Work the integral out on each of those regions. Then add them up. You'll get a very different answer.
 
Last edited:

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