How to Evaluate the Absolute Value Integration?

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SUMMARY

The discussion focuses on evaluating the integral of the absolute value function, specifically the integral \int_{-1}^{3} \left|x^2 -4\right| dx. The user applies the theorem for integrable functions over a closed interval, selecting c = 2 to split the integral into two parts. The correct formulation is \int_{-1}^{2} (4-x^2)dx + \int_{2}^{3}(x^2 -4)dx, as this ensures the integrand remains non-negative across the specified intervals.

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  • Understanding of definite integrals and the Fundamental Theorem of Calculus
  • Familiarity with absolute value functions in calculus
  • Knowledge of piecewise functions and their integration
  • Ability to identify points of discontinuity in functions
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  • Learn about piecewise integration techniques
  • Explore the Fundamental Theorem of Calculus in depth
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Students and educators in calculus, mathematicians focusing on integral calculus, and anyone seeking to enhance their understanding of absolute value integration techniques.

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1. Evaluate

\int_{-1}^{3} \left|x^2 -4\right| dx




3. The Attempt at a Solution

This is the first time I'm trying this type of question & I think I need to use the following theorem for such questions;

f is integrable on a closed interval a to b.

\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx

(c; any point in between, no matter how the points are ordered)

Therefore in my problem \int_{-1}^{3} \left|x^2 -4\right| dx, I choose c = 2

Now since the integrand is absolute value we have two cases:

1) x2-4
2) 4-x2

I'm not which one of the following is correct;

\int_{-1}^{2} (4-x^2)dx + \int_{2}^{3}(x^2 -4)dx

OR

\int_{-1}^{2} (x^2-4)dx + \int_{2}^{3}(4-x^2)dx

Which one is correct and why?

Thanks!

 
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|x^2-4| is nonnegative everywhere. So the pair of integrals which have a nonnegative integrand over the region of integration is the correct one. Which is it? The first or the second?
 
The first one is non-negative!
I think I understand the idea now, thank you very much!
 

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