Definite integrals of absolute values

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Homework Help Overview

The discussion revolves around evaluating the definite integral of the absolute value of the function \( f(x) = x^2 + x - 2 \) over the interval from -2 to 2. Participants are exploring how to handle the absolute value in the context of definite integrals.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to determine where the function \( x^2 + x - 2 \) is positive or negative to correctly apply the definition of absolute value. There is mention of splitting the integral based on these intervals and considering the symmetry of the function.

Discussion Status

There is an ongoing exploration of how to set up the integral correctly, with some participants questioning the implications of the absolute value on the integral's value. Guidance has been offered regarding the need to split the integral based on the sign of the function in different intervals.

Contextual Notes

Participants note specific points where the function changes sign, such as at \( x = -2 \) and \( x = 1 \), which are critical for determining the setup of the integral. There is also a recognition of the function's symmetry, which may influence the evaluation of the integral.

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Homework Statement


\int|x^{2}+x-2|dx from -2 to 2


Homework Equations


The integral of f(x) from a to b = F(b) - F(a)
|x| = { x if x >= 0; -x if x < 0

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Ok, I don't know how to do the definite integrals of absolute values.. was never shown an example of it in class, but I kind of have an idea...

we know that
|x| = { x if x >= 0; -x if x < 0,

and then I am lost from here...

btw, x= 1, -2
 
Last edited:
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If you want to apply your definition of abs, you'll first have to figure out where x^2+x-2 is positive and negative.
 
quick sketches of the function will help you...
 
ok so from -2 to 1, it is negative and from 1 to 2 it is positive...

do i just split up the integral from here?
like from -2 to 1, its negative, so the integral would be negative, and from 1 to 2, it is positive so the integral would be positive...

and just from inspection, would the integral be equal to 0 because its symmetric and one is a negative the other is positive...
 
The integral of x^2+x-2 from -2 to 1 IS negative. The integral of |x^2+x-2| is not. Do you see why?
 
nvm. i got 19/3...
 
Last edited:
Dick said:
The integral of x^2+x-2 from -2 to 1 IS negative. The integral of |x^2+x-2| is not. Do you see why?

I don't know if I do..

is it because the absolute value of any function is always positive?
 
Well from 2 to one it is +ve so you want to get
\int_1 ^{2} (x^2+x-2)dx

and from 1 to -2 it is -ve so you want to find

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

and sum them up
 

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