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Integral calculation using areas

  • Thread starter mech-eng
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Problem Statement
The problem requires interpreting integrals as areas when calculating. ##\int_1^2 (1-2x)dx##
Relevant Equations
Are of a triangle is base X height.
Evaluate the integral using the properties of definite integral and interpreting integrals as areas.

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

I need to see there are two areas and these are the same but one is under x-axis the other is above x-axis so the value of the integral is zero. To see this is difficult to me.

Source: Calculus A Complete Course by Robert A. Adams

Thanks.
 

LCKurtz

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Evaluate the integral using the properties of definite integral and interpreting integrals as areas.

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

I need to see there are two areas and these are the same but one is under x-axis the other is above x-axis so the value of the integral is zero. To see this is difficult to me.

Source: Calculus A Complete Course by Robert A. Adams

Thanks.
In the right hand triangle the height is negative because ##y## is negative. That's why the integral for that part gives a negative area. If you want the geometric area use ##y_{\text{upper}}-y_{\text{lower}} = 0 -(1-2x)## in your integrand for the right hand integral.
 

fresh_42

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One area is oriented: first in direction +x then in direction -y (green), and the other area is oriented: first in direction +x then in direction +y (red), which results in a different sign, because the orientation has changed.
To calculate the area, the absolute values of both integrals have to be added (split att x=1/2), and to calculate the integral, the areas will cancel out to zero.
 

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