How to calculate areas betweens functions?

[EudecioGabriel]

I'm using the book Calculus, 6th Edition (Stewart's Calculus Series), and I'm having some trouble to figure out

why he calculates the difference between two integrals to find the area between then. It's ok for me to understand when the y values of both functions are positive. But when one of the y values in one of the functions is negative he uses the same formula (integral of A) - (integral of B) {A and B are two functions}!

 

What about that rule that when the y value of the function is negative the integral is also negative?

[fiveworlds]

The example he gives says in the case where both are positive

 

Anyway in cases where the curve is below the x-axis you have to split the function into parts and use the absolute value for the area below the x-axis. Obviously if both are on the same side of the x-axis then you take one away from the other and if one is above the x-axis you add.

Edited April 15, 2017 by fiveworlds

[Country Boy]

If f(x)> g(x) for all x between a and b (the graph of y= f(x) is always higher than the graph of g(x)) then the area between the graphs is given by [math]\int_a^b f(x)- g(x) dx[/math]. It doesn't matter whether f or g are positive or negative.

[EudecioGabriel]

The example he gives says in the case where both are positive

 

Anyway in cases where the curve is below the x-axis you have to split the function into parts and use the absolute value for the area below the x-axis. Obviously if both are on the same side of the x-axis then you take one away from the other and if one is above the x-axis you add.

Ok. BUT ( integral of (x^2 + 5 ) - integral of (x^2 + 3 ) calculates the same way as (integral of (-x^2 +5) - integral of (-x^2+5) ?

[fiveworlds]

Ok. BUT ( integral of (x^2 + 5 ) - integral of (x^2 + 3 ) calculates the same way as (integral of (-x^2 +5) - integral of (-x^2+5) ?​

 

 

Here explains it a bit better than the book. I had to study it in college and it's a bit hard to follow. http://tutorial.math.lamar.edu/Classes/CalcI/AreaBetweenCurves.aspx

 

 

 

 

Integrals under the curve always give a negative value so you need to get the absolute value by adding the extra minus sign e.g. for the root 2x+6

[Country Boy]

Ok. BUT ( integral of (x^2 + 5 ) - integral of (x^2 + 3 ) calculates the same way as (integral of (-x^2 +5) - integral of (-x^2+5) ?

 

No, why should it? (x^2+ 5)- (x^2+ 3)= 2 while (-x^2+ 5)- (-x^2+ 5)= 0.

 

If you meant "(integral of (-x^2+ 5)- (integral of (-x^2+ 3))" that is the same as "integral of ((-x^2+ 5)- (-x^2+ 3))= integral of 2" the same as the first integral.