An Area Problem

mathelord

I want to know how the area between two curves can be determined,do i just multiply the functions and then equate everything to 0,so i can get the limits,and the integrate the multiplied function within those limits

saltydog

Homework Helper
Mathelord, your description indicates some confusion. I plotted two functions:

$$y1(x)=x^2$$

$$y2(x)=-x^2+4x$$

To find the area between them, in this particular case, you would subtract them:

$$A=\int_0^2 [y2(x)-y1(x)] dx$$

$$=\int_0^2[(-x^2+4x)-x^2] dx$$

You can do the rest right?

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HallsofIvy

Homework Helper
mathelord said:
I want to know how the area between two curves can be determined,do i just multiply the functions and then equate everything to 0,so i can get the limits,and the integrate the multiplied function within those limits
NO, you don't "multiply the functions" OR "equate everything to 0"! I wonder where you would have gotten the idea that you should multiply the two functions. The limits of integration are the values of x where the area "ends"- where the two curves intersect. To find where the curves y= f(x) and y= g(x) intersect, solve y= f(x)= g(x).

Don't "integrate the multiplied function". Remember the "Riemann sums" that become the integral? Each term is the area of a skinny rectangle with width &Delta;x and height the difference between the two functions: f(x)- g(x). You integrate the difference between the two functions.

mathelord

do i just subtract one from the other,which is the exact on to be subtracted from

whozum

Subtract the lower function from the higher function.

In Saltydog's example the lower function is x^2 and the upper function is 4x-x^2.

mathelord

in cases like ax^2+bx+c,and -ax^2-bx-c.which is the lower function so i can get one integrated

whozum

Just graph them and check, or evaluate a test point, f(x) and g(x) to see which is lower.

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