Finding area between 2 curves with 3 intersections

[jgoff14]

Homework Statement

I need help getting the area between 2*x^3-x^2-5*x and -x^2+2*x. I found the intersections at +/- sqrt(7/2) and 0.

Homework Equations

f(x)=2*x^3-x^2-5*x
g(x)=-x^2+2*x

The Attempt at a Solution

This is the problem, I know that for something simple like x^2 and sqrt(x)+14 they intersect at x=4, so it is 2/3x^(3/2)-x^3/3 from 0 to 4 giving 40 square units, so generally speaking I remember how to do this. I know that it should be integral of top function - integral of bottom function, I also know that f(x)>g(x) on the initial interval (-sqrt(7/2),0) and then it switches to g(x)>f(x) from (0,sqrt(7/2)) so my work reflected that by splitting the integrals but I keep getting negative areas and just plain non-sense. I can't remember what I need to do exactly and in what order for something like this.

Thanks in advance for any help you can give.

 

[SammyS]

Homework Statement

I need help getting the area between 2*x^3-x^2-5*x and -x^2+2*x. I found the intersections at +/- sqrt(7/2) and 0.

Homework Equations

f(x)=2*x^3-x^2-5*x
g(x)=-x^2+2*x

The Attempt at a Solution

This is the problem, I know that for something simple like x^2 and sqrt(x)+14 they intersect at x=4, so it is 2/3x^(3/2)-x^3/3 from 0 to 4 giving 40 square units, so generally speaking I remember how to do this. I know that it should be integral of top function - integral of bottom function, I also know that f(x)>g(x) on the initial interval (-sqrt(7/2),0) and then it switches to g(x)>f(x) from (0,sqrt(7/2)) so my work reflected that by splitting the integrals but I keep getting negative areas and just plain non-sense. I can't remember what I need to do exactly and in what order for something like this.

Thanks in advance for any help you can give.

In the interval -√(7/2) < x < 0, f(x) > g(x) , so you should integrate f(x)-g(x) .

In the interval 0 < x < √(7/2), g(x) > f(x) , so you should integrate g(x)-f(x) .

 

[lanedance]

you've probably just got the functions mixed up, your method sounds reasonable, the actual integral is
$$Area = \int_{\sqrt{7/2}}^{\sqrt{7/2}} |f(x) - g(x)|dx$$

say you know f(x)>g(x) on (-\sqrt{7/2}} ,0) and g(x)>f(x) on (0,\sqrt{7/2}} ) then the intergal can be written
$$Area = \int_{\sqrt{7/2}}^{\sqrt{7/2}} |f(x) - g(x)|dx$$
$$= \int_{-\sqrt{7/2}}^{0}(f((x) - g(x))dx + \int_{0}^{\sqrt{7/2}} (g(x) - f(x))dx<br />$$

 

[jgoff14]

Thanks for the clarification. It turns out I did know what I was doing and I was doing the Calculus correct I just made a boo boo in my algebra, -x^2-(-x^2) is clearly not -2x^2 lol thanks so much though for taking the time to re-educate me :)!