How do I find the area between a curve and the x-axis using integration?

[Peter G.]

Hi,

Curve: x(x+1)(x-2)

Find the area between the interval: (-1 to 2)

Answer:

The first thing I did was to find the x-intercepts: x=0, x=-1 and x=2

I then integrated the curve. I got: (x⁴/4)-(x³/3)-(x²)

I then found the area between 0 and -1 and the area between 0 and 2. I added the absolute of both areas and got 37/12. The book, however, claims it is 2.25.

Can anyone help me please?

Thanks!

 

[kjohnson]

I believe your problem is in the absolute value? The area that is below the x-axis should be counted as negative.

 

[Peter G.]

Oh, I thought we used the absolute value because we want the total area under the curve. The total area under the curve would be the area below the x-axis plus the are above the x axis, no?

 

[kjohnson]

Oh, I thought we used the absolute value because we want the total area under the curve. The total area under the curve would be the area below the x-axis plus the are above the x axis, no?

That is a good question...from a purely geometric standpoint then yes. But mathematically the area below the x-axis is taken to be negative because the differential area is defined as:

dA=f(x)dx

and when f(x) is negative then that value of dA will also be negative since dx is positive. This will become second nature to you as you work more problems and see real world integration applications in other courses.

 

[kjohnson]

Also another tip...you don't need to worry about splitting the interval up according to the roots, the sign of f(x) will take care of everything. There is nothing wrong with splitting the integral up, but it is not necessary for most simple continuous functions.

 

[Peter G.]

Oh, I see. Thanks a lot for the help! I figured now that when the book wants me to perform what I described in the first post they ask for the total area. This question, however, simply asks me to evaluate the integral.

Thanks once again,
PeterG.