# Thread: area between 2 curves using |f(x)dx-g(x)dx|

1. ## area between 2 curves using |f(x)dx-g(x)dx|

problem:

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.

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Here is what the 2 lines look like:

In doing these types of area problems when part of the curve is negative, would i set up the integral to be from -1 to 0 of 6x*dx-(x^2-7)*dx and add it to the integral of 0 to 7 of 6x*dx-(x^2-7)*dx?

thanks.

2. First, if y= |6x| then your graph is wrong. You have y= 6x.

Because both |6x| and $x^2- 7$ are even functions, the graphs are symmetric about the y-axis and I would recommend just integrating on one side then doubling. For x> 0, |6x|= 6x so the two graphs will cross, on the right, when $x^2- 7= 6x$ which is the same as $x^2- 6x- 7= (x- 7)(x+ 1)= 0$. The only positive root of that is x= 7 so the area is $2\int_0^7 6x- (x^2- 7) dx$.

If it really is y= 6x rather than y= |6x|, just integrate $\int_{-1}^7 6x- (x^2- 7)dx$. You don't need to worry about changing signs because 6x is always larger than $x^2- 7$ on that interval so $6x- (x^2- 7)$ is always positive.

3. thank you, i spotted my error.