Question

Area between graphs. I thought this would be simple until I tried to calculate it. The area between f(x) = x^2 - 6 and g(x) = |x|

Answer 1

What have you tried ?

Answer 2

Did you solve your problem with the absolute values by observing symmetry about the y-axis?

Answer 3

Here is a graph of the two curves

Answer 4

I tried calculating each area (between graph and x-axis) and subtracting but I don't think that will work

Answer 5

Correction. it should be -|x|

Answer 6

if we look at the right half for x≥0
the area of a thin rectangle will have height g(x)- f(x) or
x-( x^2 - 6)
and width dx

do you know how to integrate from x=0 to 3 this expression ?

Answer 7

if g(x) = -|x|

then the graph is this

Answer 8

and the height of each rectangle is -x - (x^2 -6)

notice that the integration should go from 0 to x= 2

Answer 9

and then double the area of the right side for the total area.

Answer 10

I tried Wolfram Alpha but they don't show how to do it.
\[\int\limits_{-2}^{2} x ^{2}-6-|x| dx = \int\limits_{-2}^{0} x ^{2}-6-(-x) dx + \int\limits_{0}^{2} x ^{2}-6-(x) dx \]
Is that right? Then I work out one half and double?

Answer 11

yes, you have to split the integral because |x| is really two different functions depending on what x is.

So you can do what you posted, or integrate one or the other and double the result, because the two parts have identical area (by symmetry)

but I would make the integral for x=0 to 2:
-x - (x^2 -6) = -x^2 -x + 6

Answer 12

I get 22/3 which will give me 44/3 and that is one of my options. Multiple choice.

Answer 13

yes. The only detail is whether we should make the area negative. Area below the x-axis is often made negative... on the other hand, if all the choices area positive, then go with +44/3