Question

sketch the region enclosed by the given curves and find its area: y=abs(x), y=x^2-2

Answer 1

Hmm ok, let's start by finding out where the two functions intersect. This will tell where we should integrate from, and to.

Answer 2

\[\large |x|=x^2-2\]

This will have two solutions,\[\large -x=x^2-2\]and\[\large x=x^2-2\]

Ummmm So we have some quadratic business going on.
With the first equation, let's add \(\large x\) to both sides,\[\large x^2+x-2=0\]
Looks like this will factor nicely :) As will the other one.
Can you factor it?

Answer 3

The two functions intersect at (-2,2) and (2,2)

Answer 4

Ok cool c:
So to find the area, we should probably try to correctly figure out which one is the upper function and which is the lower.

We want to subtract the lower boundary function from the upper to find the area.

Answer 5

the upper function is the abs(x)

Answer 6

Ok good times.

So we'll be subtracting the quadratic from the abs function.

\[\large \int\limits_{-2}^{2}|x|-(x^2-2)\;dx\]

Answer 7

This is the part that might seem a little tricky.
We have to use our brains for this part! :O

Answer 8

Just to make sure we're on the same page, here's what our graph looks like.
https://www.desmos.com/calculator/e5oixet6sc

Answer 9

Yep that is the graph I have

Answer 10

Notice that the function |x| is composed of two straight lines?

We're going to split our integral into two parts and take advantage of this.

Answer 11

\[\large \int\limits\limits_{-2}^{2}|x|-(x^2-2)\;dx \qquad \\ \large = \qquad \int\limits_{-2}^0|x|-(x^2-2)\;dx+\int\limits_0^{2}|x|-(x^2-2)\;dx\]

Understand what I did so far? I chopped it up into two areas.

Answer 12

Yes that makes sense. So now that we have two integrals that are the same, can we put a 2 out front and make it \[2\int\limits\limits_{0}^{2}\left| x \right|-\left( x ^{2}-2 \right)dx\]

Answer 13

No no no. We certainly could, but that would negate what we're trying to accomplish here. :)

Answer 14

Oh, okay

Answer 15

Notice that the function y=|x| is simply y=x when we're on the positive side of 0 correct?

Answer 16

Yes, I see that.

Answer 17

\[\large \int\limits\limits_{-2}^0|x|-(x^2-2)\;dx+\int\limits\limits_0^{2}\color{orangered}{|x|}-(x^2-2)\;dx\]
\[\large =\int\limits\limits_{-2}^0|x|-(x^2-2)\;dx+\int\limits\limits_0^{2}\color{orangered}{x}-(x^2-2)\;dx\]

We can replace the |x| with x in our second integral (The integral which goes from 0 to positive 2).

Answer 18

We can do something similar on the left integral. What does the function y=|x| look like when we're on the negative side? :O

Answer 19

so are we replacing the |x| with -x in our first integral from -2 to 0.

Answer 20

Yes, very good.\[\large =\int\limits\limits\limits_{-2}^0\color{orangered}{-x}-(x^2-2)\;dx+\int\limits\limits\limits_0^{2}x-(x^2-2)\;dx\]

Answer 21

And it's FINALLY to a point where it is solvable! Yay!

Answer 22

So now, do I anti-differentiate both integrals?

Answer 23

Yes. Bunch of power rule it looks like.
Then just be sure to plug in the correct values. Lot of negative's floating around so it's easy to make a mistake.

Answer 24

My anti-differentiation looks like...\[(-1/2) x ^{2} - (1/3) x ^{3}-2x | \left(\begin{matrix}0 \\ -2\end{matrix}\right) + (1/2) x ^{2} - (1/3) x ^{3}-2x | \left(\begin{matrix}2 \\ 0\end{matrix}\right) \]

Answer 25

Woops, small mistake.
Looks like in both cases, you forgot to distribute the negative to the -2 terms.

Answer 26

So instead its +2x in both rather than -2x

Answer 27

Yah :)

Answer 28

So for my answer should I have \[\frac{ 20 }{ 3 }\]

Answer 29

mm I'm not sure, lemme do the work on paper real quick and compare c:

Answer 30

Yah that's what I'm coming up with also, so it's probably the correct answer. Yay team!

Answer 31

Thank you for the help!

Answer 32

no probs c: