Question

graph the curves y=IxI-1, y=1-x^2. Identify the points of intersection and hence the area enclosed by the 2 curves.

Answer 1

|x|-1=1-x^2 set equations = to find where they intersect
|x|-1-1+x^2=0
|x|-2+x^2=0
x^2-|x|-2=0
if x>0, |x|=x
if x<0, |x|=-x
so lets assume the fist x>0

x^2-x-2=0
(x-2)(x+1)=0
x=2,x=-1


now lets assume the second x<0

x^2-(-x)-2=0
x^2+x-2=0
(x+2)(x-1)=0
x=-2, x=1

Answer 2

so we have four points on intersection

Answer 3

maybe i made a mistake

Answer 4

yes

Answer 5

when you say consider x<0 , you should take the negative x soln

Answer 6

so x=2 and x=-2

Answer 7

are the ones u want

Answer 8

wait, they arent even correct :|

Answer 9

ok we can check all of our possible points of intersection

we have the equation |x|-1=1-x^2
lets check x=1
|1|-1=1-(1)^2
0=0
so x=1 works

Answer 10

x=-1

|-1|-1=1-(-1)^2
0=0
so -1 works

x=-2

|-2|-1=1-(-2)^2
1=-3

-2 doesn't work

|2|-1=1-2^2
1=-1

-2 doesn't work

Answer 11

ok so after you get your possible points of intersection plug them back into the both equations to see if you get the same thing if you don't then they don't intersect there

Answer 12

so they intersect at -1 and 1 according to what i did above

Answer 13

now to find the area you have to find out which of the functions are greater at the intersection points

Answer 14

now to find area
\[\int\limits_{-1}^{1}(topfunction-bottomfunction)dx=\int\limits_{-1}^{1}((1-x^2)-(|x|-1) )dx\]

Answer 15

to integrate
\[\int\limits_{-1}^{1}|x| dx=\int\limits_{-1}^{0}-x dx+\int\limits_{0}^{1}x dx\]

Answer 16

just in case you didn't know about that part

Answer 17

do you need more help?
do you understand?
are there any questions?
what are more ways to ask this question?
lol

Answer 18

yes,need help

Answer 19

A=1 if my math is correct

Answer 20

on what part

Answer 21

after integrating...

Answer 22

sub the values to get the area?

Answer 23

i need guidance plz

Answer 24

\[\int\limits_{-1}^{1}|x|dx=\frac{-x^2}{2} ,x=-1..0 +\frac{x^2}{2}, x=0..1=\frac{-(0)^2}{2}-\frac{-(-1)^2}{2}+\frac{1^2}{2}-\frac{0^2}{2}\]
\[=\frac{1}{2}+\frac{1}{2}\]
\[=1\]
okay now lets look at the whole thing
\[\int\limits_{-1}^{1}((1-x^2)-(|x|-1))dx=\int\limits_{-1}^{1}(2-x^2-|x|) dx=\int\limits_{-1}^{1}2dx-\int\limits_{-1}^{1}x^2dx-\int\limits_{-1}^{1}|x|dx\]

Answer 25

\[(2x, x=-1..1)-(\frac{x^3}{3}, x=-1..1)-1\]

Answer 26

\[[2(1)-2(-1)]-[\frac{1^3}{3}-\frac{(-1)^3}{3}]-1\]

Answer 27

\[2+2-\frac{1}{3}+\frac{-1}{3}-1=4-\frac{2}{3}-1=3-\frac{2}{3}=\frac{3*3}{3}-\frac{2}{3}=\frac{9-2}{3}=\frac{7}{3}\]
if i didn't make a mistake

Answer 28

dang i made a mistake lol

Answer 29

oh what did i let me look again

Answer 30

looks good :)

Answer 31

any questions? :)

Answer 32

oh that's the final answer

Answer 33

yes 7/3 is the final correct answer

Answer 34

brilliant,thanks a lot.God bless