Calculating the Area Bounded by Two Curves

[the white sou]

Homework Statement

find the Area Bounded by the two curves, y=|x+1|, y= - ( x+1)2 + 6

Homework Equations

y=|x+1|, y= - ( x+1)2 + 6

The Attempt at a Solution

A= Integration of | f (x) - g(x) |

x+1= f(x)
-(x+1)2 + 6= g(x)


getting the limit of integration:

x+1= - (x+1)2 + 6

x2 + 3x - 4=0

(X+4) ( x-1)

so x=-4, and x=1

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now when dividing the absolute valuey= x+1 ; x<-1
y=-(x+1) ; x>-1Integration is denoted by { ( upper limit , lower limit ) |f(x)|

so the Area= {(-1,-4) |(-x-1) + (x+1)2 -6 | - {(-1,1) |(x+1)+(x+1)2 - 6|what's wrong in this solution, I think that we should use -x-1 to get the limits as well so the area will be (the integration from -4 to -1) - (the integration from -1 to 3)

 

[tiny-tim]

welcome to pf!

hi the white sou!

find the Area Bounded by the two curves, y=|x+1|, y= - ( x+1)2 + 6
…getting the limit of integration:

x+1= - (x+1)2 + 6

x2 + 3x - 4=0

(X+4) ( x-1)

so x=-4, and x=1

No … you need to use |x+1|, not x+1.

 

[rootX]

Graph it first.

so the Area= {(-1,-4) |(-x-1) + (x+1)2 -6 | - {(-1,1) |(x+1)+(x+1)2 - 6|

Looks like you are using -4 as one of your limit however,
at -4
y=|x+1|, y= - ( x+1)2 + 6
y = 3, y = -3
They don't intersect at -4

(tiny-tim already pointed out)

 

[the white sou]

hi the white sou!


No … you need to use |x+1|, not x+1.

okay |x+1| consists of x+1, -x-1


once we use x+1=-(x+1)2 +6


x= 1, x= -4

and when we are using -x-1= -(x+1)2 + 6

we get x=-2, x=3

now we are having 4 points of intersection


soo how would we get the limits

 

[the white sou]

Graph it first.

so the Area= {(-1,-4) |(-x-1) + (x+1)2 -6 | - {(-1,1) |(x+1)+(x+1)2 - 6|

Looks like you are using -4 as one of your limit however,
at -4
y=|x+1|, y= - ( x+1)2 + 6
y = 3, y = -3
They don't intersect at -4

(tiny-tim already pointed out)

I already graphed it


for |x+1| there will be two lines intersecting at the point -1 and for the other curve we are having the the head of the curve at (-1,6) and it is opening downword

 

[rootX]

I already graphed it


for |x+1| there will be two lines intersecting at the point -1 and for the other curve we are having the the head of the curve at (-1,6) and it is opening downword

When you used x+1, you got x= 1, x= -4
When you used -(x+1), you got x=-2, x=3

Does x+1 is right or left line in the curve |x+1|. Label |x+1| with x+1 and -(x+1) and you should be able to identify which two of the above four solutions is valid

 

[tiny-tim]

An alternative method:

both curves are obviously symmetric about the line x + 1 = 0, ie x = -1,

so the solutions must also be symmetric about x = -1.

 

[the white sou]

When you used x+1, you got x= 1, x= -4
When you used -(x+1), you got x=-2, x=3

Does x+1 is right or left line in the curve |x+1|. Label |x+1| with x+1 and -(x+1) and you should be able to identify which two of the above four solutions is valid

so Area= {(-2,-1) |f(x)-g(x)| - {(-1,1) |f(x)-g(x)|


am i right now?

 

[rootX]

so Area= {(-2,-1) |f(x)-g(x)| - {(-1,1) |f(x)-g(x)|


am i right now?

1)

and when we are using -x-1= -(x+1)^2 + 6
we get x=-2, x=3

Should have been
x= 2, x=-3

 

[the white sou]

okay now I knew my mistakethank you bro :)