Calc 2 Integration Area Problem

  • #1
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Homework Statement


Please help me solve the calc problem pictured!

Homework Equations


y=3-x^2 and y=x+1

The Attempt at a Solution


My attempt is in one of the photos!
 

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Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Please help me solve the calc problem pictured!

Homework Equations


y=3-x^2 and y=x+1

The Attempt at a Solution


My attempt is in one of the photos!
Type the problem statement, and your solution. Your images are not readable on my devices, and so I am unable to help or give hints.

For more on this issue, see the post "Guidelines for students and helpers", by Vela.
 
  • #3
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The region R is bounded by y=3−x^2 and y=x+1.
The area of the region can be found by integrating: integral from 1 to 2 ______dy + integral from 2 to 3 ______dy
For the first blank I had (sqrt(3-y))-(y-1) and for the second I had (sqrt(3-y)-2)
These are both wrong according to the system
 
  • #4
Ray Vickson
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The region R is bounded by y=3−x^2 and y=x+1.
The area of the region can be found by integrating: integral from 1 to 2 ______dy + integral from 2 to 3 ______dy
For the first blank I had (sqrt(3-y))-(y-1) and for the second I had (sqrt(3-y)-2)
These are both wrong according to the system

So, the problem statement is asking you to find the area the hard way; integrating with respect to ##x## would be a lot easier.

Anyway, to see the ##x##-limits in the ##y##-integrals, you should start by drawing a picture of your region. The first (vertically lower) region goes from a negative value of ##y## to a positive value, and for each such ##y##, from a smaller (sometimes negative, sometimes positive) value of ##x## to a larger value of ##x##---giving a positive ##x##-length. You have your first integral going from a large value of ##x## to a smaller one---giving a negative ##x##-length.

You seem to be assuming that ##x## must be positive, but that is not stated anywhere in the problem as you wrote it.
 
  • #5
vela
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The region R is bounded by y=3−x^2 and y=x+1.
The area of the region can be found by integrating: integral from 1 to 2 ______dy + integral from 2 to 3 ______dy
For the first blank I had (sqrt(3-y))-(y-1) and for the second I had (sqrt(3-y)-2)
These are both wrong according to the system
Could you explain to us how you came up with your attempt?
 

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