• Want to help us with this year's BoS Trials?
    Let us know before 30 June. See this thread for details
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

Urgent Help! Question on Sums and Differences of Areas? (1 Viewer)

theodore0307

Active Member
Joined
Oct 31, 2013
Messages
221
Gender
Male
HSC
2014
Can someone show me how to do this with working out cause I'm getting the wrong answer.
3.8 qu5.png
 

dracster

New Member
Joined
Oct 19, 2013
Messages
5
Gender
Male
HSC
2014
Area enclosed by both curves
Find the Point of intersections of both graphs
y=x^2,y=x+6

x^2=x+6
x^2 -x -6 =0
(x + 2) , (x - 3)

x=-2 , x=3

3
∫ (x+6) - (x^2) dx Area under y=x+6 is larger than area under y=x^2
-2

3
∫ x+6 - x^2 dx
-2
3
= (x^2)/2 +6x - (x^3)/3 ]
-2

= (3^2)/2 +6(3) - (3^3)/3 - [ ((-2)^2)/2 +6(-2) - ((-2)^3)/3 ]

= 27/2 - ( - 22/3 )
= 125/6 u^2

Is that right? maybe i did something wrong :/
 

suika

Member
Joined
Mar 2, 2013
Messages
106
Location
NSW
Gender
Female
HSC
2014
Area enclosed by both curves
Find the Point of intersections of both graphs
y=x^2,y=x+6

x^2=x+6
x^2 -x -6 =0
(x + 2) , (x - 3)

x=-2 , x=3

3
∫ (x+6) - (x^2) dx Area under y=x+6 is larger than area under y=x^2
-2

3
∫ x+6 - x^2 dx
-2
3
= (x^2)/2 +6x - (x^3)/3 ]
-2

= (3^2)/2 +6(3) - (3^3)/3 - [ ((-2)^2)/2 +6(-2) - ((-2)^3)/3 ]

= 27/2 - ( - 22/3 )
= 125/6 u^2

Is that right? maybe i did something wrong :/
I got that answer as well. But I did 65/2 - 35/3 (different method from yours, but yours is good).

Seems like the right answer anyways.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top