A few integration problems.. (1 Viewer)

[Coookies]

Oh thanks! My teacher skipped the odd and even stuff. Maybe I should read over it lol. Thanks guys!

 

[Timske]

Also, you know how its big area - small area, if I accidentally do the other way around, can i just absolute value the answer and just say its positive? Or do I have to absolute the whole working out, or start again? you absolute the area that is negative in your working out because remember area cannot be negative. You have to absolute the area that is negative

 

[Coookies]

Also, you know how its big area - small area, if I accidentally do the other way around, can i just absolute value the answer and just say its positive? Or do I have to absolute the whole working out, or start again? you absolute the area that is negative in your working out because remember area cannot be negative. You have to absolute the area that is negative

Yeah but from which point do I have to add the absolute sign?
Just the answer or the whole way through my working?

 

[nightweaver066]

Yeah but from which point do I have to add the absolute sign?
Just the answer or the whole way through my working?

The moment you write the expression for the area.

E.g.

 

[Coookies]

The moment you write the expression for the area.

E.g.

Oh thanks! I didn't know you could do it like that lol

 

[Timske]

Q8. y = x^2 , y = -6x + 16
Find point of intersection to find the area bounded by the curves.

x^2 = -6x + 16
x^2 + 6x - 16 = 0
(x+8)(x-2) = 0 , x = -8 and x = 2

top curve - bottom curve
-6x + 16 - x^2
integrate that
- 3x^2 + 16x - x^3/3
sub in 2 and -8
[ -12 + 32 - 8/3] - [ -192 - 128 + 512/3]
= 52/3 + 448/3 = 166 2/3

 

[Coookies]

Thanks!
How come you guys are so smart? lol

 

[nightweaver066]

Thanks!
How come you guys are so smart? lol

When doing these questions, first thing you must do is draw a graph.

Then identify what area you are trying to find by shading in the regions on the graph.

Think about what approach you'll take.

How will i find the area of this? Will i need to add/subtract the areas? Is it below the x-axis (resulting a negative area)?

 

[Timske]

x^2 = y and x = y^2

x = sqrt(y) and x = y^2
sqrt(y)=y^2 - get rid of sqrt
y = y^4 - divide by y
1 = y^3
therefore y = 1
sub into y = x^2
1 = x^2
therefore x = 1

not sure about this one

 

[Timske]

nightwaver can u try doing 10 i keep getting 2.25...

 

[nightweaver066]

Simultaneously solving y = x^2 and x = y^2,

 

[Timske]

13. y=x^2+2x-8 and y=2x+1
A:36 units squared
Remember step 1 find points of intersection

2x + 1 - x^2 - 2x + 8 = 0
x^2 - 9 = 0
x^2 = 9
therefore x = plus/minus 3

top curve - bottom
2x + 1 - x^2 - 2x + 8 = 1 - x^2 + 8
now integrate that
x - x^3/3 + 8x
sub in +-3
[3 - 9 + 24] - [ -3 + 9 - 24 ]
= 18 + 18 = 32 units squared

 

[Drongoski]

nightwaver can u try doing 10 i keep getting 2.25...

Q10

the 2 parabolae intersect at x=3. Therefore the required area:

 

[Coookies]

Simultaneously solving y = x^2 and x = y^2,

So smart... haha
So how would I go about integrating it?
I did from 0 to 1 x^2 + x^2 which gives me 2/3 which is wrong lol.

 

[Timske]

i see where i went wrong lol

 

[Coookies]

Q10

the 2 parabolae intersect at x=3. Therefore the required area:

I expanded the brackets.... grrrr!

 

[Timske]

14. y = 1 - x^2 , y = x^2 - 1
1 - x^2 - x^2 + 1 = 0
2 - 2x^2 = 0
-2x^2 = -2
x^2 = 1
therefore x = +-1
if u actually drew the curves you see they are parabolas intersecting at - 1 to 1 on the x axis and -1 to 1 on the y axis
integrate one of the curve, i chose y = 1 -x^2
x - x^3/x sub in 0 and 1
1 - 1/3 = 2/3
i found 1/4 of the area bounded so now i times 4 for the whole area
2/3 * 4 = 2 2/3 units squared

hope this make sense

 

[Coookies]

14. y = 1 - x^2 , y = x^2 - 1
1 - x^2 - x^2 + 1 = 0
2 - 2x^2 = 0
-2x^2 = -2
x^2 = 1
therefore x = +-1
if u actually drew the curves you see they are parabolas intersecting at - 1 to 1 on the x axis and -1 to 1 on the y axis
integrate one of the curve, i chose y = 1 -x^2
x - x^3/x sub in 0 and 1
1 - 1/3 = 2/3
i found 1/4 of the area bounded so now i times 4 for the whole area
2/3 * 4 = 2 2/3 units squared

hope this make sense

Can you legit do that? The times by 4 bit? And how come you 'chose' to do 1 curve? Im really confused, I thought you had to use both

 

[SpiralFlex]

Will someone tell me what they need help in? The front page is too far for my fat hands.

 

[Coookies]

Simultaneously solving y = x^2 and x = y^2,

I need help finishing this question!

Also, a few volume Qs...
Q8. The parabola y=(x+2)^2 is rotated about the x-axis from x=0 to x=2. Find the volume of solid formed
Q19. Find the volume of solid formed when line x+3y-1=0 is rotated about x-axis from x=0 to x=8
Q20. I need help changing the subject. y=x^3, rotated about the y-axis. (So I need to get x^2)