A few integration problems.. (3 Viewers)

[Coookies]

This is from Maths in focus

Areas enclosed by x-axis
Q 19. Find the area bounded by the curve y=sqrt(4-x^2), the x-axis and the y-axis in the first quadrant

Areas enclosed by the y-axis
Q5. Find the area bounded by the line y=2x+1, the y-axis and the lines y=3 and y=4
Q9. Find the area enclosed by the curve y=sqrt(3x-5), the y-axis and the lines y=2 and y=3
Q10. Find the area enclosed between the curve y=1/x^2, the y-axis and the lines y=1 and y=4 in the first quadrant

Thanks in advance!

 

[Coookies]

Areas enclosed by x-axis
Q 19. Find the area bounded by the curve y=sqrt(4-x^2), the x-axis and the y-axis in the first quadrant

this curve is a semi circle that cuts y and x at 2
To find the volume at x make y the subject and square it so u get y^2 = 4 - x^2
from x= 0 and x = 2
volume at x axis is v = pi*integral of y^2 dx
integrate 4 - x^2 you get 4x - x^3/3 sub in 2 and 0 u get
pi*[ 8 - 8/3 ] = 16pi/3 cubic units

volume at y axis
y = 0 and y = 2
v = pi*integral x^2 dy
x^2 = 4 - y^2
integrate 4 - y^2 you get 4y - y^3/3
sub in your limits you get 8 - 8/3 = 16pi/3 cubits units

Im doing area so would it still be the same?
Answer is pi = 3.14 square units

 

[Aysce]

Im doing area so would it still be the same?
Answer is pi = 3.14 square units

Easier way is to think logically.

The equation is of a semi circle where it it has the x-intercepts of -2 and 2, and the y-intercept of 2. So now we have have a semi-circle! Since you now have the limits of 0 and 2 within the first quadrant, you can integrate. Alternatively you can find the area of a whole circle using the radius of 2 and divide it by 1/4 since it is only within the first quadrant. Hence giving you pi as your answer.

 

[Timske]

soz i accidentally did volume for the first post ignore that

 

[Timske]

Q5. Find the area bounded by the line y=2x+1, the y-axis and the lines y=3 and y=4
make x the subject
x = ( y - 1 )/2
integrate that you get (y^2/4 - y/2)
sub in limits 3 and 4
(2) - (3/4) = 5/4 units squared
Q9. Find the area enclosed by the curve y=sqrt(3x-5), the y-axis and the lines y=2 and y=3
make x the subject, x = (y^2 + 5) / 3
Integrate that you get x^3/9 +5x/3
sub in limits 2 and 3
(8) - (38/9) = 34/9 units squared
Q10. Find the area enclosed between the curve y=1/x^2, the y-axis and the lines y=1 and y=4 in the first quadrant
x = 1/sqrt(y)
integrate that you get 2sqrt(y)
sub in 1 and 4
2(2) - (2) = 2
Area = 2 units

 

[Coookies]

For Q10, could you give me the steps to changing the subject?
Thanks A LOT for this btw!

 

[Coookies]

Could someone give me the proper steps for Q19? Im still not sure (mostly on changing the subject)

Also I have a few more questions (I just need help factorising)
Q6. y= -2x^2 - 5x +3
Q8. x = -y^2 -5y - 6

They're from different exercises in case Im confusing you lol

and Q15. Find the area bounded by the curve y=x^4+1, the y-axis and the lines y=1 and y=3 in the first quadrant

Thanks everyone!

 

[Timske]

Q.10
y = 1/x^2 - Time both sides by x^2
x^2*y=1 - Divide both sides by y
x^2 = 1/y - square root both sides
x = sqrt(1)/sqrt(y)
therefore x = 1/sqrt(y)



Also I have a few more questions (I just need help factorising)
Q6. y= -2x^2 - 5x +3 - factor negative 1 out
y = -(2x^2 + 5x - 3) - factor normally
= -(x+3)(2x-1)

Q8. x = -y^2 -5y - 6 : same thing factor -1 out
x = -(y^2 + 5y + 6)
x = -(y+2)(y+3)

Q15. Find the area bounded by the curve y=x^4+1, the y-axis and the lines y=1 and y=3 in the first quadrant
limits are 1 and 3
make x the subject
x^4=y-1
x=(y-1)^(1/4)

integrate x=(y-1)^(1/4)

= [(y-1)^5/4]/5/4

= 4/5*(y-1)^5/4
sub in 3 and 1
4/5*(2)^5/4 - 4/5*0
Area = 4/5*(2)^5/4
= 1.903 ( 3.d.p) units squared.

sorry if its messy, is that what the answer says?

Q19. y = sqrt(4-x^2) - square both sides
y^2 = 4 - x^2 - move x^2 and y^2 to other side
x^2 = 4 - y^2 - root both sides
x = sqrt(4-y^2)

 

[Coookies]

You're amazing!

But for Q19, I got 16/3(5.333333) but its meant to be 3.14

 

[Carrotsticks]

Q 19. Find the area bounded by the curve y=sqrt(4-x^2), the x-axis and the y-axis in the first quadrant

 

[RealiseNothing]

For this question, I'll show you a quick shortcut that requires no integration:

Q5. Find the area bounded by the line y=2x+1, the y-axis and the lines y=3 and y=4

The area enclosed it clearly a trapezium, so we can use the formula (h/2)(a+b).

The height is 1 as you just minus the y co-ordinates (4-3).

a and b are just the x co-ordinates when you put y as 3 and 4, so:

x = (y-1)/2

x = 1 or 3/2

Hence (1/2)(1+3/2)

= 5/4

 

[zeebobDD]

For q10, you could change the 1/x^2 to x^-2, and then work from there

 

[Coookies]

Thanks for the help everyone!

Continuing on.... Im now onto sums and differences of areas, or areas between 2 curves.

I can't seem to get this one...
Q6. Find area bounded by curve y=x^3 and y=4x
Answer is 8 but I keep getting -4

 

[nightweaver066]

Thanks for the help everyone!

Continuing on.... Im now onto sums and differences of areas, or areas between 2 curves.

I can't seem to get this one...
Q6. Find area bounded by curve y=x^3 and y=4x
Answer is 8 but I keep getting -4

Your answer tells me you attempted to find the area bounded between the two curves between x = -2 and x = 0.

Since this area is below the x-axis, you must apply the absolute value sign as you cannot get negative areas.

Now you need to consider the other region bounded by both curves, between x = 0 and 2.

Add the two areas together and you get your answer.

Find intersection points.

x^3 = 4x
x^3 - 4x = 0
x(x + 2)(x - 2) = 0

As it isn't the origin, x = 2, x = -2

 

[Timske]

anymore problems im keen to help

 

[Coookies]

Thanks! But what I don't get is that why you're not adding the areas from -2 to 0 and 2 to 0

@Timske I have plenty more questions I need help with! lol

 

[Coookies]

Q8. Find area enclosed between curve y=x^2 and line y=-6x+16
A: 166 2/3

Im going to stop typing the whole area enclosed thing, you know what I mean...

Q10. y=(x-2)^2, y=(x-4)^2 and x-axis
A: 2/3

I get close to the answer, but close won't get me a mark. This is so frustrating.

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?

Im asking so many questions lol.
How do you simultaneously solve y=x^2 and x=y^2
I have changed the x=y^2 to y=x^1/2. But Im not sure how to take out the common factor

 

[nightweaver066]

Thanks! But what I don't get is that why you're not adding the areas from -2 to 0 and 2 to 0

@Timske I have plenty more questions I need help with! lol

The area from -2 to 0 is exactly the same as the area from 0 to 2.

It is an odd function with point symmetry about (0, 0) so the two sides of the graphs (x > 0 and x < 0) are essentially the same, just that one is flipped over.

 

[Timske]

Thanks! But what I don't get is that why you're not adding the areas from -2 to 0 and 2 to 0

@Timske I have plenty more questions I need help with! lol

he did add the areas but he did it an easier way, he did this by multiplying 2 because the area of -2 to 0 and 0 to 2 is the same right so all he did was find the area from 0 to 2 and times 2.

 

[Coookies]

A few more I can't seem to do -_-

13. y=x^2+2x-8 and y=2x+1
A:36 units squared

14. y=1-x^2 and y=x^2-1
A:2 2/3 units squared

15. Find exact area: y=sqrt(4-x^2) and x-y+2=0
Api - 2) units squared

And thats the end of this exercise! yay!