integration area question (1 Viewer)

[velox]

its been ages since ive done this ive kinda forgotten how to do this question:

Find the area of the region enclosed by the area of y = 2x, the parabola y = x² and the line x = 2.

 

[haboozin]

its been ages since ive done this ive kinda forgotten how to do this question:

Find the area of the region enclosed by the area of y = 2x, the parabola y = x² and the line x = 2.

2
S 2x dx - x² dx
0
.....................2
[x^2 - X^3/3]
.....................0

4 - 8/3 = 1 1/3 units²

 

[velox]

how did we get the zero?

 

[haboozin]

well this question kind've gave you a clue... by saying between the like x=2

however if the question just asked you to find the area enclosed between any 2 lines... what you need to do is to find out where they meet and between those 2 points will be the enclosed space..

if you arn't very confident i suggest you draw a graph!


to find where they meet:

(1) y = x^2
(2) y= 2x


sub (1) in (2)

x^2 = 2x
x^2 - 2x = 0

x(x-2) = 0

x= 0 or x = 2

 

[velox]

thanks heaps. Now its coming back

EDIT : Yeh i should draw graphs it does help.

 

[velox]

next question I keep screwing them up =/
Find the area enclosed between y = 4x - 3 and y = X² + 1

 

[KFunk]

y = 4x - 3 is actually a tangent to the curve y=x² +1 at the point (2,5) so they don't really 'enclose' any area at all.

 

[Slidey]

y=4x-3
y=x^2+1

4x-3=x^2+1
0=x^2-4x+4
(x-2)^2=0 --> x=2

Now, I suppose you mean area enclosed between the curves AND the positive x-axis?

If so then we go from 0 to 2, but it's important to tell that sort of thing in the question.

Int[ (x-2)^2 ] from 0 to 2:
(x-2)^3/3 from 0 to 2
(0)-(-8/3)=8/3

Area is 8/3

 

[velox]

Yeah Slide ur right about the x axis thing. But not about the answer. Maybe i typed it out wrong, ill re do it.
Find the equation of the tangent to the parabola y=x²+1 at the point where x=2 and find the area enclosed by:
(i) the parabola, the tangent and the Y-axis.

 

[haboozin]

y=4x-3
y=x^2+1

4x-3=x^2+1
0=x^2-4x+4
(x-2)^2=0 --> x=2

Now, I suppose you mean area enclosed between the curves AND the positive x-axis?

If so then we go from 0 to 2, but it's important to tell that sort of thing in the question.

Int[ (x-2)^2 ] from 0 to 2:
(x-2)^3/3 from 0 to 2
(0)-(-8/3)=8/3

Area is 8/3

Can you do that?


wouldnt the area between x= 0 and x=2 and the 2 lines be:

2...........................2
S x^2 + 1 dx - S 4x -3 dx
0...........................3/4

then you work it out from there.

 

[KFunk]

Can you do that?


wouldnt the area between x= 0 and x=2 and the 2 lines be:

2...........................2
S x^2 + 1 dx - S 4x -3 dx
0...........................3/4

then you work it out from there.

Slide rule's method is correct, yours doesn't account for the area of the tangent below the X-axis. Note that the question says "area enclosed by (i) the parabola, the tangent and the Y-axis."

EDIT: Yours would be right if you were then to add the area between y = 4x-3 and the x-axis between 0 and 3/4. It's probably easier to deal with triangles if you want to get the area under a line or you can just combine it all to find the area between two curves as slide rule did which is much quicker.

 

[velox]

yeah thats the way i usually do them. Because it saves time as uve already combined them to find the point of intersection. Still - no-one has got the question right yet.....

 

[KFunk]

8/3 seemed right, what does the book say?

 

[shafqat]

yes i get 8/3 too

 

[haboozin]

EDIT: Yours would be right if you were then to add the area between y = 4x-3 and the x-axis between 0 and 3/4. It's probably easier to deal with triangles if you want to get the area under a line or you can just combine it all to find the area between two curves as slide rule did which is much quicker.

yea thats what i was doing

but ooo the method slide_rule used was the method i used accidently when my teacher was teaching us how to do an area between 2 curves and it was so much easier but for some reason my teacher told me not to use it...

 

[velox]

that is right. I read it wrong. thanks guys. Sometimes ur mind doesnt think straight =/

 

[KFunk]

Good0, another day, another mystery solved. I'm starting to sound like a wanker so I think I shall sleep. Adieu.

 

[velox]

What about this question? I cant find my error in my working.

Find the area encolsesd between the curve y = -2x²-5x+3 and the x axis.

AND

Find the area enclosedd beteen the curve y= 2/(x-3)² the x axis and the lines x = 0 and x = 1.

 

[KFunk]

What about this question? I cant find my error in my working.

Find the area encolsesd between the curve y = -2x²-5x+3 and the x axis.

just factorise it to find the x-intercepts so y=(1-2x)(x+3) i.e y=0 when x=-3, 1/2 so you just integrate y = -2x²-5x+3 between -3 and 1/2

{-3-->1/2} ∫-2x²-5x+3 dx
=[-2/3.x³ -5/2.x² +3x]{-3-->1/2)
= [19/24 - (-13 1/2)]
=14 7/24

I hope I worked that right but that's the gist of it.

 

[KFunk]

Find the area enclosedd beteen the curve y= 2/(x-3)² the x axis and the lines x = 0 and x = 1.

Just integrate the curve between 0 and 1? Which you can do by substitution or the automated: ∫(ax+b)ⁿdx = (ax+b)ⁿ⁺¹/a(n+1) +C

{0-->1} ∫2(x-3)^-2dx
= [-2/(x-3)] {0-->1}
= 1 - 2/3 = 1/3