Mathematics: Post your doubts here!

[abcde]

^No! The only thing given in the question was the equation of the curve i.e. y = 1 / (2x + 2) ^2 !
If you have that book, Pure Mathematics 1, you can see the question for yourself! Exercise 16 d, Question 5, part c !

The line x=1 is clearly mentioned there.
Find the area under the curve using limits x=0 and x=1. Subtract this from the area of the entire rectangle.
Try solving it now. You can let us know if you still find it troublesome.

 

[ffaadyy]

1. Find the area of shaded region View attachment 3503

I think you didn't clearly look at the diagram which is printed in the book. The point at which the curve and the vertical line intersects is x=1. Therefore, we'll first integrate y = 1 / (2x + 1)^2 using '1' as the upper limit and '0' as the lower limit. After integrating, we'll obtain the answer '1/3' which is the area under the curve. Now to find the area of the shaded region, we need to subtract the area under the curve from the area of the rectangle. To find the area of the rectangle, we'll use the formula Length x Height. We know that the length is '1' but we don't know it's height. The height can be found out by putting x=0 in the curve equation and obtaining y=1. The height is '1' and the overall area of the rectangle is '1 x 1=1'. Now to find the area of the shaded region, we'll simply subtract 'Area under the Curve' from the 'Area of the Rectangle. Area of the shaded region comes out to be '1 - (1/3) = 2/3'

 

[VelaneDeBeaute]

^I guess the line ain't visible because of the pirated version! I'll solve it though! Thankyou!
What about the other questions though?

 

[ffaadyy]

^I guess the line ain't visible because of the pirated version! I'll solve it though! Thankyou!
What about the other questions though?

Okay so this is how you'll do the first question you've mentioned. You need to find the upper and the lower limit first because without it you wont be able to calculate the area enclosed between the curves. You'll find the first limit by keeping both the curves equal to each other (in other words, the point of intersection). (x-2)^4=(x-2)^3. From this, you'll get x=3. This is our first limit. Now to find the second limit, we'll equate any of the two curve equations to 0. For example, (x-2)^4=0 or (x-2)^3=0. From this, we'll get x=2. Now we have two limits, '3' being the upper limit and '2' being the lower limit. Having obtained both the limits, we'll first integrate (x-2)^4 using both these limits. The area obtained would be '1/5'. We'll do the same with the second curve, (x-2)^3. We'll integrate it using the limits '3' and '2' and obtain area '1/4'. Now that we have both the areas, we'll simply subtract the smaller area from the bigger area. This'll be (1/4) - (1/5) and we'll get the answer (1/20)

 

[abcde]

AoA!
Had some problems with integration!

1. Find the area of the region enclosed by y = ( x - 2 ) ^ 4 and y = ( x - 2 ) ^ 3
2. Draw the graph of the function f ( x ) when the graph of the inverse function f ' ( x ) is given ! http://www.mediafire.com/imageview.php?quickkey=sllcr94vcjmb0bz

2. It will be the reflection of the given graph in the line y=x.
The inverse function is like the parabola y = x^2.
f^-1 (x) = x^2
=> f(x^2) = x
=> f = y^(1/2)
So your sketch would be of the curve y = x^(1/2). It looks like this: http://www.wolframalpha.com/input/?i=sketch x^1/2&t=crmtb01
Btw, f '(x) is not the inverse function. It is the derived function. f^-1 (x) is the inverse function.

 

[VelaneDeBeaute]

AH Thankyou abcde and ffaady ( i hope i got the spellings right )

 

[ffaadyy]

AH Thankyou abcde and ffaady ( i hope i got the spellings right )

you sure did except for missing out one 'y' in my username anyways, you are welcome.

 

[ShudyShab]

Hi Guys. I have a question from the P2 and P3 book. Exercise 3C Number 6.

Q. How many terms of the geometric series 2+1/3=1/18+1/108+ . . . must be taken for its sum to differ from its sum to infinity by less than 10^-5 .

Thanks

 

[YousufIGCSE]

Two circles of radii 5 cm and 12 cm are drawn partly overlapping. Their centers are 13 cm apart. Find the area common to two circle
....This is my doubt....

 

[VelaneDeBeaute]

The distance is 13 cm, which shows the triangles are right angled and you can use trigonometric ratios to find the angles! Calculate angles A and B and then use the area of segment formula (Area of segment = 1/2 * r^2 * [ Theta - sin theta ] to find the areas of the grey and peach regions! Remember to work in radians and once both areas are there, sum them up!
DIY now! Good Luck!

 

[M Z 7]

Assalaamoualaikum...

i cannot find november 2011 papers...
could anyone please upload them or give a link where to get them?
جزاك اللهُ خيراً‎

 

[CaptainDanger]

Assalaamoualaikum...

i cannot find november 2011 papers...
could anyone please upload them or give a link where to get them?
جزاك اللهُ خيراً‎

http://www.xtremepapers.com/community/threads/oct-nov-2011-papers.10997/#post-142235

 

[ShudyShab]

Hi Guys. I really need help with this question. I have a question from the P2 and P3 book. Exercise 3C Number 6.

Q. How many terms of the geometric series 2+1/3=1/18+1/108+ . . . must be taken for its sum to differ from its sum to infinity by less than 10^-5 .

Thanks

 

[slayers]

Find the coordinates of two points A and B on the plane 2x+3y+4z=4. Verify that the vector AB is Perpendicular to normal to the plane..
Pls solve this

 

[BeeBee]

really need help in trigonometry..it has got me so confused

 

[ppaayas]

Hey, could someone upload mathematics notes?

 

[ffaadyy]

Find the coordinates of two points A and B on the plane 2x+3y+4z=4. Verify that the vector AB is Perpendicular to normal to the plane..
Pls solve this

To find A and B, you can take any values of x, y and z such that if you put those values in the equation of the plane, their sum equals to '4'.

For example:

Take A = (2, 0, 0) & Take B = (0, 0, 1)

Clearly A and B are on the plane since each point satisfies the equation of the plane.

The vector AB = -2i + k

The normal to the plane is N = 2i + 3j + 4k

AB . N = (-2)(2) + (1)(4) = 0, so AB is perpendicular to N

Similarly, you can take A = (1, 2, -1) & B = (6, 0, -2)

Again, if you substitute each point in the plane equation, their sum equals to '4' which means that these points are also on the plane.

The vector AB = 5i - 2j - k

The normal to the plane is N = 2i + 3j + 4k

AB . N = (5)(2) + (-2)(3) + (-1)(4) = 0, so AB is perpendicular to N

 

[ffaadyy]

really need help in trigonometry..it has got me so confused

AS Mathematics Trig. or A2 Mathematics Trig.? And are there any specific questions in which you are having problems?

 

[CaptainDanger]

Hey, could someone upload mathematics notes?

Search around you will find them!

 

[BeeBee]

AS Mathematics Trig. or A2 Mathematics Trig.? And are there any specific questions in which you are having problems?

A2 trig eg no.5