4U Revising Game (1 Viewer)

[conics2008]

tommy undalay did the same working out as me,, read my old post i hinted perpendicular distance.

 

[undalay]

does any one have some integration question, i just finished that chapter...

can some one post some please....

From my class test.

Prove: [file attached]

 

[conics2008]

the last intergal is something tan^-1 something something right ??

can you tell me if im on the right track.

 

[undalay]

pi
|
| sin x / (a + bcos^2 x) dx
|
0

Is this the answer:

Spoiler

1/rt( ab) tan^-1 (rt (b/a))

 

[conics2008]

ill post the final answer

it goes something like this


-root of b / b root of a tan^-1 u root of b / root of a } 1 and -1

 

[conics2008]

those 1 and -1 are my limits

tke of b from cos then use u=cosx then, change into tan^-1 intergal thing.. then simplify... soo whats the answer

 

[conics2008]

From my class test.

Prove: [file attached]

you just sub in x= pi/2 -x into x

then prove its an odd function then its = 0 xD

 

[tommykins]

remember to sub back your u. also limits can be changed to 1, 0 even function. dont have an answer, but youre doing it right

Do you have to revert u back to cos x? I just changed the limits.

Heres my working -
http://img145.imageshack.us/img145/9559/12li9.jpg

PS. ignore the "a1"² bit, that was just a mistake for a when doing 1/a²+x² integrals.

 

[qqmore]

Volumes question:
The top face of a container is a rectangle of sides 3m and 4m respectively.
The bottom is a rectangle of sides 2m and 3m respectively.
(i.e. side of TOP 4m // BOT 3m and side of TOP 3m // BOT 2m)

If the height of the container is 1.5 m, find the volume of this container by slicing.

 

[Mark576]

prove:

pi/4
|
| 1 / (1 + sinx) dx
|
-pi/4

=

pi/4
|
| 2 (sec x)^2 dx
|
0

1/(1+sinx) = (1-sinx)/cos²x = sec²x - secxtanx

I = [-pi/4 --> pi/4]∫sec²xdx - [-pi/4 --> pi/4]∫secxtanxdx

Second integral is odd, therefore = 0. And the first integral is even:

I = [0 --> pi/4]∫2sec²xdx

Next Question:

Find:

[0 --> pi/2] ∫sinⁿxdx/(sinⁿx + cosⁿx)

Not very hard, but interesting nonetheless.

 

[nottellingu]

RONKEE please post up the solution to the polynomial q...i giv up lol

 

[ngogiathuan]

Next Question:

Find:

[0 --> pi/2] ∫sinⁿxdx/(sinⁿx + cosⁿx)

Not very hard, but interesting nonetheless.

Yeah pretty nice question. It looks tricky at a glance though.

 

[Mark576]

Yeah, that's correct. Post up a question ngogiathuan.

 

[Undermyskin]

Sorry to ask this but are we supposed to remember that rule for x and (a-x)? I've seen many questions but they always, kind of, giving hints by asking us to proof it first.

Q: Find the volume obtained by rotating the region enclosed by the circle

(x-b)^2 + y^2 = a^2

Where b>a, about the y-axis

If you have the Cambridge solution, don't look at it!

 

[ronnknee]

(x-root2)(x+root2)(x-2)(x+1)

Correct, but reasoning/working out?

 

[tommykins]

Sorry to ask this but are we supposed to remember that rule for x and (a-x)? I've seen many questions but they always, kind of, giving hints by asking us to proof it first.

Only takes 3-4 lines max to prove, I'd say prove it once and then you can use it as much as you want.

 

[ngogiathuan]

Currently unsolved:

Question 10
A solid figure has as its base, in the xy plane, the ellipse x²/16 + y²/4 = 1

Cross sections perpendicular to the x-axis are parabolas with latus rectums in the xy plane. Find the volume of the solid

Im currently working on this question. Havent figured it out yet. It looks pretty hard cause i think we have to find the area bounded the the parabola and its latus rectum. Anyway, give it a try.

 

[conics2008]

3units my answer to that intgeral quesiton was correct. who cares if he put it back in terms of cos(x) i left it in terms of u with the limits..

it still makes my answer correct... next question..

i haven't done volumes, i just started it today, looks not bad xD

post some more integration quesiton...

 

[ronnknee]

prove:

pi/4
|
| 1 / (1 + sinx) dx
|
-pi/4

=

pi/4
|
| 2 (sec x)^2 dx
|
0

Alternative method:

Let f(x) = 1 / (1 + sin x)
f(-x) = 1 / (1 - sin x) [note: sin -x = -sin x]

Therefore the integral would equal to:

/ pi/4
|
| [1 / (1 + sin x)] + [1 / (1 - sin x)] dx
|
/ 0

=

/ pi/4
| 2
| ------------ dx
| 1 - sin² x
/ 0

=

/ pi/4
| 2
| ------------ dx
| cos² x
/ 0

=

/ pi/4
|
| 2 sec² x dx
|
/ 0

 

[ronnknee]

find what f'(x) is equal to, if the function f is defined on all reals greater or equal to 0, and:

f (x) =

x^2
|
| cos [sqrt(t)] dt
|
0

Show that the roots of z⁶ + z³ + 1 = 0 are among the roots of z⁹ - 1 = 0. Hence find the roots of z⁶ + z³ + 1 = 0 in modulus/argument form