rhydex 247: una dey decline my question @ my maths general.

Mbahchiboy: my generals::
hw can one know wat to use as 'u' & 'dv' when solving integration by part??
is there a particular format or what?
answers will b appreciated

benbuks: ....it depends on hw d function appears...
E.g integral of (xarcsinx)dx ,u =arcsinx[since du/dx will easily give (1-x^2)^-1] , dv=x , integral of (x lnx)dx u=lnx dv=x .and so on.. U should always check b4 chosin ur 'u' or 'dv' ...bt there are functions , where you are free 2chose any of ur 'u' or 'dv' . E.g integral of (sinx.cosx)dx
hope my little explanations helps....other generals will throw more light....
Al z wel.
1love.

Mbahchiboy: my generals::
hw can one know wat to use as 'u' & 'dv' when solving integration by part??
is there a particular format or what?
answers will b appreciated

benbuks: Q1 compute the derivative of cosecx wrt.x from first principle ]

benbuks: Q1 compute the derivative of cosecx wrt.x from first principle

Q2 evaluate lim. as x-->0 of [(1/x +2)^1^2 - (1/x)^1/2 ]

Mr Calculus: i'm not clear with ur no 2 first bracket,is it ^1^2 or ^1/2?
all dsame i got 2/root2

smurfy:

Use U for what you can easily differentiate and DV for what you can easily integrate.

Mr Calculus: divide through by ¥x.
¥y/¥x=-[2cos(x+{¥x/2})sin(¥x/2)]/¥x/sinxsin(x+¥x).
¥y/¥x=-cos(x+{¥x/2})sin(¥x/2)/(¥x/2)/sinxsin(x+¥x).
recall::lim x-->0.¥x=0,sin(¥x/2)/(¥x/2)=1,¥y/¥x=dy/dx.
dy/dx=-cos(x+0)/sinxsin(x+0)
dy/dx=-cosx/sin^2x.
dy/dx=-cosx/sinx*1/sinx.
dy/dx=-cotxcosecx

benbuks: ^^ modified..sori it's ^1/2 ...check it again.

Mr Calculus: in dat case i got 2/root2 or rather root2

benbuks: ...beautifully wrong.

Samcathe: I have been on nairaland for more than two yrs.. I dont knw that a thread like this exist..api to be hear

use LAET-
L for log, A for algebra, E for exp n T for trig..
Anyone that come first will be 'U'

smurfy:

Let me welcome you to this outstanding thread by giving you something to chew and digest.

Find the integral of x * root(x-1) dx.

Mikebis: time to shw ma talent..here is d solutn
1.four vectors a,b,c nd d are said to b coplanar if nd only if there xist scalars p,q,v,s(not all zero)
such dat pa+qb+vc+sd=0
where p+q+v+s=0
since we are nt gvn d equatn of d vector,to shw dat d equatn is equal to zero.we slv 4 dat
3i-2j+4k=a
6i+3j+k=b
5i+3j+3k=c
2i+2j+6k=d
d vector eqn is
-a+b-c+d=0
sub 4 a,b,c nd d
-(3i-2j+4k)+(6i+3j+k)-(5i+7j+3k)+(2i+2j+6j)=0
-3i+2j-4k+6i+3j+k-5i-7j-3k+2i+2j+6k=0
(-3i+6i-5i+2i)+(2j+3j-7j+2j)+(-4k+k-3k+6k)=0
(0+0+0)=0
hence a,b,c nd d are complanar.
2.loadin tinz.......
3.prvn dat d identity of(a^b).(c^d)=(a.c)(b.d)-(a.d)(b.c)
the identity of
(a^b).(c^d) are knwn as commutatv
so let assume dat
p.q=p+q-pq
where p=(a^b)
q=(c^b)
(a^b).(c^b)=(a^b)+(c^b)-(a^b)(c^b)
(a^b).e=(a^b)+e-(a^b)e
(a^b)-(a^b)=e[1-(a^b)]
0=e[1-(a^b)]
e=0
d identity of LHS is equal to zero
(a.c)(b.d)-(a.d)(b.c)
(a.c)(b.d)^(a.d)(b.c)=(a.c)(b.d)-(a.d)(b.c)
(a.c)(b.d)^(b.d)^e=(a.c)(b.d)-e
(a.c)(b.d)=(a.c)(b.d)-e
(a.c)(b.d)-(a.c)(b.d)=-e
0=-e
e=0
d identity RHS is equal to zero
dazzal thank u

factorial1: Gud evening here...it been a while, hope you all are doing great...?

This's 1 or 2 qt for the gurus in the house.

1. All possible arrangements of the letters of the word REARRANGE are made. If one of these arrangements is chosen at random, what's the probability that it will end in (a) G, (b) R?

2. A man fires at a target, let a = P (he hits the target) = 1/5. Then b = P (he misses the target) = 1 - 1/5 = 4/5. If he fires four times at the target, what is the probability of his obtaining (a) exactly 2 hits, (b) at least 2 hits?


Do have a wonderful night rest.
#Shalom

factorial1: Gud evening here...it been a while, hope you all are doing great...?

This's 1 or 2 qt for the gurus in the house.

1. All possible arrangements of the letters of the word REARRANGE are made. If one of these arrangements is chosen at random, what's the probability that it will end in (a) G, (b) R?

2. A man fires at a target, let a = P (he hits the target) = 1/5. Then b = P (he misses the target) = 1 - 1/5 = 4/5. If he fires four times at the target, what is the probability of his obtaining (a) exactly 2 hits, (b) at least 2 hits?


Do have a wonderful night rest.
#Shalom

factorial1: Gud evening here...it been a while, hope you all are doing great...?

This's 1 or 2 qt for the gurus in the house.

1. All possible arrangements of the letters of the word REARRANGE are made. If one of these arrangements is chosen at random, what's the probability that it will end in (a) G, (b) R?

2. A man fires at a target, let a = P (he hits the target) = 1/5. Then b = P (he misses the target) = 1 - 1/5 = 4/5. If he fires four times at the target, what is the probability of his obtaining (a) exactly 2 hits, (b) at least 2 hits?


Do have a wonderful night rest.
#Shalom