Fanyeni hii intergration waungwana

[JPM605]

Nisaidieni 'intergral of the Squareroot of Sin X with respect to X' Ahsanteni.

 

[ProBook]

Pure 2......

 

[King Kong III]

Cosx/2

 

[Mbwiga88]

jaribu kutumia concept ya half angle

 

[Boniface92]

Nisaidieni 'intergral of the Squareroot of Sin X with respect to X' Ahsanteni.

taylor theorem

 

[Kituko]

Duh ndonga yangu imeshajaa maisha ambayo hayana uhusiano kabisa na hiyo intergration, enzi za mwalimu hiyo nilikuwa nakushukia kwa kichwa tu, haaaa ,,, SO MUCH SIMPLE

 

[Autorun]

=-cos x+c

 

[Moyibi]

Nilikuwa CBG na nilikuwa nakula BAM Ila uwezo wa kupiga Banda la Pure Maths nilikuwa nao Xo jibu mpango mzma tumia taylor serie na jibu ni =x2/4-x4/64+c
Note: x2 is x power 2 and x4 is x power 4

 

[Kijakazi]

Nisaidieni 'intergral of the Squareroot of Sin X with respect to X' Ahsanteni.

Nafikiri unaweza ukatumia Substitution method

S (sinx) X (SinX) dx (S naitumia kama integral sign) hapo nimeexpand tu sin x squared
halafu fanya substitution z= sinx
z'= cos
dz/dx=cosx
dx= 1/cos dz
Halafu unaweka kwenye sinx = z; S z (sinx)dx

Baada ya hapo substutute dx; S z(sinx)1/sinx dz
sinx na sinx zinajicancel unabakia na S z dz hapo sasa kazi imeisha unaintegrate tu z dz
S z dz=1/2 z(squared) +C halafu unafanya resubstitution yaani pale pa z = sin x
(sinx)squared/2 + C na hilo ndilo jibu lake (kumbuka nimetumia S kama integral sign!)

 

[fortunho]

nimeamin hesabu sio mapishi.dogo fanya hivi,hauwezi ukapata exact answer na ndio maana wakilitoa huwaga wanaweka limits ilimradi utumie simpson au trapezoidal,akikuuliza aproxmat ans bila limit ndo unamkumbuka taylor.

 

[NingaR]

kamawewe unajua hesabu tumia By-Part method then acha jibu katika 3 terms

 

[Kurzweil]

nimeamin hesabu sio mapishi.dogo fanya hivi,hauwezi ukapata exact answer na ndio maana wakilitoa huwaga wanaweka limits ilimradi utumie simpson au trapezoidal,akikuuliza aproxmat ans bila limit ndo unamkumbuka taylor.

Kumbe ulikua huamini.

 

[fortunho]

Kumbe ulikua huamini.

haya bhana

 

[paul kitereja]

tatizo hilo!

 

[JPM605]

Halina limits wala number of recquired ordinates hivyo naona trapezoidal na simpson zinafail..., haf tailor inaaply if f(x)=(x+a) kitu ambayo hii function haina.., kuhusu by parts nisaidieni ni ipi itakuwa U na dv?

 

[NingaR]

S√sinx.dx
Hapa hakuna ujanja zaidi ya kutumia By-Part method.
ie. SVdU=UV-SUdV----eqn(a)
let dU=dx
Integrating both side
=>U=x------eqn(i)
and V=(sinx)^½
differentiating both sides
=>dV=(cosx)/(2√sinx)-------eqn(ii)
Substituting ---eqn(i)&(ii) into --eqn(a)
SVdU=x√sinx - 1/2S(xcosx)dx/(√sinx)
For S(xcosx)dx/(√sinx)
By By-Part tena
SUdV=UV-SVdU
Hapa let
U=x
=>dU=dx
And
dV=(cosx)/(√sinx)
Integrating both sides
=>V=2√sinx (kama hujaelewa hapo basi wewe ni kilaza)



NIMECHOKA KUANDIKA
NATHANI MTAKUA MMEPATA MWANGA
(KAMA HUJUI HESABU KAA KUSHOTO)

 

[Moyibi]

S√sinx.dx
Hapa hakuna ujanja zaidi ya kutumia By-Part method.
ie. SVdU=UV-SUdV----eqn(a)
let dU=dx
Integrating both side
=>U=x------eqn(i)
and V=(sinx)^½
differentiating both sides
=>dV=(cosx)/(2√sinx)-------eqn(ii)
Substituting ---eqn(i)&(ii) into --eqn(a)
SVdU=x√sinx - 1/2S(xcosx)dx/(√sinx)
For S(xcosx)dx/(√sinx)
By By-Part tena
SUdV=UV-SVdU
Hapa let
U=x
=>dU=dx
And
dV=(cosx)/(√sinx)
Integrating both sides
=>V=2√sinx (kama hujaelewa hapo basi wewe ni kilaza)



NIMECHOKA KUANDIKA
NATHANI MTAKUA MMEPATA MWANGA
(KAMA HUJUI HESABU KAA KUSHOTO)

teh teh kaka ulipata ngapi? Namba

 

[NingaR]

teh teh kaka ulipata ngapi? Namba

Mkuu kwani VP??

 

[tindikalikali]

Nilikuwa CBG na nilikuwa nakula BAM Ila uwezo wa kupiga Banda la Pure Maths nilikuwa nao Xo jibu mpango mzma tumia taylor serie na jibu ni =x2/4-x4/64+c
Note: x2 is x power 2 and x4 is x power 4

Huna lolote wewe, ulijuaje kama ulikuwa na uwezo wa kupata banda? Pure siyo hizo BAM zenu

 

[NingaR]

Nafikiri unaweza ukatumia Substitution method

S (sinx) X (SinX) dx (S naitumia kama integral sign) hapo nimeexpand tu sin x squared
halafu fanya substitution z= sinx
z'= cos
dz/dx=cosx
dx= 1/cos dz
Halafu unaweka kwenye sinx = z; S z (sinx)dx

Baada ya hapo substutute dx; S z(sinx)1/sinx dz
sinx na sinx zinajicancel unabakia na S z dz hapo sasa kazi imeisha unaintegrate tu z dz
S z dz=1/2 z(squared) +C halafu unafanya resubstitution yaani pale pa z = sin x
(sinx)squared/2 + C na hilo ndilo jibu lake (kumbuka nimetumia S kama integral sign!)

mkubwa swali ni integral ya √(sinX) sio integral ya sin²X