yeah i didlolokay said:just do i.b.p
let x=u, (4-x)1/2 = dvshaon0 said:yeah i did
but i got some weird answer
oh, i got what you got but i didn't simpligfy it any further from:lolokay said:let x=u, (4-x)1/2 = dv
du = dx
v = -2/3(4-x)3/2
so I x(4-x)1/2.dx
= -2/3 x(4-x)3/2 + I 2/3(4-x)3/2
= -2/3 x(4-x)3/2 - 4/15(4-x)5/2
= (4-x)3/2(-10/15 x - 16/15 + 4x/15)
= -2/15 (4-x)3/2(3x + 8) [+C]
Yes! thanks for your helplolokay said:I assume you would, unless it said "in its simplest form" or something
You can avoid integration by parts using the following:shaon0 said:sorry but i need some help;
Integrate: xsqrt(4-x) dx
sorry, thats too pro for meTrebla said:You can avoid integration by parts using the following:
∫ x√(4 - x) dx = - ∫ ([4 - x] - 4)√(4 - x) dx
= - ∫ [4 - x]√(4 - x) dx + ∫ 4√(4 - x) dx
= - ∫ (4 - x)3/2 dx + 4 ∫ (4 - x)1/2 dx
= 2(4 - x)5/2/5 - 8(4 - x)3/2/3 + c
That should be the same answer if you factorise (4 - x)3/2 and put over a common denominator.
alrite, i'll take that on board. I wasn't thinking of substitution because it was in the i.b.p section of fitzpatrick.tacogym27101990 said:yeah i wouldn't use i.b.p for a question like this as its not two different functions really
u use ibp for things like e^x.sinx and stuff
you can just use a suitable substitution such as u=4-x
its alot quicker and easier