integration Qs

[still hemlock]

hi can someone help with this

[pulsar fiber]

rewrite it as ln x times x^(-1/3)

u = lnx

v' = x^(-1/3)

and go from there

[still hemlock]

thankss

[pulsar fiber]

nw, lmk if you need anything else

[still hemlock]

there was one more question its rlly difficult tho

[pulsar fiber]

go for it

[pulsar fiber]

okay so part a is partial fractions, all good with them?

[still hemlock]

is it asking me to find B and C

[pulsar fiber]

yeah

[still hemlock]

let me try one minute

[copper pendant]

You would use comparing co-efficients or cross multiplication

[still hemlock]

ummm i got both B and C to be 2

[pulsar fiber]

yh that's right

part b now

[still hemlock]

i dont know what its asking

😭

[pulsar fiber]

have you done separating variables before?

[chilly gull]

is this further

[still hemlock]

idk i dont recognise what that means 💀

[still hemlock]

no

[chilly gull]

i think it just asking u to solve the differential equation

[pulsar fiber]

seperating variables is where we just move the e.g dy/dx to opposite sides of the equation

so something dy = something dx

so we can integrate both sides with respect to each variable

so we want to get anything involving lambda to the side with d(lambda)

and anything involving t with the dt

[chilly gull]

but when you solve dont u get ln(something) + ln(something) so how does it become ln(something)/ln(something)

[pulsar fiber]

in the actual integration, youll find it works out to ln(something) - ln(something) then by the log law, lnA-lnB=ln(A/B)

[chilly gull]

wait how though unless im missing something, if you integrate that you get ln(1+2lamda) + ln(1-2 lambda)

[still hemlock]

ohhh no ive came across that yh

[chilly gull]

unless B or C is minus Idk what they are yet

[pulsar fiber]

you will be integrating 2/1-2lambda which is -ln(1-2lambda)

[pulsar fiber]

B,C = 2

[chilly gull]

oh i forgot the -1 oh i get it now thanks

[pulsar fiber]

nw

[still hemlock]

im so confused wtf

[pulsar fiber]

ill write some stuff out

[still hemlock]

thank you

[chilly gull]

also @pulsar fiber do u think i should bother remembering this

havent seen a past exam question on this

[pulsar fiber]

ive made a quizlet set on everything for large data set, ive put the different types and descriptive terms on there jsut in case

doesnt hurt not to

[chilly gull]

oo do you think u can send me the quizlet link?

so i can add it to my flashcards

[pulsar fiber]

in part a, we worked out the partial fraction part so we might as well try implement it into our equation

[still hemlock]

okk so yk the left hand side shall i write that out in the B and C form

[pulsar fiber]

yh

[still hemlock]

tysmm

[pulsar fiber]

you all good with the rest of the integration?

[still hemlock]

ill have a go and let you knoww

[pulsar fiber]

okay

for when you are rly stuck, but ill try get back to you before then

[still hemlock]

umm im a little stuck integrating the right side

[pulsar fiber]

integrating e^kx = 1/k x e^kx

You differentiate the power and divide by it

Then keep power the same

[still hemlock]

but theres an A infront of it

[pulsar fiber]

It’s just a constant so you can keep it there

Then multiply by it after

Like integrating 2x^2 it’s the same 2 times integral of x^2

[still hemlock]

ohhhh mbmb