Answer to q10 f? (1 Viewer)

[Thecorey0]

What did everyone get to the answer to the last question in q10? I got 17/12 -2ln2.

 

[domo.kun]

I got the same.

 

[sydneyroosters]

Yep got the same

 

[TechnIx]

N/A

Lol. Sigh.

 

[bell_9192]

What did everyone get to the answer to the last question in q10? I got 17/12 -2ln2.

I got exactly that!
i think that q 10 was easier than q8 lol

 

[bemer]

the answer is 41/12 -2ln2

 

[sydneyroosters]

the answer is 41/12 -2ln2

I think you stuffed up the substitution of 0, instead of 1 you got -1 so hence you ended up with 41/2 - 2ln2

A = Integral of x - x^2 /2 + x^3/3 - ln(1+x) from 0 to 1

= {[x^2 /2 - x^3/6 + x^4/12] - [(1+x)ln(1+x) - (1+x)]} from 0 to 1

= {[1/2 - 1/6 + 1/12)] - [2ln(2) - 2]} - {[0] - [(1ln(1) - 1)]}

= { 5/12 - 2ln(2) + 2 } - {0-[0-1]}

= 29/12 - 2ln(2) - {0--1}

= 29/12 - 2ln(2) - 1

= 17/12 - 2ln(2)

 

[boxhunter91]

Woot i did everything write except i used the x^3/(1+x) integral lol.
I got an answer of 2 ln 2 rekn ill get the 1 mark?

 

[papa2020]

i got the same

 

[gunne]

i screwed up the very simple derivation... at the beginning of q10
hopefully ecf!

 

[daniieee]

I think you stuffed up the substitution of 0, instead of 1 you got -1 so hence you ended up with 41/2 - 2ln2

A = Integral of x - x^2 /2 + x^3/3 - ln(1+x) from 0 to 1

= {[x^2 /2 - x^3/6 + x^4/12] - [(1+x)ln(1+x) - (1+x)]} from 0 to 1

= {[1/2 - 1/6 + 1/12)] - [2ln(2) - 2]} - {[0] - [(1ln(1) - 1)]}

= { 5/12 - 2ln(2) + 2 } - {0-[0-1]}

= 29/12 - 2ln(2) - {0--1}

= 29/12 - 2ln(2) - 1

= 17/12 - 2ln(2)

Got to about here, and then PENS DOWN! !#%^!!@%&#!%#@$%! Stupidly wasted too much time fixing up 9b.

 

[heavenfalling]

I stuffed up part d (the graphs) but still managed to do it. I got something along those lines. Lol 10 e & f is easier compared to other questions.