Serious admaths help

[xhizors]

http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_2.pdf

Q9 part 3

i have tried thrice , doubt on mark scheme lol any one plz help

 

[donor]

whats the answer

 

[123add1]

no the mark scheme is absolutely correct and i have done the same.when u integrate 20/2t+4square u get -10/2t+4 +c
u know that when t=0 ,s also equals to 0.hence put the value of t=0 and equate it to 0.u will get the value of c as 2.5.so ur
s=-10/2t+4 +2.5.put t=8 and u will get 2.
another method is to put limits from 0 to 8 and u will get the same answer.

 

[xhizors]

no the mark scheme is absolutely correct and i have done the same.when u integrate 20/2t+4square u get -10/2t+4 +c
u know that when t=0 ,s also equals to 0.hence put the value of t=0 and equate it to 0.u will get the value of c as 2.5.so ur
s=-10/2t+4 +2.5.put t=8 and u will get 2.
another method is to put limits from 0 to 8 and u will get the same answer.

tell me one thing when we integrate we divide with n+1 power over here how can we divide with just n which is -2
btw thnks

 

[donor]

the equation is in the form of (ax+b)^z
so in this type of equations we divide it by (a)(z+1)

 

[donor]

heres the equation

 

[xhizors]

thnks buddy i missed the imp concept