A-level Integration query

Original post by med123456^^

Is it important to take out the constants before integrating so for example can the integral of 2/(25-10P) not be written as 2 x (-1/10)ln(25-10P)
which is -1/5ln(25-10P) or is this wrong? If I take out the constants I get integral of 2/(5)(5-2P) is equal to (2/5) the integral of 1/(5-2P) which is (2/5)x(-1/2)ln(5-2P) this simplified is -1/5ln(5-2P). Which method is correct and why?

they are both correct as long as you put appropriate constants of integration at the end.

Reply 3

old_engineer

11

Original post by med123456^^

Is it important to take out the constants before integrating so for example can the integral of 2/(25-10P) not be written as 2 x (-1/10)ln(25-10P)
which is -1/5ln(25-10P) or is this wrong? If I take out the constants I get integral of 2/(5)(5-2P) is equal to (2/5) the integral of 1/(5-2P) which is (2/5)x(-1/2)ln(5-2P) this simplified is -1/5ln(5-2P). Which method is correct and why?

The two versions are less different than they look.

ln(25 - 10P) + C = ln(5(5 - 2P)) + C
= ln(5 - 2P) + ln(5) + C = ln(5 - 2P) + D

In other words, the apparent difference can be absorbed within the arbitrary constant of integration.

Reply 4

OP

Original post by AmazingPhysicist

Yes, you are allowed to take multiplicative constants outside integrals. The first answer you provided, -1/5ln(25-10P) is correct.

Original post by the bear

they are both correct as long as you put appropriate constants of integration at the end.

Why is my value for c wrong then if I use -1/5ln(25-10P) in this question? even more confused now

Original post by med123456^^

Why is my value for c wrong then if I use -1/5ln(25-10P) in this question? even more confused now

cannot read .docx files :redface:

Reply 6

OP

Original post by the bear

cannot read .docx files :redface:

This

77B9AC83-E381-481D-A25B-DDF6BB6133BA.jpg.jpeg

is the wrong value of c. The mark scheme says a different answer to c

Reply 7

OP

Original post by the bear

cannot read .docx files :redface:

7928675C-E092-412F-B02E-941DB5546A0D.jpg.jpeg

Reply 8

Original post by med123456^^

This

is the wrong value of c. The mark scheme says a different answer to c

you have not put in 1/2 for p in the last line

Reply 9

Notnek

21

Original post by the bear

cannot read .docx files :redface:

What year do you live in?

Reply 10

Original post by Sir Cumference

What year do you live in?

also i do not have twitta or faecebook :teehee:

Reply 11

OP

Original post by the bear

also i do not have twitta or faecebook :teehee:

😂 ah yes I forgot to sub P=1/2 but even then it is not the correct c value

Reply 12

Original post by med123456^^

😂 ah yes I forgot to sub P=1/2 but even then it is not the correct c value

you should be integrating 4/5p and 8/(5 - 2p) according to the mark scheme.

Reply 13

old_engineer

11

Original post by med123456^^

😂 ah yes I forgot to sub P=1/2 but even then it is not the correct c value

I believe the difference between your solution and the book solution is entirely due to the factor of 1/20 which you have left on the RHS. Nothing wrong with that, but it will inevitably lead to a different value for C. If you start with your complete solution (including your C), you should find that you are able to rearrange it into the book solution.

Reply 14

OP

Original post by old_engineer

I believe the difference between your solution and the book solution is entirely due to the factor of 1/20 which you have left on the RHS. Nothing wrong with that, but it will inevitably lead to a different value for C. If you start with your complete solution (including your C), you should find that you are able to rearrange it into the book solution.

So out of these which is wrong 69B669FB-8CEE-4796-954B-C82F8DB11BFA.jpg.jpeg

Reply 15

OP

The question was basically this and I rearranged the integral to this.... 76F40A92-234F-468E-93C5-F4F54FC78E47.jpg.jpeg

4351E1EC-70DC-4891-BCC8-38597CE78DF8.jpg.jpeg

so is how I’ve rearranged it wrong?

Reply 16

old_engineer

11

Original post by med123456^^

The question was basically this and I rearranged the integral to this.... so is how I’ve rearranged it wrong?

Am I right in thinking that you complete solution is (1/5)ln(P/(25 - 10P)) = (1/20)t + (1/5)ln(1/40) ? If so, I believe that is equivalent to the book solution and can be rearranged to match the book solution.

Reply 17

OP