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# C4 Integration - (Handwritten) watch

1. Can someone tell me what am I doing wrong? I keep getting wrong answer

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2. (Original post by Moa)
Can someone tell me what am I doing wrong? I keep getting wrong answer

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If you are planning on studying at Imperial, C4 should be piss easy!
3. Where you say "c stays as a constant" and therefore add c (after exponentiating both sides), you should be multiplying by e^c = C, not adding.
4. Beaten to the punch. Yeah c = 4 and you're mutiplying not adding, as above stated
5. (Original post by Swayum)
Where you say "c stays as a constant" and therefore add c (after exponentiating both sides), you should be multiplying by e^c = C, not adding.
I do not see how it makes a difference I was always taught that:

-as long as you do not perform any operations using unknowns (like dividing both sides by "x" or multiplying by "y" etc.) constant "c" stays the same.

Here I do not perform any operations of this kind, I exponentiate both sides, however "c" is not in the exponentials power - it simply stays as +c, so why would I change it to e^c.

The same way if I multiplied both sides by 2, "c" would stay as "c", not "2c"

Can you tell me why am I wrong? Thank you for help
6. Rewrite your constant as lnA, where lnA is just another constant. Then when you are rewriting it the 2+y/2-y will be multiplied by A.
7. (Original post by Moa)
I do not see how it makes a difference I was always taught that:

-as long as you do not perform any operations using unknowns (like dividing both sides by "x" or multiplying by "y" etc.) constant "c" stays the same.

Here I do not perform any operations of this kind, I exponentiate both sides, however "c" is not in the exponentials power - it simply stays as +c, so why would I change it to e^c.

The same way if I multiplied both sides by 2, "c" would stay as "c", not "2c"

Can you tell me why am I wrong? Thank you for help
The same reason for when you multiplied 1/2-y by 2+y when you brought it down from the exponential. The e^c is also multiplied.
8. By the laws of logarithims y=x+a does not imply lny=lnx+a, if you like.
9. (Original post by Moa)
Here I do not perform any operations of this kind, I exponentiate both sides, however "c" is not in the exponentials power - it simply stays as +c, so why would I change it to e^c.
It is in the power though.

a = b + c

Goes to e^a = e^(b+c) - you need to exponentiate both sides and the RHS includes c.

And if you say let e^c = d, then you get e^a = de^b.

Here a = your LHS and b = those logarithms.

The same way if I multiplied both sides by 2, "c" would stay as "c", not "2c"
That's just bad notation for convenience. Writing 2c as c works because it's like saying "Well, when I integrated, I could have added c/2 instead of c and so multiplying by 2 would give me c on this line, but I can't be bothered to cross out my previous working, so I'll continue calling it c".
10. Right, I guess from now on I will always properly write what happens with c Thanks guys!
11. Going back to my a = b + c example:

Let's say we want y = x + c, where c is a constant.

Your claim is that e^y = e^(x) + c, so c = e^y - e^x.

Now let's say my line y = x + c passes through (3, 5), then c = 2 right?

Then c = 2 = e^y - e^x, but 2 = e^5 - e^3 isn't true...
12. (Original post by Swayum)
Going back to my a = b + c example:

Let's say we want y = x + c, where c is a constant.

Your claim is that e^y = e^(x) + c, so c = e^y - e^x.

Now let's say my line y = x + c passes through (3, 5), then c = 2 right?

Then c = 2 = e^y - e^x, but 2 = e^5 - e^3 isn't true...
Thanks! That clears everything up
13. this isnt edexcel is it ?
14. (Original post by bonbon)
this isnt edexcel is it ?
it is

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