A level integrating to find area under the curve

okay so for this question i understand you first find dy/dx which i got to be lnx+1 and then i would’ve substituted e as the x coordinate bc we have no other coordinates so that i could use the negative reciprocal as the gradient when working out the equation of a line for l but the mark scheme some how finds the gradient of the curve as 2 so the gradient of the line as -1/2

i just realised is it because e is e as in exponential so ln(e) is 1 and 1+1 is 2? but how was i supposed to know it was exponential e bc is that not meant to have a power or anything 😭😭😭😭

https://imgur.com/a/5LXqpo6

(edited 1 year ago)

Reply 1

mqb2766

Original post by esha06

dy(e)/dx = ln(e)+1 = 2
and its as in the mark scheme.

Reply 2

BankaiGintoki

Original post by esha06

okay so for this question i understand you first find dy/dx which i got to be lnx+1 and then i would’ve substituted e as the x coordinate bc we have no other coordinates so that i could use the negative reciprocal as the gradient when working out the equation of a line for l but the mark scheme some how finds the gradient of the curve as 2 so the gradient of the line as -1/2
i just realised is it because e is e as in exponential so ln(e) is 1 and 1+1 is 2? but how was i supposed to know it was exponential e bc is that not meant to have a power or anything 😭😭😭😭
https://imgur.com/a/5LXqpo6

ln(X) means ‘what power do I raise e to make it X’, so ln(e) means ‘what power do I raise e to, to make it e’.That would be 1. As any number to the power of 1 is itself.

(edited 1 year ago)

Reply 3

math-path

When you look at ln and log, imagine the implied base and the general identity that log(base, number) = value means base^value = number
Screenshot 2024-05-30 142740.png

(edited 1 year ago)

Reply 4

davros

Original post by esha06

okay so for this question i understand you first find dy/dx which i got to be lnx+1 and then i would’ve substituted e as the x coordinate bc we have no other coordinates so that i could use the negative reciprocal as the gradient when working out the equation of a line for l but the mark scheme some how finds the gradient of the curve as 2 so the gradient of the line as -1/2
i just realised is it because e is e as in exponential so ln(e) is 1 and 1+1 is 2? but how was i supposed to know it was exponential e bc is that not meant to have a power or anything 😭😭😭😭
https://imgur.com/a/5LXqpo6

Even if you'd had a temporary brain fade and forgotten that ln e = 1, the question pretty much tells you that because P is the point (e, e) and since y = xlnx, then at the point P you would need e = elne so you would be forced to have ln e = 1 :smile:

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