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# Integration limit help? watch

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1. https://gyazo.com/b1cd3f8a6c96dd2e4b5c9e67480c10e3

Q5 (a) if you integrate it you would get - [integral ( e^-lambda.x) dx ] with limits
of infinity and 0.

How exactly do you tackle the question if you have limits of infinity?
What exactly do I sub in or what happens in the answer?

(b) i) you just differentiate via quotient rule but for ii) I don't have any clues
or idea how to go about solving especially when I have infinity as my limit etc
as I haven't been introduced this concept before. :s

I'd appreciate any help.
2. (Original post by XxKingSniprxX)
https://gyazo.com/b1cd3f8a6c96dd2e4b5c9e67480c10e3

Q5 (a) if you integrate it you would get - [integral ( e^-lambda.x) dx ] with limits
of infinity and 0.

How exactly do you tackle the question if you have limits of infinity?
What exactly do I sub in or what happens in the answer?

(b) i) you just differentiate via quotient rule but for ii) I don't have any clues
or idea how to go about solving especially when I have infinity as my limit etc
as I haven't been introduced this concept before. :s

I'd appreciate any help.
You do the indefinite integral, and then take limits of the result as x goes to infinity minus the result as x goes to 0
3. (Original post by XxKingSniprxX)
https://gyazo.com/b1cd3f8a6c96dd2e4b5c9e67480c10e3

Q5 (a) if you integrate it you would get - [integral ( e^-lambda.x) dx ] with limits
of infinity and 0.

How exactly do you tackle the question if you have limits of infinity?
What exactly do I sub in or what happens in the answer?

(b) i) you just differentiate via quotient rule but for ii) I don't have any clues
or idea how to go about solving especially when I have infinity as my limit etc
as I haven't been introduced this concept before. :s

I'd appreciate any help.
Since this is an improper integral, it might be useful to first determine at which points the integral isn't very nicely defined - which is infinity here, then rewrite it as and take the integral as normal inside the limit then take the limit of it.

Notice that the integral is just from 0 to . The expression is defined at but not 0, so take the limit as that goes to 0 noting that we have for small

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