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- Thread starter gl0ck
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- #2

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Sorry, I won't stand on my head to read a question.

- #3

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Hello, Is this the right way to compute this integral. I get the answer from the book, but I am not sure if I've done it right.

Thanks

PS Sorry for the upside down image :)

Now should be alright

I have no idea what those symbols around the ##x## mean nor can I follow your steps. In particular you can't take the antiderivative of just part of the integrand.

- #4

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This is how its written in the book..

- #5

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I have no idea what those symbols around the ##x## mean nor can I follow your steps. In particular you can't take the antiderivative of just part of the integrand.

This is how its written in the book..

So you are trying to work a problem where you don't know and can't tell us what the symbols mean? Better read that book a bit more closely and get back to us...

- #6

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So you are trying to work a problem where you don't know and can't tell us what the symbols mean? Better read that book a bit more closely and get back to us...

So, if the found value is 0.6 and it is surrounded by these strange brackets this is equal to 1, if the brackets are turned other way around 0.6 is equal to 0

- #7

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So, if the found value is 0.6 and it is surrounded by these strange brackets this is equal to 1, if the brackets are turned other way around 0.6 is equal to 0

I'm guessing that ##\lfloor x \rfloor## stands for the greatest integer less than or equal to ##x##. Surely your text must define that symbol. You need to look it up and get it correct to use it in your integral. Your original integration attempt was nonsense. How could you possibly expect to get a correct antiderivative of a function when you don't know what it is in the first place? Start over.

- #8

Infrared

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