Simple integration, stupid answer by textbook..or me? Watch

Destroyviruses
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So its evaluate : integral of "lambda"x/l The top is e and bottom is zero.

I get 2"lambda"e squared/l Textbook says, "lambda"x squared/2l



???????
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Ollie901
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(Original post by Destroyviruses)
So its evaluate : integral of "lambda"x/l The top is e and bottom is zero.

I get 2"lambda"e squared/l Textbook says, "lambda"x squared/2l



???????
The book is right. Where are you getting e's from? is that a typo?

Anyway, the x must become x^2. If you differentiate x^2 you get 2x, you want lamda/l lots of x, not 2. So divide lambda over l by 2 to get lambda over 2l lots of x^2.

Differentiating this now gives 2x * (lamda/2l) = Lambdax/l which is the original function
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Destroyviruses
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(Original post by Ollie901)
The book is right. Where are you getting e's from? is that a typo?

Anyway, the x must become x^2. If you differentiate x^2 you get 2x, you want lamda/l lots of x, not 2. So divide lambda over l by 2 to get lambda over 2l lots of x^2.

Differentiating this now gives 2x * (lamda/2l) = Lambdax/l which is the original function
Oh yeahhh! I forgot you divide! But the e comes from putting it into the equation! You know like x=e and x=0 ??

Thanks for your help. Why doesnt the textbook answer have an e?
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EEngWillow
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(Original post by Ollie901)
The book is right. Where are you getting e's from? is that a typo?

Anyway, the x must become x^2. If you differentiate x^2 you get 2x, you want lamda/l lots of x, not 2. So divide lambda over l by 2 to get lambda over 2l lots of x^2.

Differentiating this now gives 2x * (lamda/2l) = Lambdax/l which is the original function
I'd imagine the e comes from the upper limit of the definite integral. All you've (and the textbook by the look of it) have done is find the indefinite integral.

(Original post by Destroyviruses)
So its evaluate : integral of "lambda"x/l The top is e and bottom is zero.

I get 2"lambda"e squared/l Textbook says, "lambda"x squared/2l



???????
To clarify, do you mean \displaystyle\int_0^e \dfrac{\lambda x}{l}dx? If so then both answers are wrong - you should have a half in your answer instead of a two, and the book should have an e instead of an x.
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Destroyviruses
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(Original post by EEngWillow)
I'd imagine the e comes from the upper limit of the definite integral. All you've (and the textbook by the look of it) have done is find the indefinite integral.



To clarify, do you mean \displaystyle\int_0^e \dfrac{\lambda x}{l}dx? If so then both answers are wrong - you should have a half in your answer instead of a two, and the book should have an e instead of an x.
Thankyou!! xx Big HUG
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