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Integrals

Am I right in thinking the integral of 0 is 0? I`m trying to follow someones working in a fourier series and their working implies they`re taken the integral of 0 dt to be t, which ain`t right to me.

TIA

Reply 1

claret_n_blue

The integral of 0 is 'c', a constant.

Reply 2

ClaraOswald

Original post by arson_fire

Just think about what you can differentiate to get 0.

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Reply 3

ClaraOswald

Original post by arson_fire

Any constant, but when I take the interval into account I get C-C=0. Looks like the lecturers messed up!

Are you sure the constants are equal? You might just be getting two different constants which can be simplified into just a single constant.

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Reply 4

davros

Original post by arson_fire

Think so. Its the definite integral of 0 dt with the interval [T-deltaT,T]. I get 0, but the lecture notes say -deltaT which I can only get by by changing 0 for 1.

I'm inclined to agree with you - you would need to integrate 1 to get that answer :smile:

Reply 5

user2020user

Original post by arson_fire

Think so. Its the definite integral of 0 dt with the interval [T-deltaT,T]. I get 0, but the lecture notes say -deltaT which I can only get by by changing 0 for 1.

\displaystyle \int_{t + \Delta t}^{t} 0 \text{ d}t = \mathcal{C} \Big|_{t + \Delta t}^{t} = \mathcal{C} - \mathcal{C} = 0

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