Converting derivative into integral

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To convert the equation d(B)/dt = Q/K into integral form, it's essential to clarify that Q is solely a function of time and does not depend on B. This simplification allows for a more straightforward integration process. The integral form can be expressed as B(t) = (1/K) ∫ Q(t) dt + C, where C is the constant of integration. Understanding the relationship between Q and B is crucial for solving similar problems in calculus. This discussion highlights the importance of recognizing the dependencies in equations when transitioning from derivatives to integrals.
zealous131
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Hi,

I need to ask a very basic question and am very weak at calculus. I have an equation d(B)/dt=Q/K. Where K is a constant and Q is a function of time. B is the variable, which when differentiated w.r.t. time gives the Q. I want to convert this equation into integral form. Can anyone help me out on this? I will really really appreciate any help! Thanks!
 
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zealous131 said:
Hi,

I need to ask a very basic question and am very weak at calculus. I have an equation d(B)/dt=Q/K. Where K is a constant and Q is a function of time. B is the variable, which when differentiated w.r.t. time gives the Q. I want to convert this equation into integral form. Can anyone help me out on this? I will really really appreciate any help! Thanks!

Hey zealous131 and welcome to the forums.

Does Q only involve time (t) or does it also involve B? If the answer is no then JJacquelin has provided a very good outline of what to do. If not, then you will need to state how B is involved in with Q. If you don't understand what I'm saying and you've only been told that Q is a function of t (i.e. Q(t)) then don't worry about my advice (but keep it in mind later on if you have to solve things like this).
 
Thanks a lot JJacquelin for providing me the solution. I really apreciate it! Thanks Chiro for your valuable comment, Q is only a function of time and doesn't depend on B so that makes ife easier. I will definitely keep the point you've mantioned. Thanks!
 

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