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If I have the definite integral of the derivative of a function, is it the same as having the derivative of the definite integral of a function?

For the latter, it would be the derivative of the upper bound*the inside function at that upper bound.

But if the $d/dx$ is inside the integral symbol, is it the same process? I would assume it would be since moving terms in or out of the integral is flexible

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  • $\begingroup$ Do you know the Fundamental Theorem of Calculus? $\endgroup$ – Viktor Glombik Dec 16 '18 at 15:21
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If you perform the definite integral of a function, you will get a constant and differentiating a constant leads to zero.

Also having the definite integral of the derivative function is just the going to be the average value of the function times the length of the interval.

They are not equal.

Perhaps you meant the indefinite integral?

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  • $\begingroup$ Actually what I meant was that you have d/dx(integral of f(t)) with the upper bound of the definite integral being something like x^2. So it would be 2x*f(x^2). But what if the d/dx was inside the integral symbol? $\endgroup$ – J.W. Dec 16 '18 at 15:27
  • $\begingroup$ This might be of help. en.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$ – harshit54 Dec 16 '18 at 15:28

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