A general question on integrals and derivatives?

Is the following always true:

d/dx (∫ f(x) dx) = f(x).

I cant think of a case that this doesn't hold.

3 Answers

  • 8 years ago
    Favorite Answer

    The Fundamental Theorem of Calculus is a little more precise, but that's the gist of it, that integration and differentiation are inverse operations.

  • Anonymous
    4 years ago

    permit F(0,x) = ?[61968d34834b1dff53be4676ffef3e61968d34834b1dff53be4676ffef3e61968d34834b1dff53be4676ffef3e] (u^2 - a million)/(u^2 + a million) du. Then g(0,x) = F(361968d34834b1dff53be4676ffef3e) - F(261968d34834b1dff53be4676ffef3e) ? g'(0,x) = 3F'(361968d34834b1dff53be4676ffef3e) - 2F'(261968d34834b1dff53be4676ffef3e)0,x utilising the chain rule. via the elementary theorem of calculus61968d34834b1dff53be4676ffef3e F'(361968d34834b1dff53be4676ffef3e) = ((361968d34834b1dff53be4676ffef3e)^2 - a million)/((361968d34834b1dff53be4676ffef3e)^2 + a million) and F'(261968d34834b1dff53be4676ffef3e) = ((261968d34834b1dff53be4676ffef3e)^2 - a million)/((261968d34834b1dff53be4676ffef3e)^2 + a million). as a effect g'(0,x) = 3(961968d34834b1dff53be4676ffef3e^2 - a million)/(961968d34834b1dff53be4676ffef3e^2 + a million) - 2(461968d34834b1dff53be4676ffef3e^2 - a million)/(461968d34834b1dff53be4676ffef3e^2 + a million).

  • 8 years ago

    It will always hold.

    I am pretty sure thats the fundamental theorem of calculus

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