Quick Integral/Derivative question.

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If a function f(x) satisfies the condition f(x) = f'(x), then it is established that -f(x) equals -∫f'(x) dx, disregarding arbitrary constants. This relationship confirms that the negative of a derivative corresponds to the negative of the original function when integrated. The conclusion is that the integral of the derivative, when negated, directly relates back to the original function.

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synthetic.
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Is it always true that if; [tex]f(x) = f '(x)[/tex], then; [tex]-f (x) = - \int f '(x)[/tex] ?

That is, the negative of a derivative has an integral which is simply the negative of the original function? (ignoring arbitiary constants)
 
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Yes - the right side is just [tex]f(x)[/tex], up to a constant.
 

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