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$$\displaystyle \int_{0}^{f(x)} g'(t) \, \mathrm{d}t$$

Can someone explain why this equals $g(f(x))-g(f(0))$? I'm really confused since the derivative is different and the fundamental theorem of calculus only supports integral from $0$ to $x$. Thanks after a little bit of thinking and rechecking the fundamental theorem of calculus, i got it in my head its 3 am here so i'm a bit dizzy, everything is clear now , thanks guys.

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    $\begingroup$ It does not matter, the limits are treated as constant during the integration. Also, that should be $g(f(x))-g(0)$ (or otherwise the lower limit should be $f(0)$) $\endgroup$ – L. F. Aug 24 '13 at 1:46
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    $\begingroup$ The definition of the integral from $a$ to $b$ of $f'(x)$ is $f(b)-f(a)$. The result you show is a simple substitution matching this definition. $\endgroup$ – abiessu Aug 24 '13 at 1:47
  • $\begingroup$ if the limits are treated as constants during the integration , then why the derivative of $$\displaystyle \int_{0}^{f(x)} g'(t) \, \mathrm{d}t$$ isn't g'(x)? if we put f(x)=n , then we get F(n) ?I know it must be treated as fog composite of two functions when deriving but why this logic yields a wrong answer $\endgroup$ – Steve Aug 24 '13 at 1:48
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    $\begingroup$ That's an entirely different operation (it is differentiating w.r.t. one of the limits, rather than performing the integration). I don't understand your second sentence. $\endgroup$ – L. F. Aug 24 '13 at 1:50
  • $\begingroup$ Were you asked for the derivative? The derivative of $g(f(x))-g(f(0))$ with respect to $x$ is ${d(g(f(x))) \over dx} = f'(x)g'(f(x))$. $\endgroup$ – abiessu Aug 24 '13 at 1:52
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$$\int_a^bg'(x)dx=g(b)-g(a)$$ $$\int_{\text{lion}}^{\text{elephant}}g'(x)dx=g(\text{elephant})-g(\text{lion})$$ $$\int_{\text{sun}}^{\text{moon}}g'(x)dx=g(\text{moon})-g(\text{sun})$$ $$\int_0^{f(x)}g'(x)dx=g(f(x))-g(0)$$

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The integral of a derivative f'(a) is clearly the function f(a). Thus the answer to the integral would be g(t), t1 = 0 and t2 = f(x).

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