Calc2.63

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Suppose F(y)=\int_a^y f(t) \, dt
We know that there exist real numbers x \, and x+\delta \, in [a,b] \,, where a and b are non-equal real numbers.

F(x+\delta)-F(x) = \int_a^{x+\delta} f(t) \, dt - \int_a^{x} f(t) \, dt

It can be shown that \int_a^{x+\delta} f(t) \, dt - \int_a^{x} f(t) \, dt = \int_x^{x+\delta} f(t) \, dt

By the mean value theorem: \int_x^{x+\delta} f(t) \, dt = f(c) \delta for some c in [x,x+ \delta] \,

Hence F(x+\delta)-F(x) = f(c) \delta \,

\Rightarrow \frac{F(x+\delta)-F(x)}{\delta} = f(c)

\Rightarrow \lim_{\delta \to 0} \frac{F(x+\delta)-F(x)}{\delta} = \lim_{\delta \to 0} f(c)

But as \delta \to 0 , \, c \to x by the Sandwich Theorem

Hence \lim_{\delta \to 0} \frac{F(x+\delta)-F(x)}{\delta} = f(x)

Therefore, relabelling y as x: \, \frac{d}{dx} \int_a^x f(t) \, dt = f(x)

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