# Proof

(related to Theorem: Indefinite Integral, Antiderivative)

We want to show that on the close interval $$[a,x]$$, the indefinite integral $$F(x):=\int_a^x f(t)dt$$ of a continuous real function $$f$$ is differentiable, and that its derivative is $$F'=f$$. Note, that continuity is a sufficient condition for $$f$$ to be Riemann integrable. Therefore, the integral $$F(x):=\int_a^x f(t)dt$$ exists.

Let $$h\neq 0$$. By definition, the derivative is given by

$F'(x):=\lim_{h\to 0}\frac {F(x+h)-F(x)}{h}.$

It follows from the linearity of the Riemann integral. $\frac {F(x+h)-F(x)}{h}=\frac 1h\left(\int_a^{x+h}f(t)dt-\int_a^xf(t)dt\right)=\frac 1h\int_x^{x+h}f(t)dt.$

Because $$f$$ is continuous, the mean value theorem for Riemann integrals tells us that there exists $$\xi_h\in[x,x+h]$$ (or $$\xi_h\in[x+h,x]$$ for $$h < 0$$) with

$\int_x^{x+h}f(t)dt=hf(\xi_h).$

Note that $$\xi_h$$ tends to $$x$$, as $$h$$ tends to $$0$$. Because $$f$$ is continuous, we get the required result: $F'(x)=\lim_{h\to 0}\frac 1h\int_x^{x+h}f(t)dt=\lim_{h\to 0}\frac 1h(hf(\xi_h))=f(x).$

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983