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Continuous Riemann Multiple Integral Well Defined Additive Function Integral Mean Value Theorem Domain Antiderivative Order Preserving Map

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Proof Of The Second Fundamental Theorem Of Calculus

Recall that a continuous function is Riemann integrable, so the integral

$\displaystyle F(x) = \int_c^x f(t)\, dt$

is well defined.

Consider the increment of :

$\displaystyle F(x+h)-F(x) = \int_c^{x+h} f(t)\, dt - \int_c^x f(t)\, dt = \int_x^{x+h} f(t)\, dt$

(we have used the linearity of the integral with respect to the function and the additivity with respect to the domain).

Now let be the maximum of on and be the minimum. Clearly we have

$\displaystyle m h \le \int_x^{x+h} f(t)\, dt \le M h$

(this is due to the monotonicity of the integral with respect to the integrand) which can be written as

$\displaystyle \frac{F(x+h)-F(x)}{h} = \frac{\int_{x}^{x+h}f(t)\, dt}{h} \in [m,M]$

Since is continuous, by the mean-value theorem, there exists $\xi_h\in [x,x+h]$ such that $f(\xi_h) = \frac{F(x+h)-F(x)}{h}$ so that

$\displaystyle F'(x)= \lim_{h\to 0} \frac{ F(x+h)-F(x)}{h} = \lim_{h\to 0} f(\xi_h) = f(x)$

since $\xi_h\to x$ as $h\to 0$ . This proves the first part of the theorem.

For the second part suppose that is any antiderivative of , i.e. . Let be the integral function

$\displaystyle F(x)= \int_a^x f(t) \, dt.$

We have just proven that

. So

for all $x\in [a,b]$ or, which is the same,

. This means that

is constant on

that is, there exists

such that

. Since

we have

and hence

for all $x\in[a,b]$ . Thus

$\displaystyle \int_a^b f(t)\, dt = F(b) = G(b) - G(a).$

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