Fundamental Theorem of Calculus

First Part
Let $$f$$ be a real function which is continuous on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Let $$F$$ be a real function which is defined on $$\left[{a \,. \, . \, b}\right]$$ by $$F \left({x}\right) = \int_a^x f \left({t}\right) dt$$.

Then $$F$$ is a primitive of $$f$$ on $$\left[{a \,. \, . \, b}\right]$$.

Second Part
Let $$f$$ be a real function which is continuous on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Then:
 * $$f$$ has a primitive on $$\left[{a \, . \, . \, b}\right]$$;
 * If $$F$$ is any primitive of $$f$$ on $$\left[{a \, . \, . \, b}\right]$$, then $$\int_a^b f \left({t}\right) dt = F \left({b}\right) - F \left({a}\right) = \left[{F \left({t}\right)}\right]_a^b$$.

Proof of First Part
To show that $$F$$ is a primitive of $$f$$ on $$\left[{a \,. \, . \, b}\right]$$, we need to establish the following:


 * $$F$$ is continuous on $$\left[{a \, . \, . \, b}\right]$$;
 * $$F$$ is differentiable on the open interval $$\left({a \, . \, . \, b}\right)$$;
 * $$\forall x \in \left[{a \, . \, . \, b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$$.

Proof that F is Continuous
Since $$f$$ is continuous on $$\left[{a \,. \, . \, b}\right]$$, it follows from the Continuity Property that $$f$$ is bounded on $$\left[{a \,. \, . \, b}\right]$$.

Suppose that $$\forall t \in \left[{a \,. \, . \, b}\right]: \left|{f \left({t}\right)}\right| < \kappa$$.

Let $$x, \xi \in \left[{a \,. \, . \, b}\right]$$.

From Sum of Integrals on Adjacent Intervals‎, we have that $$\int_a^x f \left({t}\right) dt + \int_x^\xi f \left({t}\right) dt = \int_a^\xi f \left({t}\right) dt$$, that is, $$F \left({x}\right) = \int_x^\xi f \left({t}\right) dt + F \left({\xi}\right)$$.

So $$F \left({x}\right) - F \left({\xi}\right) = \int_\xi^x f \left({t}\right) dt$$.

From the corollary to Upper and Lower Bounds of Integral, $$\left|{F \left({x}\right) - F \left({\xi}\right)}\right| < \kappa \left|{x - \xi}\right|$$.

Thus it follows that $$F$$ is continuous on $$\left[{a \,. \, . \, b}\right]$$.

Proof that F is Differentiable and f is its Derivative

 * Second, that $$F$$ is differentiable on $$\left({a \, . \, . \, b}\right)$$ and that $$\forall x \in \left[{a \, . \, . \, b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$$:

Let $$x, \xi \in \left[{a \,. \, . \, b}\right]$$ such that $$x \ne \xi$$.

Then:

$$ $$ $$

Now, let $$\epsilon > 0$$.

If $$\xi \in \left({a \, . \, . \, b}\right)$$, then $$f$$ is continuous at $$\xi$$.

So for some $$\delta > 0$$, $$\left|{f \left({t}\right) - f \left({\xi}\right)}\right| < \epsilon$$ provided $$\left|{t - \xi}\right| < \delta$$.

So provided $$\left|{x - \xi}\right| < \delta$$ it follows that $$\left|{f \left({t}\right) - f \left({\xi}\right)}\right| < \epsilon$$ for any $$t$$ in an interval whose endpoints are $$x$$ and $$\xi$$.

So from the corollary to Upper and Lower Bounds of Integral, we have:

$$ $$ $$

provided $$0 < \left|{x - \xi}\right| < \delta$$.

But that's what this means:


 * $$\frac {F \left({x}\right) - F \left({\xi}\right)} {x - \xi} \to f \left({\xi}\right)$$ as $$x \to \xi$$

So $$F$$ is differentiable on $$\left({a \, . \, . \, b}\right)$$ and $$\forall x \in \left[{a \,. \, . \, b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$$.

Proof of Second Part
Let $$G$$ be defined on $$\left[{a \,. \, . \, b}\right]$$ by $$G \left({x}\right) = \int_a^x f \left({t}\right) dt$$.

We have:
 * $$G \left({a}\right) = \int_a^a f \left({t}\right) dt = 0$$ from Integral on Zero Interval;
 * $$G \left({b}\right) = \int_a^b f \left({t}\right) dt$$ from the definition of $$G$$ above.

From Sum of Integrals on Adjacent Intervals, we have $$\int_a^b f \left({t}\right) dt = \int_a^a f \left({t}\right) dt + \int_a^b f \left({t}\right) dt$$.

Thus $$\int_a^b f \left({t}\right) dt = G \left({b}\right) - G \left({a}\right)$$.

By the first part, $$G$$ is a primitive of $$f$$ on $$\left[{a \,. \, . \, b}\right]$$.

By Primitives which Differ by a Constant‎, we have that any primitive $$F$$ of $$f$$ on $$\left[{a \,. \, . \, b}\right]$$ satisfies $$F \left({x}\right) = G \left({x}\right) = c$$, where $$c$$ is a constant.

Thus $$\int_a^b f \left({t}\right) dt = \left({G \left({b}\right) + c}\right) - \left({G \left({a}\right) + c}\right)$$

and so $$\int_a^b f \left({t}\right) dt = F \left({b}\right) - F \left({a}\right)$$.