Proof of the Fundamental Theorem of Calculus 📂Calculus

Proof of the Fundamental Theorem of Calculus

Theorem¹

Assume that the function $f$ is continuous on the closed interval $[a,b]$.

(1) The function $\displaystyle F(x) = \int_{a}^{x} f(t) dt$ is continuous on $[a,b]$, differentiable on $(a,b)$, and satisfies $\displaystyle {{dF(x)} \over {dx}} = f(x)$.

(2) For any antiderivative $F$ of $f$, $\displaystyle \int_{a}^{b} f(x) dx = F(b) - F(a)$

Explanation

Of course, we use the words differentiation and integration so we can easily guess the relationship between them. However, in English, they are called differential and integral, which seem totally unrelated, and even the concepts don’t really resemble each other.

The Fundamental Theorem of Calculus shows that differentiation and integration are, indeed, inverse operations of each other.

Proof

(1)

By the Mean Value Theorem of Integration, there exists $c$ between $x, x+h$ that satisfies $\displaystyle f(c) = {{1} \over {h}} \int_{x}^{x+h} f(t) dt$.

When $h \to 0$, it becomes $c \to x$, thus

$$ \lim_{h \to 0} {{1} \over {h}} \int_{x}^{x+h} f(t) dt = \lim_{h \to 0} f(c) = f(x) $$

Meanwhile, because of $\displaystyle F(x+h) - F(x) = \int_{a}^{x+h} f(t) dt - \int_{a}^{x} f(t) dt = \int_{x}^{x+h} f(t) dt$, it is also

$$ {{1} \over {h}} \int_{x}^{x+h} f(t) dt = { {F(x+h) - F(x)} \over {h} } $$

Therefore,

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

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(2)

Since $F$ is an antiderivative of $f$, it follows $\displaystyle \int_{a}^{b} f(t) dt = F(b) + C$, thus

$$ \int_{a}^{a} f(t) dt = F(a) + C $$

Subtracting the two sides from each other results in

$$ \int_{a}^{b} f(x) dx = F(b) - F(a) $$

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