Theorem
Let \(f: [a,b] \to \mathbb{R}\). Then
  1. Suppose \(f\) is Riemann integrable on \([a,b]\) and \(f\) is continuous on the point \(x_0 \in [a,b]\). Define $$ F(x) = \int_a^x f(t)dt $$ Then \(F\) is differentiable at \(x = x_0\) and $$ F'(x_0) = f(x_0). $$ Moreover, if \(f\) is continuous on \([a,b]\). Then \(F'(x)=f(x)\) for all \(x \in [a,b]\).
  2. Suppose \(f\) is differentiable on \([a,b]\) and \(f'\) is Riemann integrable on \([a,b]\). Then $$ \int_a^b f'(t)dt = f(b) - f(a) $$

Notes + Proof (a)

For (a) if \(f\) is Riemann integrable on an interval and in that interval, \(f\) is continuous at one specific point \(x_0\). Then we can build a new function \(F(x)\) by accumulating the area

$$ \begin{align*} F(x) = \int_a^x f(t)dt \end{align*} $$

This new function \(F\) represents the signed area under \(f\) from \(a\) up to the point \(x\). The claim is that this area function is differentiable at \(x_0\) and its derivative at \(x_0\) is the original function at \(x_0\). For \(F\) to be differentiable at \(x_0\) and for this value to be \(f(x_0)\), then by definition we want to show that

$$ \begin{align*} F'(x_0) = \lim_{h \to 0} \frac{F(x_0 + h) - F(x_0)}{h} = f(x_0) \end{align*} $$

But what is \(F(x_0 + h) - F(x_0)\)? Since we defined \(F(x) = \int_a^x f(t)dt\), then

$$ \begin{align*} F(x_0 + h) &= \int_{a}^{x_0+h} f(t)dt\\ F(x_0) &= \int_{a}^{x_0} f(t)dt \\ \end{align*} $$

Therefore

$$ \begin{align*} F(x_0 + h) - F(x_0) &= \int_{a}^{x_0+h} f(t)dt - \int_{a}^{x_0} f(t)dt \end{align*} $$

But by the additivity property of Riemann integrable functions, then

$$ \begin{align*} \int_{a}^{x_0+h} f(t)dt = \int_{a}^{x_0} f(t)dt + \int_{x_0}^{x_0+h} f(t) dt \end{align*} $$

Hence

$$ \begin{align*} \frac{F(x_0 + h) - F(x_0)}{h} = \frac{1}{h} \int_{x_0}^{x_0+h} f(t) dt \end{align*} $$

Recall our claim was that the limit of the quotient is \(f(x_0)\). Let \(\epsilon > 0\) be given. We want to show that

$$ \begin{align*} \left| \frac{F(x_0 + h) - F(x_0)}{h} - f(x_0) \right| = \left| \frac{1}{h} \int_{x_0}^{x_0+h} f(t) dt - f(x_0) \right| < \epsilon \end{align*} $$

To see this, recall that by assumption \(f\) is continuous at \(x_0\). This means that there exists a \(\delta > 0\) such that when \(|h| < \delta\), then

$$ \begin{align*} \left| f(t) - f(x_0) \right| < \epsilon \end{align*} $$

for all \(t \in [x_0, x_0+h]\). Then by the definition of absolute value

$$ \begin{align*} f(x_0) - \epsilon < f(t) < f(x_0) + \epsilon \end{align*} $$

If we integrate this inequality over \([x_0, x_0+h]\), then

$$ \begin{align*} \int_{x_0}^{x_0+h} (f(x_0) - \epsilon) dt< \int_{x_0}^{x_0+h} f(t)dt < \int_{x_0}^{x_0+h} (f(x_0) + \epsilon)dt \end{align*} $$

But \(\int_{x_0}^{x_0+h} (f(x_0) - \epsilon) = (f(x_0) - \epsilon) \int_{x_0}^{x_0+h} 1dt = (f(x_0)- \epsilon)h\). Hence

$$ \begin{align*} h(f(x_0) - \epsilon) &< \int_{x_0}^{x_0+h} f(t)dt < h(f(x_0) + \epsilon) < h(f(x_0) + \epsilon)\\ f(x_0) - \epsilon &< \frac{1}{h}\int_{x_0}^{x_0+h} f(t)dt < f(x_0) + \epsilon \end{align*} $$

Thus

$$ \begin{align*} \left| \frac{1}{h}\int_{x_0}^{x_0+h} f(t) - f(x_0) \right| < \epsilon \end{align*} $$

Hence \(F\) is differentiable at \(x_0\) and \(F'(x_0) = f(x_0)\) as we wanted to show. \(\blacksquare\)

Proof (b)

Since \(f'\) is Riemann integrable on \([a,b]\). Then for all \(\epsilon > 0\), there exists a partition \(\mathcal{P}_{\epsilon}\) such that for all \(\mathcal{P} \supset \mathcal{P}_{\epsilon}\) (any refinement), we have

$$ \begin{align*} \left| \sum_{j=1}^n f'(t_j)(x_j - x_{j-1}) - \int_a^b f'(t)dt \right| < \epsilon \end{align*} $$

for choices of all tags \(t_j\). Now, fix the refinement \(\mathcal{P}\)

$$ \begin{align*} \mathcal{P} = \{x_0 < x_1 < \cdots < x_n\} \end{align*} $$

Since \(f\) is differentiable, then on each subinterval \([x_{j-1}, x_j]\), the Mean Value Theorem applies to \(f\). Therefore, there exists some \(t_j \in (x_{j-1}, x_j)\) such that

$$ \begin{align*} f(x_j) - f(x_{j-1}) = f'(t_j)(x_j - x_{j-1}) \end{align*} $$

Using \(t_j\) as our tags,

$$ \begin{align*} \left| \sum_{j=1}^n f'(t_j)(x_j - x_{j-1}) - \int_a^b f'(t)dt \right| &< \epsilon \\ \left| \sum_{j=1}^n (f(x_j) - f(x_{j-1})) - \int_a^b f'(t)dt \right| &< \epsilon \\ \left| f(x_n) - f(x_0) - \int_a^b f'(t)dt \right| &< \epsilon\\ \left| f(b) - f(a) - \int_a^b f'(t)dt \right| &< \epsilon \\ \end{align*} $$

Since this holds for any \(\epsilon > 0\), then

$$ \begin{align*} \int_a^b f'(t)dt = f(b) - f(a) \end{align*} $$

References