Relaxing the Fundamental Theorem

Problem 1. Let $f : [a, b] \to\mathbb R$ be differentiable such that $f'$ is Riemann-integrable. Prove that

$\displaystyle \int_a^b f' = f(b) - f(a),\quad \frac{\mathrm d}{\mathrm d x} \int_a^x f' = f'(x).$

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Solution. Fix $\epsilon > 0$ . For any $k > 0$ , let $P \subseteq [a, b]$ be any partition satisfying

$\displaystyle \sum_{i=1}^n (M_i(f', P) - m_i(f', P)) \Delta x_i < k \cdot \epsilon.$

For each $i = 1, \dots, n$ , use the mean value theorem to find $c_i \in (x_{i-1}, x_i)$ such that

$f(x_i) - f(x_{i-1}) = f'(c_i)\Delta x_i,$

and furthermore $f'(c_i) \in [m_i(f', P), M_i(f', P)]$ . This implies

$\displaystyle \begin{aligned} f(b) - f(a) - L(f', P)&= \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \Delta x_i - L(f', P) \\ &= \sum_{i=1}^n (f'(c_i) - m_i(f',P))\Delta x_i \\ &\leq \sum_{i=1}^n (M_i(f', P) - m_i(f',P))\Delta x_i \in [0, k \cdot \epsilon).\end{aligned}$