Let \(f:[a,b]\to\mathbb R\) be differentiable and suppose \(f'\) is Riemann integrable. Prove that \[ \int_a^b f'(x)\,dx=f(b)-f(a). \]
Solution
For each positive integer \(N\), let \[ x_i \coloneqq a+\frac{i}{N}(b-a), \qquad i=0,\ldots,N. \]
1. For every \(i=1,\ldots,N\), there is a point\[ c_i\in(x_{i-1},x_i) \]such that\[ f(x_i)-f(x_{i-1}) = f'(c_i)(x_i-x_{i-1}). \]
The differentiable function \(f\) is continuous. Apply the mean value theorem to \(f\) on each interval \([x_{i-1},x_i]\).
2. For every \(N\),\[ f(b)-f(a) = \sum_{i=1}^N f'(c_i)(x_i-x_{i-1}). \]
1. The left-hand side telescopes:\[ \sum_{i=1}^N \bigl(f(x_i)-f(x_{i-1})\bigr) = f(x_N)-f(x_0) = f(b)-f(a). \]
3. The right-hand side in step 2 is a tagged Riemann sum for \(f'\)whose mesh tends to zero as \(N\to\infty\).
The points \(c_i\) are tags satisfying \[ c_i\in[x_{i-1},x_i]. \] The partition is uniform, with mesh \[ \max_i(x_i-x_{i-1}) = \frac{b-a}{N} \longrightarrow0. \] Thus the displayed sum is precisely a tagged Riemann sum.
4. One has\[ \lim_{N\to\infty} \sum_{i=1}^N f'(c_i)(x_i-x_{i-1}) = \int_a^b f'(x)\,dx. \]
The function \(f'\) is Riemann integrable by hypothesis. For a Riemann integrable function, every sequence of tagged Riemann sums whose mesh tends to zero converges to the Riemann integral.
5. Therefore\[ \boxed{ \int_a^b f'(x)\,dx = f(b)-f(a) }. \]
The sums in step 2 are equal to the fixed number \(f(b)-f(a)\) for every \(N\), while step 4 says that the same sums converge to the integral.
6. Q.E.D.
Step 5 is the required identity.