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**1. Homework Statement**

Prove that if f(x) is a differentiable real-valued function, and f'(x) is continuous and integrable. Then:

[tex]\int_a^bf'(x)\,dx = f(b) - f(a)[/tex]

**2. Hint Provided**

Use the Cauchy Criterion for integrals, the mean value theorem applied to subintervals of a partition, and the relationship between Riemann integrals, Riemann sums, and Darboux sums.

**3. The Attempt at a Solution**

Let [tex]\epsilon > 0[/tex] be given. By the Cauchy Criterion for integrals, since f'(x) is integrable, there exists a partition P such that

(1) [tex]U(f'(x), P) - L(f'(x), P) < \epsilon[/tex]

where[tex]U(f'(x), P), L(f'(x), P)[/tex] represent the upper and lower darboux sums.

Now consider each subinterval [tex][t_{k-1}, t_k][/tex] of [tex]P = \{t_1 \leq t_2 \leq \cdots t_n\}[/tex], each of width [tex]t_{k} - t_{k-1}[/tex]. Since [tex]f(x)[/tex] is differentiable, by the mean value theorem, there exists [tex]x_k \in (t_{k-1}, t_k)[/tex] such that

[tex]f'(x_k) = \frac{f(t_k) - f(t_{k-1})}{t_k - t_{k-1}}[/tex]

or equivalently,

[tex]f'(x_k)(t_{k} - t_{k-1}) = f(t_k) - f(t_{k-1})[/tex]

Taking the sum of both sides and noting that since f(x) is continuous and thus [tex]\bigcup [f(t_k) - f(t_{k-1})] = f(b) - f(a)[/tex] we obtain:

[tex]\sum f'(x_k)(t_{k}-t_{k-1}) = f(b) - f(a)[/tex]

Since the sum on the left is a Riemann sum, we now know:

(2) [tex]L(f'(x), P) \leq f(b) - f(a) \leq U(f'(x), P)[/tex]

We also know that

[tex]L(f'(x), P) \leq \int_a^b f'(x)\,dx \leq U(f'(x), P)[/tex]

Equivalently,

(3) [tex]-U(f'(x), P) \leq -\int_a^b f'(x)\,dx \leq -L(f'(x), P)[/tex]

Adding inequalities (2) and (3) gives:

(4) [tex]L(f'(x), P) - U(f'(x), P) \leq (f(b) - f(a)) - \int_a^bf'(x)\,dx \leq U(f'(x), P) - L(f'(x), P)[/tex]

Equation (4) along with equation (1) implies that:

[tex]\left|(f(b) - f(a) - \int_a^bf'(x)\,dx\right| \leq U(f'(x), P) - L(f'(x), P) < \epsilon[/tex]

Since [tex]\epsilon[/tex] is very small depending on the partition, it might as well be zero, and from this it follows that [tex]\int_a^bf'(x)\,dx = f(b) - f(a)[/tex]

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