Fundamental Theorem of Calculus

[james121515]

1. Homework Statement

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

$$\int_a^bf'(x)\,dx = f(b) - f(a)$$

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

(1) $$U(f'(x), P) - L(f'(x), P) < \epsilon$$

where$$U(f'(x), P), L(f'(x), P)$$ represent the upper and lower darboux sums.
Now consider each subinterval $$[t_{k-1}, t_k]$$ of $$P = \{t_1 \leq t_2 \leq \cdots t_n\}$$, each of width $$t_{k} - t_{k-1}$$. Since $$f(x)$$ is differentiable, by the mean value theorem, there exists $$x_k \in (t_{k-1}, t_k)$$ such that

$$f'(x_k) = \frac{f(t_k) - f(t_{k-1})}{t_k - t_{k-1}}$$

or equivalently,

$$f'(x_k)(t_{k} - t_{k-1}) = f(t_k) - f(t_{k-1})$$

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

$$\sum f'(x_k)(t_{k}-t_{k-1}) = f(b) - f(a)$$

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

(2) $$L(f'(x), P) \leq f(b) - f(a) \leq U(f'(x), P)$$

We also know that

$$L(f'(x), P) \leq \int_a^b f'(x)\,dx \leq U(f'(x), P)$$

Equivalently,

(3) $$-U(f'(x), P) \leq -\int_a^b f'(x)\,dx \leq -L(f'(x), P)$$

Adding inequalities (2) and (3) gives:

(4) $$L(f'(x), P) - U(f'(x), P) \leq (f(b) - f(a)) - \int_a^bf'(x)\,dx \leq U(f'(x), P) - L(f'(x), P)$$

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

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

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

 

[Mark44]

No, this is the right forum section.