Fundamental Theorem of Calculus
Our main goal in this section will be to prove the Fundamental theorem of Calculus, which states that integration and derivation are inverse operations.
If f be integrable on [a,b], and for each x \in [a,b], we set
F(x) = \int^x_a f(t)\,dt,
then F is uniformly continuous on [a,b]. Furthermore if f is continuous at c \in [a,b], then F is differentiable at c and
F^\prime(c) = f(c)
Direct Proof
Since f is Riemann integrable on [a,b], it is bounded by some B. We know f is uniformly continuous on [a,b], therefore given any x,y \in [a,b] such that x < y and |x-y| < \delta = \varepsilon/B, then
\begin{align*} |F(x) - F(y)| &= \left| \int^y_a f - \int^x_a f \right|\\ &= \left| \int^y_a f + \int^a_x f \right| \text{ (Theorem $4.2.4$)}\\ &= \left| \int^y_x f \right| \leq \int^y_x |f| \leq \int^y_x B\\ &\leq \frac{\varepsilon}{B}B = \varepsilon \end{align*}
Thus F is uniformly continuous on [a,b].
Furthermore, suppose f is continuous at some point c \in [a,b]. Given any \varepsilon > 0, there exists \delta > 0 such that |f(t) - f(c)| < \varepsilon whenever t \in [a,b] and |t-c| < \delta. Since f(c) is a constant, we write
f(c) = \frac{1}{x-c}\int^x_c f(c) \,dt, \quad \text{for } x \not = c.
Since f(c) is just a number, the integral evaluates to (x-c)f(c). Then for any x \in [a,b] such that 0 < |x-c| < \delta, we have
\begin{align*} \left|\frac{F(x) - F(c)}{x-c} - f(c)\right| &= \left| \frac{1}{x-c} \left[ \int^x_a f - \int^c_a f\right] - f(c) \right|\\ &= \left| \frac{1}{x-c} \int^x_c f - \frac{1}{x-c} \int^x_c f(c) \right| \text{ (Theorem $4.2.4$)}\\ &= \left| \frac{1}{x-c} \right| \left| \int^x_c [f - f(c)] \right| \\ &\leq \left| \frac{1}{x-c} \right| \int^x_c \left| f - f(c) \right|\\ &<\left| \frac{1}{x-c} \right| \varepsilon |x-c| = \varepsilon \end{align*}
In the last step, because t is inbetween x and c, we know that |t - c| < \delta, and thus |f(t) - f(c)| < \varepsilon. Since \varepsilon > 0 was arbitrary, we conclude that
F^\prime(c) = \lim_{x \to c} \frac{F(x) - F(c)}{x - c} = f(c)
Let f be continuous on [a,b] and g be differentiable on [c,d], where g([c,d]) \subseteq [a,b]. Define
F(x) = \int^{g(x)}_a f, \quad \text{for all $x \in [c,d]$.}
Then F is differentiable on [c,d] and F^\prime(x) = g(f(x))g^\prime(x).
Direct Proof
Let G(x) = \int^x_a f for x \in [a,b] so that F = G \circ g on [c,d].
Let f be differentiable on [a,b] and f^\prime is Riemann integerable on [a,b]. Then
\int^b_a f^\prime = f(b) - f(a).
Direct Proof
Let P = \{x_0,\dots,x_n\} be a partition of [a,b]. On every subinterval [x_{i-1},x_i], apply Mean Value Theorem to obtain t_i \in (x_{i-1},x_i) such that
f(x_i) - f(x_{i-1}) = f^\prime(t_i)(x_i - x_{i-1}).
Thus
f(b) - f(a) = \sum^n_{i=1} f(x_i) - f(x_{i-1}) = \sum^n_{i=1} f^\prime(t_i)\Delta x_i.
Then for all i, we have that m_i(f^\prime) \leq f^\prime(t_i) \leq M_i(f^\prime). It follows that
L(f^\prime,P) \leq f(b) - f(a) \leq U(f^\prime,P).
Since this holds for every partition P, we also have
L(f^\prime) \leq f(b) - f(a) \leq U(f^\prime).
And since f^\prime is assumed to be Riemann integrable on [a,b], we have
\int^b_a f^\prime = f(b) - f(a).