The relation between integration and differentiation constitutes the key study of calculus.

Fundamental Theorem of Calculus

  1. If f:[a,b]→f : [a, b] \rightarrow is bounded and Riemann integrable, and F(x)=∫axf(t) dtF(x) = \int_a^x f(t) \ dt for x∈[a,b]x \in [a, b], then F(x)F(x) is uniformly continuous and F′(x)F'(x) exists for all xx where f(x)f(x) is continuous and F′(x)=f(x)F'(x) = f(x).
  2. If F:[a,b]→RF: [a, b] \rightarrow R is differentiable and f:[a,b]→Rf: [a, b] \rightarrow R is integrable. Suppose F′(x)=f(x)F'(x) = f(x) for all x∈[a,b]x \in [a, b], then
F(b)−F(a)=∫abf(t) dt F(b) - F(a) = \int_a^b f(t) \ dt

Remark:

  1. Claim one require the condition that ff is bounded. Can we omit the condition?

  2. The condition ff is integrable is necessary.

Proof:

  1. As ff is bounded, we assume that sup⁡t∈[a,b]∣f(t)∣≤M\sup_{t \in [a, b]} |f(t)| \le M. For any x∈[a,b]x \in [a, b] and x+h∈[a,b]x + h \in [a, b],

    ∣F(x+h)−F(x)∣=∣∫xx+hf(t) dt∣≤M∣h∣ \left| F(x + h) - F(x) \right| = \left| \int_x^{x + h} f(t) \ dt \right| \le M|h|

    This show that F(x)F(x) is uniformly continuous.

    Now, assume that f(x)f(x) is continuous at point xx. ∀ϵ>0\forall \epsilon > 0, ∃δ>0\exists \delta > 0, s.t., ∀t∈(x±δ)\forall t \in (x \pm \delta), ∣f(t)−f(x)∣≤ϵ|f(t) - f(x)| \le \epsilon. We see that

    ∣F(x+h)−F(x)−f(x)h∣=∣∫xx+hf(t) dt−∫xx+hf(x) dt∣=∣∫xx+h[f(t)−f(x)] dt∣≤∫xx+h∣f(t)−f(x)∣ dt≤ϵh \begin{aligned} \left| F(x + h) - F(x) - f(x) h \right| &= \left| \int_x^{x + h} f(t) \ dt - \int_x^{x + h} f(x) \ dt \right| \\ &= \left| \int_x^{x + h} [f(t) - f(x)] \ dt \right| \\ &\le \int_x^{x + h} \left| f(t) - f(x) \right| \ dt \\ &\le \epsilon h \end{aligned}

    As desired.

  2. For partition P={a=x0<x1<x2<...<xn=b}P = \{ a = x_0 < x_1 < x_2 < ... < x_n = b \} of [a,b][a, b], apply the mean value theorem to FF on a sub-interval.

    F(xk)−F(xk−1)=F′(tk)(xk−xk−1)tk∈[xk−1,xk]∀ 1≤k≤n, F(x_k) - F(x_{k - 1}) = F'(t_k) (x_k - x_{k -1}) \quad t_k \in [x_{k - 1}, x_k] \quad \forall \ 1 \le k \le n,

    Let U(f,P),L(f,)U(f, P), L(f, ) be the upper sum and lower sum of the partition respectively,

    F(b)−F(a)=∑k=1nF(xk)−F(xk−1)=∑k=1nF′(tk)(xk−xk−1)=∑k=1nf(tk)(xk−xk−1)∈[L(f,P),U(f,P)] \begin{aligned} F(b) - F(a) &= \sum_{k = 1}^n F(x_k) - F(x_{k - 1}) \\ &= \sum_{k = 1}^n F'(t_k) (x_k - x_{k -1}) \\ &= \sum_{k = 1}^n f(t_k) (x_k - x_{k -1}) \\ &\in [L(f, P), U(f, P)] \end{aligned}

    This is independent of the choice of PP. Taking the limit of upper sum and lower sum, we conclude that F(b)−F(a)∈[sup⁡PL(f,P),inf⁡PU(f,P)]=[∫abf(t) dt,∫abf(t) dt]F(b) - F(a) \in [\sup_P L(f, P), \inf_P U(f, P)] = \left[ \int_a^b f(t) \ dt, \int_a^b f(t) \ dt \right].