累積函數
設 f 在 [a,b] 上可積。定義累積函數 F(x) = ∫ 從 a 到 x 的 f(t) dt。這個 F 度量右端從 a 移到 x 時掃出的累計面積。微積分基本定理第一部分說：若 f 在某點 x 處連續，則 F 在該點可導且 F′(x) = f(x)。先掃出面積、再求導，就還原出原來的 f。
FTC, Part 1. f integrable on [a,b], continuous at x. Then F'(x) = f(x),
where F(x) = integral from a to x of f.
Proof. Look at the difference quotient. For h > 0 (h < 0 is symmetric),
F(x+h) - F(x) = integral from a to x+h - integral from a to x
= integral from x to x+h of f(t) dt (additivity).
Subtract the constant f(x), written as integral from x to x+h of f(x) dt = f(x)*h:
F(x+h) - F(x) - f(x)*h = integral from x to x+h of ( f(t) - f(x) ) dt.
Let ε > 0. Continuity at x: there is δ>0 with |t - x| < δ => |f(t) - f(x)| < ε.
For 0 < h < δ every t in [x, x+h] satisfies |t - x| < δ, so by |∫g| <= ∫|g|:
| F(x+h) - F(x) - f(x)*h | <= integral from x to x+h of |f(t)-f(x)| dt
<= ε * h.
Divide by h: | (F(x+h) - F(x))/h - f(x) | <= ε.
Since ε was arbitrary, the difference quotient -> f(x). Hence F'(x) = f(x). QED用原函數求值
第二部分才是你真正用來計算的。若 f 在 [a,b] 上可積，且 G 是 f 的任意一個原函數——即在 [a,b] 上處處 G′ = f——則 ∫ 從 a 到 b 的 f = G(b) − G(a)。你完全不必碰任何分割；只需找一個原函數，再把兩端點的值相減。
FTC, Part 2. f integrable on [a,b], G'(x) = f(x) for all x in [a,b].
Then integral from a to b of f = G(b) - G(a).
Proof. Take ANY partition P: a = x_0 < ... < x_n = b. Telescope G across it:
G(b) - G(a) = sum_{k=1}^n ( G(x_k) - G(x_{k-1}) ).
Apply the Mean Value Theorem to G on each [x_{k-1}, x_k]: there is a point
t_k in (x_{k-1}, x_k) with G(x_k) - G(x_{k-1}) = G'(t_k) Δx_k = f(t_k) Δx_k.
So G(b) - G(a) = sum f(t_k) Δx_k = a Riemann sum of f for P.
But on each piece m_k <= f(t_k) <= M_k, hence
L(f,P) <= G(b) - G(a) <= U(f,P) for EVERY partition P.
The only number squeezed between all lower and all upper sums is the integral:
integral from a to b of f = G(b) - G(a). QED一個有用的夥伴：積分中值定理
一個近親是積分中值定理：若 f 在 [a,b] 上連續，則存在點 c ∈ [a,b]，使 ∫ 從 a 到 b 的 f = f(c)·(b − a)。從幾何上看，這塊面積等於一個矩形的面積，矩形高度恰是函數在某個內點處的值——平均值 f(c) 確實被取到。
Integral MVT. f continuous on [a,b]. Let m = min f, M = max f on [a,b]. Monotonicity of the integral gives m*(b-a) <= integral from a to b of f <= M*(b-a), so the AVERAGE value A = (1/(b-a)) * integral from a to b of f lies in [m, M]. f is continuous and attains m and M (Extreme Value Thm), so by the Intermediate Value Theorem it attains every value in between, including A. Hence there is c in [a,b] with f(c) = A, i.e. integral from a to b of f = f(c)*(b - a). QED