从上一节可知，对于一个连续函数在区间[a, b]的平均值，我们有公式：
a v g = 1 b − a ∫ a b f ( x ) d x avg = \frac{1}{b - a}\int_a^b f(x)dx a vg = b − a 1 ∫ a b f ( x ) d x 那么，是否存在一个自变量c ∈ [ a , b ] c \in [a, b] c ∈ [ a , b ]
f m i n ⋅ ( b − a ) ≤ ∫ a b f ( x ) d x ≤ f m a x ( b − a ) f_{min} \cdot (b - a) \leq \int_a^b f(x)dx \leq f_{max}(b - a) f min ⋅ ( b − a ) ≤ ∫ a b f ( x ) d x ≤ f ma x ( b − a ) 即
f m i n ≤ 1 b − a ∫ a b f ( x ) d x ≤ f m a x f_{min} \leq \frac{1}{b - a}\int_a^b f(x)dx \leq f_{max} f min ≤ b − a 1 ∫ a b f ( x ) d x ≤ f ma x 这里引用一下中间值定理：
"A function y = f(x) that is continuous on a close interval [a, b] takes on every value between f(a) and f(b). In other words, if y 0 y_0 y 0 y 0 = f ( c ) y_0 = f(c) y 0 = f ( c )
在这里y 0 = 1 b − a ∫ a b f ( x ) d x y_0 = \frac{1}{b - a}\int_a^b f(x)dx y 0 = b − a 1 ∫ a b f ( x ) d x c ∈ [ a , b ] c \in [a, b] c ∈ [ a , b ]
f ( c ) = 1 b − a ∫ a b f ( x ) d x f(c) = \frac{1}{b - a}\int_a^b f(x)dx f ( c ) = b − a 1 ∫ a b f ( x ) d x 这就是定积分的中值定理 。
如果将求定积分的区间作为自变量，比如区间[a, b]，固定从a开始，到x，那么可以构建一个函数：
F ( x ) = ∫ a x f ( t ) d t F(x) = \int_a^x f(t)dt F ( x ) = ∫ a x f ( t ) d t 注意这里的变量x和t有不同的意义。那么F(x)和原函数f(t)有什么关系呢？这里给出结论，即：
d d x F ( x ) = d d x ∫ a x f ( t ) d t = f ( x ) \frac{d}{dx}F(x) = \frac{d}{dx}\int_a^x f(t)dt = f(x) d x d F ( x ) = d x d ∫ a x f ( t ) d t = f ( x ) 我们知道根据导数的定义：
F ′ ( x ) = lim h → 0 F ( x + h ) − F ( x ) h F'(x) = \lim_{h \to 0}\frac{F(x + h) - F(x)}{h} F ′ ( x ) = h → 0 lim h F ( x + h ) − F ( x ) 根据函数F的定义，可知：
F ( x + h ) − F ( x ) = ∫ a x + h f ( t ) d t − ∫ a x f ( t ) d t = ∫ x x + h f ( t ) d t F(x + h) - F(x) = \int_a^{x + h}f(t)dt - \int_a^x f(t)dt = \int_x^{x + h}f(t)dt F ( x + h ) − F ( x ) = ∫ a x + h f ( t ) d t − ∫ a x f ( t ) d t = ∫ x x + h f ( t ) d t 两边除以h，得到：
F ( x + h ) − F ( x ) h = 1 h ∫ x x + h f ( t ) d t \frac{F(x + h) - F(x)}{h} = \frac{1}{h}\int_x^{x + h}f(t)dt h F ( x + h ) − F ( x ) = h 1 ∫ x x + h f ( t ) d t 根据不定积分的中值定理，存在一个c ∈ [ x , x + h ] c \in [x, x + h] c ∈ [ x , x + h ]
f ( c ) = 1 x + h − x ∫ x x + h f ( t ) d t = 1 h ∫ x x + h f ( t ) d t f(c) = \frac{1}{x + h - x} \int_x^{x + h} f(t)dt = \frac{1}{h} \int_x^{x + h} f(t)dt f ( c ) = x + h − x 1 ∫ x x + h f ( t ) d t = h 1 ∫ x x + h f ( t ) d t 那么：
F ( x + h ) − F ( x ) h = f ( c ) lim h → 0 F ( x + h ) − F ( x ) h = lim h → 0 f ( c ) \frac{F(x + h) - F(x)}{h} = f(c) \\
\lim_{h \to 0} \frac{F(x + h) - F(x)}{h} = \lim_{h \to 0} f(c) h F ( x + h ) − F ( x ) = f ( c ) h → 0 lim h F ( x + h ) − F ( x ) = h → 0 lim f ( c ) 当h → 0 h \to 0 h → 0 c → x c \to x c → x
lim h → 0 F ( x + h ) − F ( x ) h = f ( x ) \lim_{h \to 0} \frac{F(x + h) - F(x)}{h} = f(x) h → 0 lim h F ( x + h ) − F ( x ) = f ( x ) 也就是：
F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) 这就是微积分基本定理 ，引用书本原文：
"If f is continuous on [a, b] then F(x) = \int_a^x f(t)dt is continuous on [a, b] and differentiable on (a, b) and its derivative is f(x);
F ′ ( x ) = d d x ∫ a x f ( t ) d t = f ( x ) . F'(x) = \frac{d}{dx}\int_a^xf(t)dt = f(x). F ′ ( x ) = d x d ∫ a x f ( t ) d t = f ( x ) . "
从这一定理我们可知，如果：
F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) 那么
∫ a x f ( t ) d t = F ( x ) \int_a^x f(t)dt = F(x) ∫ a x f ( t ) d t = F ( x ) 我们设G为f的任意一个原函数，那么根据拉格朗日中值定理：
G ( x ) = F ( x ) + C G(x) = F(x) + C G ( x ) = F ( x ) + C 其中C为常数。假设定积分区间为[a, b]，那么：
G ( a ) = F ( a ) + C = ∫ a a f ( t ) d t + C = 0 + C G(a) = F(a) + C = \int_a^af(t)dt + C = 0 + C G ( a ) = F ( a ) + C = ∫ a a f ( t ) d t + C = 0 + C G ( b ) = F ( b ) + C = ∫ a b f ( t ) d t + C G(b) = F(b) + C = \int_a^bf(t)dt + C G ( b ) = F ( b ) + C = ∫ a b f ( t ) d t + C 两式相减可得：
G ( b ) − G ( a ) = ∫ a b f ( t ) d t + C − C = ∫ a b f ( t ) d t G(b) - G(a) = \int_a^bf(t)dt + C - C = \int_a^b f(t)dt G ( b ) − G ( a ) = ∫ a b f ( t ) d t + C − C = ∫ a b f ( t ) d t 所以得出微积分基本定理的推论 ：
"If f is continuous at every point of [a, b] and F is any antiderivative of f on [a, b], then
∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^bf(x)dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) "
以上便是本节最重要的内容，它是表明微分和积分关系的重要定理。本节还讨论了面积的求解方法，本质上就一个点，注意图像正负的意义，以及导致的分段求解需求即可。
练习
∫ − 2 0 ( 2 x + 5 ) d x = [ x 2 + 5 x ] − 2 0 = 0 − ( 4 − 10 ) = 6 \int_{-2}^0(2x + 5)dx = \Big[x^2 + 5x\Big]_{-2}^0 = 0 - (4 -10) = 6 ∫ − 2 0 ( 2 x + 5 ) d x = [ x 2 + 5 x ] − 2 0 = 0 − ( 4 − 10 ) = 6
2.
∫ − 3 4 ( 5 − x 2 ) d x = [ 5 x − x 2 4 ] − 3 4 = ( 20 − 4 ) − ( − 15 − 9 4 ) = 133 4 \int_{-3}^4(5 - \frac{x}{2})dx = \Big[5x - \frac{x^2}{4}\Big]_{-3}^4 = (20 - 4) - (-15 - \frac{9}{4}) = \frac{133}{4} ∫ − 3 4 ( 5 − 2 x ) d x = [ 5 x − 4 x 2 ] − 3 4 = ( 20 − 4 ) − ( − 15 − 4 9 ) = 4 133
3.
∫ 0 4 ( 3 x − x 3 4 ) d x = [ 3 x 2 2 − x 4 16 ] 0 4 = ( 24 − 16 ) − 0 = 8 \int_0^4(3x - \frac{x^3}{4})dx = \Big[\frac{3x^2}{2} - \frac{x^4}{16}\Big]_0^4 = (24 - 16) - 0 = 8 ∫ 0 4 ( 3 x − 4 x 3 ) d x = [ 2 3 x 2 − 16 x 4 ] 0 4 = ( 24 − 16 ) − 0 = 8
4.
∫ − 2 2 ( x 3 − 2 x + 3 ) d x = [ x 4 4 − x 2 + 3 x ] − 2 2 = ( 4 − 4 + 6 ) − ( 4 − 4 − 6 ) = 12 \int_{-2}^2(x^3 - 2x + 3)dx = \Big[\frac{x^4}{4} - x^2 + 3x\Big]_{-2}^2 = (4 - 4 + 6) - (4 - 4 - 6) = 12 ∫ − 2 2 ( x 3 − 2 x + 3 ) d x = [ 4 x 4 − x 2 + 3 x ] − 2 2 = ( 4 − 4 + 6 ) − ( 4 − 4 − 6 ) = 12 ...
27.
d d x ∫ 0 x cos t d t = d d u ∫ 0 u cos t d t ⋅ d d x x = sin x ⋅ 2 x 3 \frac{d}{dx}\int_0^{\sqrt{x}}\cos tdt = \frac{d}{du}\int_0^u \cos tdt \cdot \frac{d}{dx}\sqrt{x} = \sin \sqrt{x} \cdot \frac{2\sqrt{x}}{3} d x d ∫ 0 x cos t d t = d u d ∫ 0 u cos t d t ⋅ d x d x = sin x ⋅ 3 2 x
28.
d d x ∫ 1 sin x 3 t 2 d t = d d u ∫ 1 u 3 t 2 d t ⋅ d d x sin x = 3 sin 2 x ⋅ − cos x \frac{d}{dx}\int_1^{\sin x}3t^2dt = \frac{d}{du}\int_1^u3t^2dt \cdot \frac{d}{dx}\sin x = 3\sin^2 x \cdot -\cos x d x d ∫ 1 s i n x 3 t 2 d t = d u d ∫ 1 u 3 t 2 d t ⋅ d x d sin x = 3 sin 2 x ⋅ − cos x
29.
d d t ∫ 0 t 4 u d u = d d x ∫ 0 x u d u ⋅ d d t t 4 = 2 t 2 3 ⋅ t 5 5 \frac{d}{dt}\int_0^{t^4}\sqrt{u}du = \frac{d}{dx}\int_0^x\sqrt{u}du \cdot \frac{d}{dt}t^4 = \frac{2t^2}{3} \cdot \frac{t^5}{5} d t d ∫ 0 t 4 u d u = d x d ∫ 0 x u d u ⋅ d t d t 4 = 3 2 t 2 ⋅ 5 t 5
30.
d d θ ∫ 0 tan θ sec 2 y d y = d d u ∫ 0 u sec 2 y d y ⋅ d d θ tan θ = sec 2 ( tan θ ) ⋅ sec 2 θ \frac{d}{d\theta}\int_0^{\tan \theta}\sec^2 y dy = \frac{d}{du}\int_0^u \sec^2 ydy \cdot \frac{d}{d\theta}\tan \theta = \sec^2 (\tan \theta) \cdot \sec^2 \theta d θ d ∫ 0 t a n θ sec 2 y d y = d u d ∫ 0 u sec 2 y d y ⋅ d θ d tan θ = sec 2 ( tan θ ) ⋅ sec 2 θ 
