定积分比较大小

\int_{0}^{\frac{π}{2}} x d x > \int_{0}^{\frac{π}{2}} sin x d x ∵ x > sin x ∴ \int_{0}^{\frac{π}{2}} x d x > \int_{0}^{\frac{π}{2}} sin x d x

图像法

比较 $\int_{0}^{1} x^{2} d x$ 与 $\int_{0}^{1} x^{3} d x$
代入法：

比较 [0, 1] 内函数的大小 f (x) = x^{2} f (\frac{1}{2}) = \frac{1}{4} g (x) = x^{3} g (\frac{1}{2}) = \frac{1}{8} f (x) > g (x) ∴ \int_{0}^{1} x^{2} d x > \int_{0}^{1} x^{3} d x

比较 $\int_{0}^{1} x^{2} d x$ 与 $\int_{0}^{1} x d x$
对于第二个

y = x y^{2} = x (开口向右的抛物线)

∴ \int_{0}^{1} x^{2} d x < \int_{0}^{1} x d x

比较积分 $\int_{1}^{2} ln x d x$ 和 $\int_{1}^{2} (ln x)^{2} d x$ 的大小

当 1 \leq x \leq 2, 0 = ln 1 \leq ln x \leq ln e = 1 即 0 \leq ln x \leq 1 我们知道当 0 \leq u \leq 1, u^{2} \leq u 而 u > 1 时， u^{2} > u ∴ ln x \geq (ln x)^{2} ，仅在 x = 1 处成立 ∴ \int_{1}^{2} ln x d x > \int_{1}^{2} (ln x)^{2} d x

定积分性质

设f(x)在[a,b]上连续，下列不正确的是
A. $\int_{0}^{x} f (t) d t$ 是f(x)的一个原函数

看 $\int_{0}^{x} f (t) d t$ 求导后是否为 $f (t)$

(\int_{0}^{x} f (t) d t)^{'} = f (x) \cdot 1 = f (x)

所以正确

B. $\int_{x}^{b} f (t) d t$ 是 $- f (x)$ 的一个原函数

(\int_{x}^{b} f (t) d t)^{'} = f (b) \cdot b^{'} - f (x) = - f (x)

所以正确

C. $\int_{a}^{b} f (x) d x$ 是 $f (x)$ 的一个原函数

$\int_{a}^{b} f (x) d x$ 是一个数，一个数求导为0，不等于f(x)
错误答案选C。

D. $f (x)$ 在[a,b]上可积

判断这句话是否正确，若在[a,b]上f(x),g(x)均连续且 $f (x) \neq = g (x)$ ，则 $\int_{a}^{b} f (x) d x \neq = \int_{a}^{b} g (x) d x$ 这句话对吗
$令 a = - 1, b = 1, f (x) = x, g (x) = - x$
$\int_{- 1}^{1} x d x 与 \int_{- 1}^{1} - x d x$ 都为0

显然不成立

如果在区间[a,b]上

g (x) = h (x) (\forall x \in [a, b])

如果函数g(x),h(x)可积，那么

\int_{a}^{b} g (x) d x = \int_{a}^{b} h (x) d x

设f(x)在区间[a,b]可积，且f(x)=1

\int_{a}^{b} 1 d x = b - a 证明： x_{a}^{b} = b - a

设f(x)在区间[a,b]可积

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x 证明： - \int_{b}^{a} f (x) d x = - [F (a) - F (b)] = F (b) - F (a)

如何估值

估计 $\int_{0}^{1} e^{x^{2}} d x$ 的大小，直接算不出来这个积分，只能估算

0 < x < 1 0 < x^{2} < 1 e^{0} < e^{x^{2}} < e^{1} 1 < e^{x^{2}} < e 两边同取定积分 \int_{0}^{1} 1 d x < \int_{0}^{1} e^{x^{2}} < \int_{0}^{1} e d x x_{0}^{1} < \int_{0}^{1} e^{x^{2}} < e x_{0}^{1} 1 < \int_{0}^{1} e^{x^{2}} < e

估计定积分 $\int_{- 1}^{1} e^{- x} d x$

- 1 < x < 1 1 > - x > - 1 - 1 < - x < 1 e^{- 1} < e^{- x} < e \int_{- 1}^{1} e^{- 1} d x < \int_{- 1}^{1} e^{- x} < \int_{- 1}^{1} e d x e^{- 1} x_{- 1}^{1} < \int_{- 1}^{1} e^{- x} < e x_{- 1}^{1} e^{- 1} x_{- 1}^{1} = e^{- 1} - (- 1) \cdot e^{- 1} = 2 e^{- 1} e x_{- 1}^{1} = e - (- e) = 2 e ∴ \frac{2}{e} < \int_{- 1}^{1} e^{- x} < 2 e

估值定理

估值定理：如果f(x)在[a,b]上的最大值为M，最小值为m，那么

m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)

证明：

m \leq f (x) \leq M \int_{a}^{b} m d x \leq \int_{a}^{b} f (x) d x \leq \int_{a}^{b} M d x m \int_{a}^{b} d x \leq \int_{a}^{b} f (x) d x \leq M \int_{a}^{b} d x m x_{a}^{b} \leq \int_{a}^{b} f (x) d x \leq M x_{a}^{b} m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)

另一种方法，用导数判断单调性，结合估值定理

令 f (x) = e^{- x} f^{'} (x) = - e^{- x} < 0 ∴ f (x) 在 [- 1, 1] 单调递减 f (x)_{ma x} = f (x)_{- 1} = e^{- (- 1)} = e f (x)_{min} = f (x)_{1} = e^{- 1} = \frac{1}{e} 根据估值定理 \frac{1}{e} \cdot (1 - (- 1)) \leq \int_{- 1}^{1} e^{- x} d x \leq e (1 - (- 1)) \frac{2}{e} \leq \int_{- 1}^{1} e^{- x} d x \leq 2 e

估计积分 $\int_{\frac{π}{4}}^{\frac{3 π}{4}} (1 + sin^{2} x) d x$ 的值

令 f (x) = 1 + sin^{2} x, x \in [\frac{π}{4}, \frac{3 π}{4}] f^{'} (x) = 2 sin cos x = sin 2 x, x \in [\frac{π}{4}, \frac{3 π}{4}] 令 sin 2 x = 0 \Rightarrow 2 x = kπ \Rightarrow x = \frac{kπ}{2} ∵ x \in [\frac{π}{4}, \frac{3 π}{4}] ∴ x = \frac{π}{2} 为驻点

x	$[\frac{π}{4}, \frac{π}{2}]$	$\frac{π}{2}$	$[\frac{π}{2}, \frac{3 π}{4}]$
$f^{'} (x)$	+	0	-
$f (x)$	递增	极大值	递减
比较驻点，和端点值

f (\frac{π}{2}) f (\frac{π}{4}) f (\frac{3 π}{4}) ∴ f (x)_{ma x} f (x)_{min} 根据估值定理 = 1 + (sin \frac{π}{2})^{2} = 1 + 1 = 2 = 1 + (sin \frac{π}{4})^{2} = 1 + (\frac{2}{2})^{2} = 1 + \frac{1}{2} = \frac{3}{2} = 1 + [sin (π - \frac{π}{4})]^{2} = 1 + (sin \frac{π}{4})^{2} = \frac{3}{2} = M = 2 = m = \frac{3}{2} m \cdot (\frac{3 π}{4} - \frac{π}{4}) = \frac{3}{2} \cdot \frac{π}{2} = \frac{3 π}{4} M \cdot \frac{π}{2} = 2 \cdot \frac{π}{2} = π ∴ \frac{3 π}{4} \leq \int_{\frac{π}{4}}^{\frac{3 π}{4}} (1 + sin^{2} x) d x \leq π

积分中值定理

设f(x)在[a,b]上连续，则至少存在一点 $ξ \in [a, b]$ (也可以是开区间 $ξ \in (a, b)$ )使

\int_{a}^{b} f (x) d x = f (ξ) (b - a)

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其中，

f (ξ) = \frac{1}{b - a} \int_{a}^{b} f (x) d x

称为函数f(x)在区间[a,b]上的平均值。

证明：在估值定理的最后一步继续

m \leq \frac{\int _{a}^{b} f ( x ) d x}{b - a} \leq M 由介值定理得，至少 \exists ξ \in (a, b) ，使 f (ξ) = \frac{\int _{a}^{b} f ( x ) d x}{b - a} 即 \int_{a}^{b} f (x) d x = f (ξ) (b - a)

mean value theorem 介值定理

设f(x)在[0,2]上连续，f(0)=-1, $\frac{1}{2} \int_{0}^{2} f (x) d x = 1$ ，证明 $\exists ξ \in (0, 2)$ ，使 $f (ξ) = 0$

f (x) 在 [0, 2] 上连续 由积分中职定理 \exists ξ_{1} \in (0, 2) 使 \int_{0}^{2} f (x) d x = 2 = 2 f (ξ_{1}) 即 f (ξ_{1}) = 1 又 ∵ f (0) = - 1 f (0) f (ξ_{1}) < 0 由零点定理得至少存在一点 ξ \in (0, 2) ，使 f (ξ) = 0

函数 $f (x) = 2 x + e^{3 x}$ 在闭区间[1,2]上的平均值为？

\frac{1}{2 - 1} \int_{1}^{2} (2 x + e^{3 x}) d x 2 \int_{1}^{2} x d x + \int_{1}^{2} e^{3 x} d x 2 \cdot \frac{x ^{2}}{2}_{1}^{2} + \frac{1}{3} e^{3 x}_{1}^{2} 3 + \frac{1}{3} e^{6} - \frac{1}{3} e^{3}

求n项和极限

\int_{0}^{1} f (x) d x = n \to \infty lim [f (\frac{1}{n}) \cdot \frac{1}{n} + f (\frac{2}{n}) \cdot \frac{1}{n} + f (\frac{3}{n}) \cdot \frac{1}{n} + \dots f (\frac{i}{n}) \cdot \frac{1}{n} + f (\frac{n}{n}) \cdot \frac{1}{n}]

Sigma notation

n \to \infty lim [f (\frac{1}{n}) \cdot \frac{1}{n} + f (\frac{2}{n}) \cdot \frac{1}{n} + f (\frac{3}{n}) \cdot \frac{1}{n} + \dots f (\frac{i}{n}) \cdot \frac{1}{n} + f (\frac{n}{n}) \cdot \frac{1}{n}] = n \to \infty lim i = 1 \sum n f (\frac{i}{n}) \cdot \frac{1}{n}

n \to \infty lim \frac{1}{n} (1 + \frac{1}{n} + 1 + \frac{2}{n} + \dots + 1 + \frac{n}{n})

第一步写求和符号

i = 1 \sum n 1 + \frac{i}{n}

第二步找 $\frac{i}{n}$ 和 $\frac{1}{n}$

对 i 进行求和, \frac{1}{n} 可以提出来 原式 = n \to \infty lim i = 1 \sum n 1 + \frac{i}{n} \cdot \frac{1}{n}

第三步对应

\frac{i}{n} \to x \frac{1}{n} \to d x lim \sum \to \int

\frac{1}{n} < \frac{i}{n} < \frac{n}{n} 当 n \to \infty 0 < \frac{i}{n} < 1 n \to \infty lim i = 1 \sum n 1 + \frac{i}{n} \cdot \frac{1}{n} = \int_{0}^{1} 1 + x d x

n \to \infty lim \frac{1}{n ^{2}} (n^{2} - 1^{2} + n^{2} - 2^{2} + \dots + n^{2} - n^{2}) (n^{2} - 1^{2} + n^{2} - 2^{2} + \dots + n^{2} - n^{2}) = i = 1 \sum n n^{2} - i^{2} 原式 = n \to \infty lim \frac{1}{n ^{2}} i = 1 \sum n n^{2} - i^{2} = n \to \infty lim \frac{1}{n} i = 1 \sum n n^{2} - i^{2} \frac{1}{n} 在这 i 和 n 齐次，考虑凑定积分定义 = n \to \infty lim i = 1 \sum n \frac{n ^{2} - i ^{2}}{n} \frac{1}{n} = n \to \infty lim i = 1 \sum n \frac{n ^{2} - i ^{2}}{n ^{2}} \frac{1}{n} = n \to \infty lim i = 1 \sum n 1 - \frac{i ^{2}}{n ^{2}} \frac{1}{n} = n \to \infty lim i = 1 \sum n 1 - (\frac{i}{n})^{2} \frac{1}{n} = \int_{0}^{1} 1 - x^{2} d x

i和n齐次，考虑凑定积分定义，如果非齐次考虑用夹逼定理。

n \to \infty lim (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + n}) n \to \infty lim i = 1 \sum n \frac{1}{n + i} n \to \infty lim i = 1 \sum n \frac{\frac{1}{n}}{\frac{n + i}{n}} n \to \infty lim i = 1 \sum n \frac{1}{1 + \frac{i}{n}} \cdot \frac{1}{n} = \int_{0}^{1} \frac{1}{1 + x} d x

对应技巧

n \to \infty lim i = 1 \sum n \frac{i}{n} sin \frac{i}{n} \cdot \frac{1}{n} = \int_{0}^{1} x sin x d x n \to \infty lim i = 1 \sum 2 n \frac{i}{n} sin \frac{i}{n} \cdot \frac{1}{n} \frac{1}{n} < \frac{i}{n} < \frac{2 n}{n} ∴ 0 < \frac{i}{n} < 2 n \to \infty lim i = 1 \sum 2 n \frac{i}{n} sin \frac{i}{n} \cdot \frac{1}{n} = \int_{0}^{2} x sin x d x n \to \infty lim i = 1 \sum n f (a + \frac{i}{n}) \cdot \frac{1}{n} 看这个积分范围 0 < \frac{i}{n} < 1 a < a + \frac{i}{n} < 1 + a ∴ n \to \infty lim i = 1 \sum n f (a + \frac{i}{n}) \cdot \frac{1}{n} = \int_{a}^{a + 1} f (x) d x n \to \infty lim i = 1 \sum n (1 + \frac{i}{n})^{2} \cdot \frac{1}{n} = \int_{1}^{2} x^{2} d x = \int_{0}^{1} (1 + x)^{2} d x

牛顿-莱布尼兹公式

如果函数F(x)是连续函数f(x)在区间[a,b]上的一个原函数，则：

\int_{a}^{b} f (x) d x = F (x)_{a}^{b} = F (b) - F (a)

\int_{1}^{2} x d x = \frac{1}{2} x^{2}_{1}^{2} = \frac{1}{2} \cdot 2^{2} - \frac{1}{2} \cdot 1^{2} = \frac{3}{2}

分部积分公式

\int_{a}^{b} u v^{'} d x = \int_{a}^{b} u d v = uv_{a}^{b} - \int_{a}^{b} v \cdot u^{'} d x

如：

\int_{0}^{1} x e^{x} d x = \int_{0}^{1} x d e^{x} = x e^{x}_{0}^{1} - \int_{0}^{1} e^{x} d x = 1 \cdot e - 0 - e^{x}_{0}^{1} = e - (e - 1) = 1

例题

\int_{0}^{π} t sin t d t - \int_{0}^{π} t d (cos t) - (t \cdot cos t_{0}^{π} - \int_{0}^{π} cos t d t) - t \cdot cos t_{0}^{π} + sin t_{0}^{π} π + 0 = π

换元法

定积分换元记得换限

\int_{1}^{4} \frac{1}{x ( 1 + x )} d x

换元，令 t = x, x = t^{2}, d x = d t^{2} = 2 t d t 上下限换元代入 t = x 上限换元 x = 4, t = 2; 下限换元 x = 1, t = 1 \int_{1}^{2} \frac{1}{t ^{2} ( 1 + t )} 2 t d t 2 \int_{1}^{2} \frac{1}{t ^{2} ( 1 + t )} t d t 2 \int_{1}^{2} \frac{1}{t ( 1 + t )} d t 2 \int_{1}^{2} \frac{1}{t} - \frac{1}{1 + t} d t 2 (ln ∣ t ∣ - ln ∣1 + t ∣)_{1}^{2} 2 ln ∣ \frac{t}{1 + t} ∣_{1}^{2} 2 ln \frac{2}{3} - 2 ln \frac{1}{2} = 2 ln \frac{4}{3}

\int_{0}^{1} x^{2} 1 - x^{2} d x

令 x = sin t, d x = d sin t = cos t d t 上下限换元代入 x = sin t x = 1, t = \frac{π}{2}, x = 0, t = 0 原式 = \int_{0}^{\frac{π}{2}} sin^{2} t cos t \cdot cos t d t \int_{0}^{\frac{π}{2}} sin^{2} t cos^{2} t d t \int_{0}^{\frac{π}{2}} (1 - cos^{2} t) \cdot cos^{2} t d t \int_{0}^{\frac{π}{2}} cos^{2} t - cos^{4} t d t \frac{1}{2} \cdot \frac{π}{2} - \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4} - \frac{3 π}{16} = \frac{π}{16}

变限积分求导

形如： $\int_{g (x)}^{f (x)} h (t) d t$ ，叫做变限积分函数
如： $\int_{0}^{x^{2}} t d t, \int_{x}^{x^{2}} 2 t^{3} d t$

求导方法

[\int_{下限}^{上限} f (t) d t]^{'} = f (上限) \cdot 上 限^{'} - f (下限) \cdot 下 限^{'}

做题中，遇到变限积分基本上就考求导

在极限中 $\Rightarrow$ 洛必达
等式（结合微分方程）

求 $(\int_{0}^{x} 2 t^{2} d t)^{'}$

(\int_{0}^{x} 2 t^{2} d t)^{'} = 2 x^{2} \cdot (x)^{'} - 2 \cdot 0^{2} = 2 x^{2}

设 $ϕ (x) = \int_{0}^{x} x cos t^{2} d t$ ，则 $ϕ^{'} (x)$

常数因子可以拿到外面去，这道题是对t积分x就是常数因子，这个方法是对此积分类型先积分后求导的简便方法所以x可外提，即
$\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x (k 与 x 无关)$

\int_{0}^{x} x cos t^{2} d t = x \int_{0}^{x} cos t^{2} d t ϕ (x) = x \int_{0}^{x} cos t^{2} d t ϕ^{'} (x) = (x)^{'} \int_{0}^{x} cos t^{2} d t + x (\int_{0}^{x} cos t^{2} d t)^{'} = \int_{0}^{x} cos t^{2} d t + x (cos x^{2} - 0) = \int_{0}^{x} cos t^{2} d t + x cos x^{2}

当 b = a 时, \int_{a}^{a} f (x) d x = 0

求极限 $lim_{x \to 0} \frac{\int _{0}^{x} s i n t d t}{( e ^{x} - 1 ) t a n x}$
求极限第一步先代入，零比零。

x \to 0 lim \frac{\int _{0}^{x} sin t d t}{( e ^{x} - 1 ) tan x} = x \to 0 lim \frac{\int _{0}^{x} sin t d t}{x \cdot x} x \to 0 lim \frac{sin x}{2 x} = x \to 0 lim \frac{x}{2 x} = \frac{1}{2}

教材上的公式：

(\int_{_{ϕ (x)}}^{g (x)} f (t) d t)^{'} = g^{'} (x) f [g (x)] - ϕ^{'} (x) f [ϕ (x)]

计算

[\int_{0}^{x} e^{- t} sin t]^{'} = 1 \cdot e^{- x} sin x = e^{- x} sin x

已知 $F (x) = \int_{x}^{0} cos (3 t + 1) d t, 求 F^{'} (x)$

F^{'} (x) = 0 - cos (3 x + 1) \cdot 1 = - cos (3 x + 1) 第二种解法 F (x) = \int_{x}^{0} cos (3 t + 1) d t = - \int_{0}^{x} cos (3 t + 1) d t ∴ F^{'} (x) = - cos (3 x + 1)

设 $y = \int_{1}^{x^{2}} 1 + t^{3} d t ，求 \frac{d y}{d x}$

\frac{d y}{d x} = [\int_{1}^{x^{2}} 1 + t^{3} d t]^{'} = 1 + (x^{2})^{3} \cdot (x^{2})^{'} - (1)^{'} = 2 x 1 + x^{6}

求极限

x \to 0 lim \frac{\int _{0}^{2 x^{2}} e ^{t^{2}} d t}{sin ^{2} x} 第一步先带入，分子 \int_{0}^{0} f (t) d t 趋于 0 分母也趋于 0 x \to 0 lim \frac{\int _{0}^{2 x^{2}} e ^{t^{2}} d t}{x ^{2}} 洛 x \to 0 lim \frac{e ^{(2 x^{2})^{2}} \cdot 4 x}{2 x} x \to 0 lim 2 e^{4 x^{4}} = 2

积分里含有x

取出x，再求导

\frac{d}{d x} \int_{1}^{x^{2}} x e^{t} d t = (\int_{1}^{x^{2}} x e^{t} d t)^{'} 对 t 进行积分， x 当作常数 (x \cdot \int_{1}^{x^{2}} e^{t} d t)^{'} x^{'} \cdot \int_{1}^{x^{2}} e^{t} d t + x \cdot (\int_{1}^{x^{2}} e^{t} d t)^{'} 1 \cdot \int_{1}^{x^{2}} e^{t} d t + x \cdot (e^{x^{2}} \cdot 2 x) = \int_{1}^{x^{2}} e^{t} d t + 2 x^{2} e^{x^{2}} = e^{t}_{1}^{x^{2}} + 2 x^{2} e^{x^{2}} = e^{x^{2}} - e + 2 x^{2} e^{x^{2}}

提不出就换元

运算技巧

偶倍奇零

条件：区间对称

\int_{- a}^{a} f (x) d x = {2 \int_{0}^{a} f (x) d x, 0, f (x) 为偶函数 f (x) 为奇函数

另一种情况，这里 $f (x)$ 不要求是奇函数还是偶函数，可以是非奇非偶函数。

\int_{- a}^{a} f (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x

证明：

非奇非偶的处理

\int_{- a}^{a} f (x) d x = \int_{0}^{a} f (x) + f (- x) d x

Proof:

\int_{- a}^{a} f (x) d x = \int_{- a}^{0} f (x) d x + \int_{0}^{a} f (x) d x Let x = - t, d x = - d t When x = - a, t = a; x = 0, t = 0 Now \int_{- a}^{0} f (x) d x = - \int_{a}^{0} f (- t) d t = \int_{0}^{a} f (- x) d x So \int_{- a}^{a} f (x) d x = \int_{0}^{a} f (- x) d x + \int_{0}^{a} f (x) d x Proof completed

题目

求 $\int_{- 1}^{1} (x^{3} cos x + 1) d x$

\int_{- 1}^{1} (x^{3} cos x + 1) d x = \int_{- 1}^{1} x^{3} cos x d x + \int_{- 1}^{1} 1 d x x^{3} 奇函数 乘上 cos x 偶函数 = 奇函数 原式 = \int_{- 1}^{1} 1 d x = x_{- 1}^{1} = 1 - (- 1) = 2

因为1也是偶函数，也可用偶倍奇零

\int_{- 1}^{1} 1 d x = 2 \int_{0}^{1} 1 d x = 2 x_{0}^{1} = 2

设f(x)在[-l,l]上连续，则 $\int_{- l}^{l} f (x) - f (- x) d x =$

令 F (x) F (- x) = f (x) - f (- x) = f (- x) - f (x) = - [f (x) - f (- x)] ∴ F (x) 为奇函数 \int_{- l}^{l} f (x) - f (- x) d x = 0

\int_{- 1}^{1} \frac{2 + sin x}{1 + x ^{2}} d x =

\int_{- 1}^{1} \frac{2}{1 + x ^{2}} d x + \int_{- 1}^{1} \frac{sin x}{1 + x ^{2}} d x 其中 \frac{2}{1 + x ^{2}} ， 2 偶函数， 1 偶函数， x^{2} 偶函数 \frac{sin x}{1 + x ^{2}}, sin x 奇函数 2 \int_{- 1}^{1} \frac{1}{1 + x ^{2}} d x + 0 2 \cdot 2 \int_{0}^{1} \frac{1}{1 + x ^{2}} d x 4 arctan x_{0}^{1} 4 (arctan 1 - arctan 0) 4 \cdot \frac{π}{4} = π

$arctan 1$ ，tan多少度等于1

\int_{- π}^{π} (1 + cos x + ∣ x ∣ \cdot sin x) d x

cos 2 x = 2 cos^{2} x - 1 cos 2 \cdot \frac{x}{2} = 2 cos^{2} \frac{x}{2} - 1

\int_{- π}^{π} 1 + cos x d x + \int_{- π}^{π} ∣ x ∣ sin x d x cos x 偶函数， ∣ x ∣ 偶函数， sin x 奇函数, 奇 \times 偶 = 奇 2 \int_{0}^{π} 1 + cos x d x （消 1 ） 2 \int_{0}^{π} 1 + 2 cos^{2} \frac{x}{2} - 1 d x 2 \int_{0}^{π} 2 cos^{2} \frac{x}{2} d x x \in (0, π), \frac{x}{2} \in (0, \frac{π}{2}) ∴ 2 \int_{0}^{π} 2 \cdot cos \frac{x}{2} d x 22 \int_{0}^{π} cos \frac{x}{2} d x 42 \int_{0}^{π} cos \frac{x}{2} d \frac{x}{2} 42 sin \frac{x}{2}_{0}^{π} = 42 sin \frac{π}{2} = 42

求 $\int_{- 1}^{1} e^{x^{2}} sin x d x$
看见区间对称，优先考虑偶倍奇零。
现在判断 $f (x) = e^{x^{2}} sin x$ 的奇偶性。

f (x) = e^{∣ x ∣} sin x 令 g (x) = e^{∣ x ∣} g (- x) = e^{∣ x ∣} ∴ e^{∣ x ∣} 为偶函数 偶函数 \times 奇函数 = 奇函数 ∴ \int_{- 1}^{1} e^{x^{2}} sin x d x = 0

另一种看法，用复合函数的奇偶行判断全奇则奇，遇偶则偶， $e^{∣ x ∣}$ 为偶函数。

\int_{- 1}^{1} [\frac{x ^{3} sin ^{2} x}{x ^{4} + 2 x ^{2} + 1} + (x^{2} + 1) 1 - x^{2}] d x x^{3} 奇 \times sin^{2} x 偶 = 奇 x^{4} + 2 x^{2} + 1 = 偶 奇 \times 偶 = 奇 ∴ \int_{- 1}^{1} \frac{x ^{3} sin ^{2} x}{x ^{4} + 2 x ^{2} + 1} d x = 0 \int_{- 1}^{1} (x^{2} + 1) 1 - x^{2} = 2 \int_{0}^{1} (x^{2} + 1) 1 - x^{2}

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华里氏公式

点火公式

\int_{0}^{\frac{π}{2}} sin^{n} x d x = \int_{0}^{\frac{π}{2}} cos^{n} x d x = ⎩ ⎨ ⎧ \frac{n - 1}{n} \cdot \frac{n - 3}{n - 2} \cdot \frac{n - 4}{n - 5} \dots \frac{1}{2} \cdot \frac{π}{2}, n 为偶数 \frac{n - 1}{n} \cdot \frac{n - 3}{n - 2} \dots \frac{2}{3} \cdot 1, n 为奇数

\int_{0}^{π} sin^{n} d x = 2 \int_{0}^{\frac{π}{2}} sin^{n} x d x

\int_{0}^{\frac{π}{2}} cos^{6} x d x = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{π}{2}

\int_{0}^{\frac{π}{2}} sin^{5} x d x = \frac{4}{5} \cdot \frac{2}{3} \cdot 1

\int_{- \frac{π}{2}}^{\frac{π}{2}} (sin^{4} θ + cos^{3} θ) d θ

对称区间优先用偶倍奇零

sin^{4} θ ，奇（偶） = 偶函数， cos^{3} θ ，偶（奇） = 偶函数 2 \int_{0}^{\frac{π}{2}} sin^{4} θ + cos^{3} θ d θ 2 (\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{π}{2} + \frac{2}{3} \cdot 1) = \frac{3 π}{8} + \frac{4}{3}

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三角函数积分变换公式

设f(x)在[0,1]上连续，则

\int_{0}^{\frac{π}{2}} f (sin x) d x = \int_{0}^{\frac{π}{2}} f (cos x) d x

可以用区间再现公式证明。

\int_{0}^{π} f (sin x) d x = 2 \int_{0}^{\frac{π}{2}} f (sin x) d x

因为sinx的图像关于 $x = \frac{π}{2}$ 对称，

∴ \int_{0}^{π} sin x d x = 2 \int_{0}^{\frac{π}{2}} sin x d x

因为一个函数本身关于一条竖线轴对称，在这个函数外面再套一层函数，整个复合函数也关于该对称轴对称。由此简单证明： $\int_{0}^{π} f (sin x) d x = 2 \int_{0}^{\frac{π}{2}} f (sin x) d x$

严格证明：

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带绝对值

当被积函数含有绝对值符号时，应用定积分的性质，把积分区间分成若干个子区间，这样就可以把绝对值符号去掉，然后分别在各个子区间求定积分

\int_{- 1}^{1} x^{2} d x

= \int_{- 1}^{1} ∣ x ∣ d x \int_{- 1}^{0} - x d x + \int_{0}^{1} x d x - \frac{1}{2} x^{2}_{- 1}^{0} + \frac{1}{2} x^{2}_{0}^{1} 0 - (- \frac{1}{2}) + \frac{1}{2} = 1

这道题也可以先用偶倍奇零

= \int_{- 1}^{1} ∣ x ∣ d x = 2 \int_{0}^{1} ∣ x ∣ d x = 2 \cdot \frac{1}{2} x^{2}_{0}^{1} = 2 (\frac{1}{2}) = 1

\int_{0}^{π} sin x - sin^{3} x d x

\int_{0}^{π} sin x (1 - sin^{2} x) d x \int_{0}^{π} sin x ∣ cos x ∣ d x \int_{0}^{\frac{π}{2}} sin x cos x d x + \int_{\frac{π}{2}}^{π} sin x \cdot (- cos x) d x \int_{0}^{\frac{π}{2}} sin x cos x d x - \int_{\frac{π}{2}}^{π} sin x cos x d x \int_{0}^{\frac{π}{2}} sin x d sin x - \int_{\frac{π}{2}}^{π} sin x d sin x \frac{2}{3} sin x^{\frac{3}{2}}_{0}^{\frac{π}{2}} - \frac{2}{3} sin x^{\frac{3}{2}}_{\frac{π}{2}}^{π} \frac{2}{3} - (0 - (\frac{2}{3})) = \frac{4}{3}

\int_{0}^{2 π} ∣ sin x ∣ d x_{i}

第一种办法分区间讨论绝对值

\int_{0}^{π} sin x d x + \int_{π}^{2 π} (- sin x) d x - cos x_{0}^{π} + cos x_{π}^{2 π} - (cos π - cos 0) + (cos 2 π - cos π) - (- 1 - 1) + (1 - (- 1)) = 4

第二种办法，直接看 $∣ sin x ∣$ 的图像化简积分

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\int_{0}^{2 π} ∣ sin x ∣ d x = 2 \int_{0}^{π} sin x 2 \cdot (- cos x)_{0}^{π} 2 \cdot - (cos π - cos 0) = - 2 \cdot (- 1 - 1) = 4

几何应用

求圆的面积

$\int_{0}^{a} a^{2} - x^{2} d x = \frac{1}{4} π a^{2}$

$\int_{0}^{2} 4 - x^{2} d x = \int_{0}^{2} 2 - x^{2} d x = \frac{1}{4} π \cdot 2^{2} = π$

\int_{- a}^{a} (b + x) \cdot a^{2} - x^{2} d x \int_{- a}^{a} b a^{2} - x^{2} + x a^{2} - x^{2} d x 令 f (x) = b a^{2} - x^{2} a^{2} - x^{2} \geq 0 a^{2} \geq x^{2} ∣ x ∣^{2} \leq ∣ a ∣^{2} ∣ x ∣ \leq ∣ a ∣ - ∣ a ∣ \leq x \leq ∣ a ∣ ∴ x \in [- ∣ a ∣, ∣ a ∣]

大于取两边，小于取中间

∣ x ∣ \geq 6 x \geq 6 或 x \leq - 6 ∣ x ∣ \leq 6 x \geq - 6 或 x \leq 6

x \in [- ∣ a ∣, ∣ a ∣] 定义域关于 0 对称，满足判断偶函数的前提条件 f (x) = b a^{2} - x^{2} f (- x) = b a^{2} - x^{2} ∴ f (x) 为偶函数

\int_{- a}^{a} b a^{2} - x^{2} + \int_{- a}^{a} x a^{2} - x^{2} d x x a^{2} - x^{2}, 奇 \times 偶 = 奇 2 \int_{0}^{a} b a^{2} - x^{2} d x 2 b \int_{0}^{a} a^{2} - x^{2} d x 2 b \cdot \frac{1}{4} π a^{2} = \frac{1}{2} a^{2} bπ

设 $f (x) = 3 x - 1 - x^{2} - \int_{0}^{1} f (x) d x$ ，求 $f (x)$
定积分为常数： $\int_{0}^{1} f (x) d x = A$
则 $f (x) = 3 x - 1 - x^{2} - A$
对方程两边同取0到1的定积分

\int_{0}^{1} f (x) d x A A 2 A A ∴ f (x) = \int_{0}^{1} 3 x d x - \int_{0}^{1} 1 - x^{2} d x - \int_{0}^{1} A d x = 3 \cdot \frac{1}{2} x^{2}_{0}^{1} - \frac{1}{4} π - A x_{0}^{1} = \frac{3}{2} - \frac{1}{4} π - A = \frac{3}{2} - \frac{1}{4} π = \frac{3}{4} - \frac{1}{8} π = 3 x - 1 - x^{2} - \frac{3}{4} + \frac{π}{8}

用几何法求简单定积分

\int_{- 2}^{2} 4 - x^{2} d x

令 y = 4 - x^{2} y^{2} = 4 - x^{2} (y \geq 0) x^{2} + y^{2} = 2^{2} (y \geq 0)

反映到几何上

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∴ \int_{- 2}^{2} 4 - x^{2} d x = \frac{1}{2} π 2^{2} = 2 π

求 $\int_{- 1}^{1} 1 - x^{2} d x$

令 y = 1 - x^{2} ，在区间 [- 1, 1] 连续 y^{2} = 1 - x^{2} (y \geq 0) x^{2} + y^{2} = 1 (y \geq 0)

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可知

\int_{- 1}^{1} 1 - x^{2} d x = \frac{1}{2} π 1^{2} = \frac{π}{2}

\int_{- 1}^{1} x^{2} d x

$= \int_{- 1}^{1} ∣ x ∣ d x$

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∴ \int_{- 1}^{1} x^{2} d x = 2 \cdot \frac{1}{2} \cdot 1 \cdot 1 = 1

直角坐标系内求所围面积

简单理解

X 型图：由上、下两个函数围成的面积（分上、下观看）

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推导面积过程

S = \int_{a}^{b} f_{上} (x) d x - \int_{a}^{b} f_{下} (x) d x S = \int_{a}^{b} f_{上} (x) - f_{下} (x) d x

Y 型图：分左、右看；左边一个函数，右边一个函数

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\int_{c}^{d} f_{右} (y) - f_{左} (y) d y

选用 Y 型；改写为 x = f(y)

严谨理解

X 型：曲线 $y = f (x)$ 及直线 $x = a, x = b (a < b)$ 与x轴所围成的曲边梯形面积

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$A = \int_{a}^{b} ∣ f (x) ∣ d x$
2. X 型： $y = f (x) ， y = g (x)$ ，及直线 $x = a, x = b (a < b)$ 与x轴所围成的曲边梯形面积

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A = \int_{a}^{b} ∣ f (x) - g (x) ∣ d x

Y 型：曲线 $x = ϕ (y)$ 及直线 $y = c, y = d (c < d)$ 与y轴所围成的曲边梯形面积

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A = \int_{c}^{d} ∣ φ (y) ∣ d y

Y 型：曲线 $x = φ (y), x = ψ (y)$ 及直线 $y = c, y = d (c < d)$ 与y轴所围成的曲边梯形面积

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A = \int_{c}^{d} ∣ φ (y) - ψ (y) ∣ d y

题目

定义

设 $y = f (x)$ 与 $y = g (x)$ 在区间[a,b]上连续，则由这两条曲线及 $x = a, x = b$ 所围成平面图形的面积为？

Circular transclusion detected: Math/lib/Integral-libs/Drawing-2026-01-08-09.48.38.excalidraw

∴ S = \int_{a}^{b} ∣ f (x) - g (x) ∣ d x

求面积

求 $y = x^{2}, y = \frac{1}{x}, x = 2$ 所围成的图形面积

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画图后，发现用 X 型图比较好算

S = \int_{1}^{2} (x^{2} - \frac{1}{x}) d x = [\frac{1}{3} x^{3} - ln ∣ x ∣]_{1}^{2} = (\frac{8}{3} - ln 2) - (\frac{1}{3} - ln 1) = \frac{7}{3} - ln 2

或者用带绝对值的公式

S = \int_{1}^{2} ∣ \frac{1}{x} - x^{2} ∣ d x \frac{1}{\frac{3}{2}} - (\frac{3}{2})^{2} = \frac{2}{3} - \frac{9}{4} < 0 去掉绝对值 S = \int_{1}^{2} (x^{2} - \frac{1}{x}) d x 也是一样的

求 $y = x^{2}, y = x$ 所围成的图形面积

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先求交点左边

{y = x^{2} y = x x^{2} = x x^{4} = x 可知交点坐标为 (0, 0) 和 (1, 1) ∴ S = \int_{0}^{1} (x - x^{2}) d x = (\frac{2}{3} x^{\frac{3}{2}} - \frac{1}{3} x^{3})_{0}^{1} = (\frac{2}{3} - \frac{1}{3}) = \frac{1}{3}

也可以用Y型来做

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y y ∴ S = x^{2} (y \geq 0) \Rightarrow x = y (y \geq 0) = x \Rightarrow x = y^{2} = \int_{0}^{1} (y - y^{2}) d y

求由曲线 $y = \frac{1}{x}, y = x, x = 2$ 及x轴，所围成的图形面积

第一种方法，把要求的面积拆成两部分，分别计算

S = S_{1} + S_{2} S_{1} = \int_{0}^{1} x d x = \frac{1}{2} x^{2}_{0}^{1} = \frac{1}{2} S_{2} = \int_{1}^{2} \frac{1}{x} d x = ln ∣ x ∣_{1}^{2} = ln 2 - ln 1 = ln 2 ∴ S = \frac{1}{2} + ln 2

第二种方法，先求大三角形的面积，再减去旁边的阴影面积（蓝色）

S_{△} = \int_{0}^{2} x d x = \frac{1}{2} x^{2}_{0}^{2} = 2 S_{阴} = \int_{1}^{2} x d x - \int_{1}^{2} \frac{1}{x} d x = \frac{1}{2} x^{2}_{1}^{2} - ln ∣ x ∣_{1}^{2} = \frac{4}{2} - \frac{1}{2} - (ln 2 - ln 1) = \frac{3}{2} - ln 2 ∴ S = 2 - \frac{3}{2} + ln 2 = \frac{1}{2} + ln 2

求曲线 $x = y^{2}$ 与直线 $y = x$ 围成的平面图形的面积。
$y^{2} = x$ 为开口向右的抛物线。

{y^{2} = x y = x 交点 (0, 0) 和 (1, 1) y^{2} = x (x \geq 0) y = x (x \geq 0) \int_{0}^{1} x d x - \int_{0}^{1} x d x = \frac{2}{3} x^{\frac{3}{2}}_{0}^{1} - \frac{1}{2} x^{2}_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = 1

求曲线 $y^{2} = - x$ 与直线y=x围成的平面图形的面积。
$y^{2} = - x$ 为开口向左的抛物线。

{y^{2} = - x (y^{2} \geq 0) y = x x^{2} = - x x^{2} + x = 0 x (1 + x) = 0 得到交点 (0, 0) 和 (- 1, - 1) y^{2} = - x ∴ y = - - x ，要曲线的下半枝 A = \int_{a}^{b} [f_{u pp er} (x) - f_{l o w er} (x)] d x A = \int_{- 1}^{0} [x - (- - x)] d x = \frac{1}{6}

用Y形图做

{y^{2} = - x (y^{2} \geq 0) y = x \Rightarrow {x = - y^{2} x = y \int_{c}^{d} f_{右} (y) - f_{左} (y) d y S = \int_{- 1}^{0} - y^{2} d y - \int_{- 1}^{0} y d y = - \frac{1}{3} y^{3}_{- 1}^{0} - \frac{1}{2} y^{2}_{- 1}^{0} = - (0 - \frac{1}{3} \cdot (- 1)) - (0 - \frac{1}{2}) = - \frac{1}{3} + \frac{1}{2} = - \frac{2}{6} + \frac{3}{6} = \frac{1}{6}

求曲线 $x = y^{2}, y = x - 2$ 所围成的图形面积

y^{2} = x ，开口向右 {y^{2} = x y = x - 2 交点 (1, - 1) 和 (4, 2)

用Y型图做

{x = y^{2} x = y + 2 \int_{- 1}^{2} (y + 2 - y^{2}) d y

用X型图做

y^{2} = x {y = + x 上支 y = - x 下支 与 y = x - 2 \int_{0}^{1} [x - (- x)] d x + \int_{1}^{4} [x - (x - 2)] d x

求椭圆面积

求椭圆 $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ 的面积

S = 4 S_{1} = 4 \int_{0}^{a} y d x

利用椭圆的参数方程

{x = a cos t y = b sin t

用定积分换元法

d x = d (a cos t) = a \cdot (- sin t) d t = - a sin t d t x = 0 时, t = \frac{π}{2} x = a 时, t = 0 ∴ S = 4 \int_{\frac{π}{2}}^{0} b sin t (- a sin t) d t = 4 \int_{0}^{\frac{π}{2}} ab sin^{2} t d t = 4 ab \int_{0}^{\frac{π}{2}} \frac{1 - cos 2 t}{2} d t = 4 ab (\int_{0}^{\frac{π}{2}} \frac{1}{2} d t - \frac{1}{2} \int_{0}^{\frac{π}{2}} cos 2 t d t) = 4 ab (\int_{0}^{\frac{π}{2}} \frac{1}{2} d t - \frac{1}{4} \int_{0}^{\frac{π}{2}} cos 2 t d (2 t)) = 4 ab [\frac{1}{2} t_{0}^{\frac{π}{2}} - \frac{1}{4} (sin 2 t)_{0}^{\frac{π}{2}}] = 4 ab (\frac{π}{4} - \frac{1}{4} \cdot 0) = abπ = πab

另一种方法

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 \frac{y ^{2}}{b ^{2}} = 1 - \frac{x ^{2}}{a ^{2}} y^{2} = b^{2} (1 - \frac{x ^{2}}{a ^{2}}) y^{2} = b^{2} (\frac{a ^{2} - x ^{2}}{a ^{2}}) y^{2} = \frac{b ^{2}}{a ^{2}} (a^{2} - x^{2}) y = \pm \frac{b ^{2}}{a ^{2}} (a^{2} - x^{2}) = \pm \frac{b}{a} (a^{2} - x^{2}) y = \frac{b}{a} (a^{2} - x^{2}) 为上半椭圆 y = - \frac{b}{a} (a^{2} - x^{2}) 为下半椭圆

∴ S = 4 \int_{0}^{a} \frac{b}{a} (a^{2} - x^{2}) d x = 4 \cdot \frac{b}{a} \cdot \frac{1}{4} π a^{2} = πab

求弧长*

参数方程情形：

{x = ϕ (t) y = f (t) α \leq t \leq β 且 ϕ (t), f (t) 在 [α, β] 上一阶连续导数 \neq = 0 S = \int_{α}^{β} [φ^{'} (t)]^{2} + [f^{'} (t)]^{2} d t

弧长公式的推导

直角坐标系的情况

{y = f (x) x = x a \leq x \leq b ∴ S = \int_{a}^{b} [f^{'} (x)]^{2} + 1^{2} d x S = \int_{a}^{b} 1 + [f^{'} (x)]^{2} d x

分段函数求定积分

\int_{a}^{b} f (x) d x = \int_{a}^{x_{0}} f (x) d x + \int_{x_{0}}^{b} f (x) d x

其中 $x_{0}$ 为分段点。

设 $f (x) = {e^{x}, x \leq 1 ln x, x > 1$ ，求 $\int_{0}^{e} f (x) d x$

\int_{0}^{1} f (x) d x + \int_{1}^{e} f (x) d x \int_{0}^{1} e^{x} d x + \int_{1}^{e} ln x d x e^{x}_{0}^{1} + (ln x \cdot x)_{1}^{e} - \int_{1}^{e} x \cdot \frac{1}{x} d x e - 1 + e - \int_{1}^{e} 1 d x 2 e - 1 - (e - 1) = e

设 $f (x) = {1 + x^{2}, x < 0 e^{x}, x \geq 0$ ，求 $\int_{\frac{1}{2}}^{2} f (1 - x) d x$
这两个函数的自变量不是同一个，采取换元法

令 1 - x = t, x = 1 - t, d x = - d t 换元上下限，代入式子 1 - x = t x = \frac{1}{2}, t = 1 - \frac{1}{2} = \frac{1}{2} x = 2, t = 1 - 2 = - 1 ∴ \int_{\frac{1}{2}}^{2} f (1 - x) d x = - \int_{\frac{1}{2}}^{- 1} f (t) d t = \int_{- 1}^{\frac{1}{2}} f (t) d t = \int_{- 1}^{\frac{1}{2}} f (x) d x

\int_{- 1}^{0} f (x) d x + \int_{0}^{\frac{1}{2}} f (x) d x \int_{- 1}^{0} (1 + x^{2}) d x + \int_{0}^{\frac{1}{2}} e^{x} d x (x + \frac{1}{3} x^{3})_{- 1}^{0} + e^{x}_{0}^{\frac{1}{2}} = - (- 1 - \frac{1}{3}) + e^{\frac{1}{2}} - 1 = 1 + \frac{1}{3} - 1 + e^{\frac{1}{2}} = \frac{1}{3} + e^{\frac{1}{2}}

定积分等式证明

区间再现公式

积分区间再现公式

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

证明

令 a + b - x = t, x = a + b - t d x = d (a + b - t) = - d t 当 x = a 时， t = b 当 x = b 时， t = a 故原式 = - \int_{b}^{a} f (t) d t = \int_{a}^{b} f (t) d t = \int_{a}^{b} f (x) d x

使用条件：等式两侧，积分区间一样

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x = \frac{1}{2} \int_{a}^{b} [f (x) + f (a + b - x)] d x

证明：

\int_{0}^{1} x^{m} (1 - x)^{n} d x = \int_{0}^{1} x^{n} (1 - x)^{m} d x

∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x 自变量 x 变为 (上限 + 下限 - x) \int_{0}^{1} x^{m} (1 - x)^{n} d x = \int_{0}^{1} (1 + 0 - x)^{m} [1 - (1 + 0 - x)]^{n} = \int_{0}^{1} (1 - x)^{m} (x)^{n} 故原式成立

证明

\int_{0}^{\frac{π}{2}} f (sin x) d x = \int_{0}^{\frac{π}{2}} f (cos x) d x

第一种方法用上面的公式 ∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x ∴ \int_{0}^{\frac{π}{2}} f (sin x) d x = \int_{0}^{\frac{π}{2}} sin (\frac{π}{2} - x) d x = \int_{0}^{\frac{π}{2}} cos x d x 故原式成立

第二种方法，令 x = \frac{π}{2} - t, d x = d (\frac{π}{2} - t) = - d t 当 x = 0, t = \frac{π}{2} 当 x = \frac{π}{2}, t = 0 ∴ \int_{0}^{\frac{π}{2}} f (sin x) d x = - \int_{\frac{π}{2}}^{0} f [sin (\frac{π}{2} - t)] d t = \int_{0}^{\frac{π}{2}} f (cos t) d t = \int_{0}^{\frac{π}{2}} f (cos x) d x 故原式成立

证明 $\int_{0}^{π} x \cdot f (sin x) d x = \frac{π}{2} \int_{0}^{π} f (sin x) d x$

∵ \int_{a}^{b} f (x) d x \int_{0}^{π} x \cdot f (sin x) d x = = ∴ \int_{0}^{π} x \cdot f (sin x) d x ∴ = \int_{a}^{b} f (a + b - x) d x = \int_{0}^{π} (π - x) \cdot f (sin (π - x)) d x \int_{0}^{π} (π - x) \cdot f (sin x) d x \int_{0}^{π} π \cdot f (sin x) d x - \int_{0}^{π} x \cdot f (sin x) d x = \int_{0}^{π} π \cdot f (sin x) d x - \int_{0}^{π} x \cdot f (sin x) d x 2 \int_{0}^{π} x \cdot f (sin x) d x = \int_{0}^{π} π \cdot f (sin x) d x 2 \int_{0}^{π} x \cdot f (sin x) d x = π \int_{0}^{π} f (sin x) d x \int_{0}^{π} x \cdot f (sin x) d x = \frac{π}{2} \int_{0}^{π} f (sin x) d x

证明：

\int_{0}^{\frac{π}{2}} \frac{sin ^{3} x}{sin x + cos x} d x = \int_{0}^{\frac{π}{2}} \frac{cos ^{3} x}{sin x + cos x} d x

并利用证明结果求 $\int_{0}^{\frac{π}{2}} \frac{s i n ^{3} x}{s i n x + c o s x} d x$

∵ \int_{a}^{a} f (x) d x = \int_{a}^{b} f (a + b - x) d x \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x}{sin x + cos x} d x = \int_{0}^{\frac{π}{2}} \frac{sin ^{3} ( \frac{π}{2} - x )}{sin ( \frac{π}{2} - x ) + cos ( \frac{π}{2} - x )} d x = \int_{0}^{\frac{π}{2}} \frac{cos ^{3} x}{cos x + sin x} d x 故原式得证

2 \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x}{sin x + cos x} d x = \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x}{sin x + cos x} d x + \int_{0}^{\frac{π}{2}} \frac{cos ^{3} x}{sin x + cos x} d x 2 \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x}{sin x + cos x} d x = \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x + cos ^{3} x}{sin x + cos x} d x ∴ \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x}{sin x + cos x} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x + cos ^{3} x}{sin x + cos x} d x

用立方和公式

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} (a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2}) a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})

\frac{sin ^{3} x + cos ^{3} x}{sin x + cos x} = \frac{( sin x + cos ) ( sin ^{2} x - sin x cos + cos ^{2} x )}{sin x + cos x} = sin^{2} x - sin x cos + cos^{2} x = 1 - sin x cos x ∴ \int_{0}^{\frac{π}{2}} \frac{sin ^{3} x}{sin x + cos x} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 - sin x cos x) d x \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d x - \int_{0}^{\frac{π}{2}} sin x d (sin x)) 令 sin x = u x = \frac{π}{2}, u = 1 x = 0, u = 0 \frac{1}{2} (\frac{π}{2} - \int_{0}^{1} u d u) \frac{π}{4} - \frac{1}{2} \cdot (\frac{1}{2} \cdot u^{2})_{0}^{1} = \frac{π}{4} - \frac{1}{2} (\frac{1}{2}) = \frac{π - 1}{4}

证明： $F (x) = \int_{0}^{π} ln (x^{2} + 2 x cos t + 1) d t$ 为偶函数

欲证 F (x) 为偶函数 则 F (x) = F (- x) F (- x) = \int_{0}^{π} ln (x^{2} - 2 x cos t + 1) d t 又 ∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x ∴ \int_{0}^{π} ln (x^{2} - 2 x cos t + 1) d t = \int_{0}^{π} ln (x^{2} - 2 x cos (π - t) + 1) d t = \int_{0}^{π} ln (x^{2} - 2 x \cdot (- cos t) + 1) d t = \int_{0}^{π} ln (x^{2} + 2 x cos t + 1) d t = F (x) ∴ F (x) = F (- x)

求数列极限

\frac{2}{π} \int_{0}^{\frac{π}{2}} ln (cos t) d t 令 \int_{0}^{\frac{π}{2}} ln (cos t) d t = J 区间再现 J = \int_{0}^{\frac{π}{2}} ln (cos t) d t = \int_{0}^{\frac{π}{2}} ln (sin t) d t J = \frac{1}{2} \int_{0}^{\frac{π}{2}} ln (sin t) d t + \int_{0}^{\frac{π}{2}} ln (cos t) d t J = \frac{1}{2} \int_{0}^{\frac{π}{2}} ln (sin t cos t) d t J = \frac{1}{2} \int_{0}^{\frac{π}{2}} ln (\frac{1}{2} sin 2 t) d t J = \frac{1}{2} \int_{0}^{\frac{π}{2}} ln (\frac{sin 2 t}{2}) d t J = \frac{1}{2} \int_{0}^{\frac{π}{2}} sin 2 t d t - \frac{1}{2} \int_{0}^{\frac{π}{2}} ln 2 d t J = \frac{1}{4} \int_{0}^{π} sin u d u - \frac{1}{2} \cdot \frac{π}{2} \cdot ln 2 J = \frac{1}{2} J - \frac{π}{4} ln 2 J = - \frac{π}{2} ln 2

The interval reappearance formula

广义积分

又叫反常积分

无穷积分

也就是在无穷区间上的反常积分。上下界有一个是无穷，或都为无穷。
若 $\int_{a}^{+ \infty} f (x) d x = lim_{t \to \infty} \int_{a}^{t} f (x) d x = lim_{t \to \infty} [F (t) - F (a)]$ 极限存在，则称 $\int_{a}^{+ \infty} f (x) d x$ 收敛，否则称为发散，计算公式：

\int_{a}^{+ \infty} f (x) d x = F (x)_{a}^{+ \infty} = F (+ \infty) - F (a)

比如：

\int_{0}^{+ \infty} \frac{1}{1 + x ^{2}} d x = arctan x_{0}^{+ \infty} = x \to + \infty lim arctan x - arctan 0 = \frac{π}{2} ∴ \int_{0}^{+ \infty} \frac{1}{1 + x ^{2}} d x 收敛

注意：
判断 $\int_{- \infty}^{+ \infty} f (x) d x$ 是否收敛，需要先拆开 $\int_{- \infty}^{a} f (x) d x + \int_{a}^{+ \infty} f (x) d x$ ，如果拆开的两个积分都收敛，就称 $\int_{- \infty}^{+ \infty} f (x) d x$ 收敛；如果拆开后的两个积分，有任何一个发散，那么原积分也就发散。

$(- \infty, + \infty)$ 并非对称区间， $\int_{- \infty}^{+ \infty} f (x) d x$ 不适用于“偶倍奇零”（需要先验证 $\int_{- \infty}^{a} f (x) d x$ 和 $\int_{a}^{+ \infty} f (x) d x$ 都收敛）
比如：

\int_{- \infty}^{+ \infty} x d x = \int_{- \infty}^{0} x d x + \int_{0}^{+ \infty} x d x = \frac{1}{2} x^{2}_{- \infty}^{0} + \frac{1}{2} x^{2}_{0}^{+ \infty} 其中 \frac{1}{2} x^{2}_{- \infty}^{0} 为 - \infty \frac{1}{2} x^{2}_{0}^{+ \infty} 为 + \infty 这两个极限都发散 所以原积分也发散

再举一个例子

\int_{- \infty}^{+ \infty} x e^{- x^{2}} d x = \int_{0}^{+ \infty} x e^{- x^{2}} d x + \int_{- \infty}^{0} x e^{- x^{2}} d x \int_{0}^{+ \infty} x e^{- x^{2}} d x = \frac{1}{2} \int_{0}^{+ \infty} e^{- x^{2}} d (x^{2}) 令 x^{2} = u = \frac{1}{2} \int_{0}^{+ \infty} e^{- u} d u = - \frac{1}{2} \int_{0}^{+ \infty} e^{- u} d (- u) = - \frac{1}{2} e^{- u}_{0}^{+ \infty} = - \frac{1}{2} e^{- x^{2}}_{0}^{+ \infty} - \frac{1}{2} (x \to + \infty lim e^{- x^{2}} - 1) = \frac{1}{2} \int_{- \infty}^{0} x e^{- x^{2}} d x = - \frac{1}{2} e^{- x^{2}}_{- \infty}^{0} = - \frac{1}{2} (0 - x \to - \infty lim e^{- x^{2}}) = - \frac{1}{2}

拆开的两个积分都收敛，原积分可以用偶倍奇零
x为奇函数， $e^{- x^{2}}$ 为偶函数，奇 x 偶 = 奇函数
$\int_{- \infty}^{+ \infty} x e^{- x^{2}} d x = 0$

\int_{- \infty}^{+ \infty} \frac{P ( x )}{e ^{a x^{2}}} d x, P (x) 为多项式

这类积分都都收敛，因为指数函数 $e^{x^{2}}$ 趋向无穷的速度太快。

\int_{1}^{+ \infty} \frac{1}{x ^{2}} d x = - \frac{1}{x}_{1}^{+ \infty} = x \to + \infty lim - \frac{1}{x} - (- \frac{1}{1}) = 0 + 1 = 1

\int_{0}^{+ \infty} x \cdot e^{- 2 x} d x

- \frac{1}{2} \int_{0}^{+ \infty} x d (e^{- 2 x}) d x - \frac{1}{2} [(x \cdot e^{- 2 x})_{0}^{+ \infty} - \int_{0}^{+ \infty} e^{- 2 x} d x] - \frac{1}{2} x e^{- 2 x}_{0}^{+ \infty} + \frac{1}{2} \int_{0}^{+ \infty} e^{- 2 x} d x - \frac{1}{2} x \to + \infty lim x \cdot \frac{1}{e ^{2 x}} - \frac{1}{4} \int_{0}^{+ \infty} e^{- 2 x} d (- 2 x) - \frac{1}{2} x \to + \infty lim \frac{x}{e ^{2 x}} - \frac{1}{4} \cdot e^{- 2 x}_{0}^{+ \infty} - x \to + \infty lim \frac{1}{e ^{2 x}} - \frac{1}{4} \cdot (e^{- \infty} - e^{0}) 0 - \frac{1}{4} (0 - 1) = \frac{1}{4}

若 $\int_{0}^{+ \infty} e^{k x} d x = \frac{1}{2}$ ，求k

先算一下原函数 \int e^{k x} d x = \frac{1}{k} \int e^{k x} d (k x) = \frac{1}{k} \cdot e^{k x} + C \int_{0}^{+ \infty} e^{k x} d x = \frac{e ^{k x}}{k}_{0}^{+ \infty} x \to + \infty lim \frac{e ^{k x}}{k} - \frac{1}{k} 如果 k > 0, e^{k x} \to + \infty 如果 k < 0, e^{k x} \to 0 ∴ k < 0 x \to + \infty lim \frac{e ^{k x}}{k} - \frac{1}{k} = \frac{0 - 1}{k} = - \frac{1}{k} = \frac{1}{2} k = - 2

\int_{- \infty}^{+ \infty} \frac{1}{x ^{2} + 2 x + 2} d x (x + 1)^{2} = x^{2} + 2 x + 1 \int_{- \infty}^{+ \infty} \frac{1}{( x + 1 ) ^{2} + 1} d x \int_{- \infty}^{+ \infty} \frac{1}{( x + 1 ) ^{2} + 1} d (x + 1) 令 u = x + 1 \int_{- \infty}^{+ \infty} \frac{1}{1 + u ^{2}} d u = arctan u ∣_{- \infty}^{+ \infty} = x \to + \infty lim arctan x - x \to - \infty lim arctan x = \frac{π}{2} - (- \frac{π}{2}) = π

瑕积分

无界函数的反常积分

教材定义（了解）

设函数f(x)在(a,b]上连续，且 $lim_{x \to a^{+}} = \infty$ 。取 $ϵ > 0$ ，如果 $lim_{ϵ \to 0^{+}} \int_{a + ϵ}^{b} f (x) d x$ 存在，则此极限值叫做函数f(x)在区间(a,b]上的广义积分，记作 $\int_{a}^{b} f (x) d x$ ，即
$\int_{a}^{b} f (x) d x = lim_{ϵ \to 0^{+}} \int_{a + ϵ}^{b} f (x) d x$
这时称广义积分收敛，否则称广义积分发散。

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考研定义

设函数f(x)在(a,b]上连续，且 $lim_{x \to a^{+}} f (x) = \infty$ ，即称x=a为f(x)的无穷间断点（又称瑕点）。

如 $\int_{0}^{1} \frac{1}{x} d x$ ， $lim_{x \to 0^{+}} \frac{1}{x} = + \infty$ ，所以x=0为 $\int_{0}^{1} \frac{1}{x} d x$ 的瑕点

设 a 为f(x)的瑕点（ $f (x)$ 在 $x = a$ 领域无界）,若 $\int_{a}^{b} f (x) d x = F (x)_{a^{+}}^{b} = F (b) - F (a^{+})$ 极限存在，称反常积分 $\int_{a}^{b} f (x) d x$ 收敛，否则称为发散。

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设 b 为f(x)的瑕点（ $f (x)$ 在 $x = b$ 领域无界）,若 $\int_{a}^{b} f (x) d x = F (x)_{a}^{b^{-}} = F (b^{-}) - F (a)$ 极限存在，称反常积分 $\int_{a}^{b} f (x) d x$ 收敛，否则称为发散。

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若瑕点x=c在[a,b]内部，则f(x)在[a,b]上的广义积分为

x	a	c	b
		瑕点

\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x = F (x) ∣_{a}^{c} + F (x) ∣_{c}^{b} = [F (c^{-}) - F (a)] + [F (b) - F (c^{+})]

做题步骤：

判断被积函数f(x)是否连续，用来排除瑕点在内部的情况（在内部就拆开）
按普通积分算
能代值则完毕，带入不了，就考虑极限

题目

$\int_{0}^{1} \frac{s i n x}{x} d x$

x \to 0^{+} lim \frac{sin x}{x} = 1 ∴ x = 0 不是瑕点

\int_{0}^{1} \frac{1}{x} d x \int_{0}^{1} \frac{1}{x} d x = 2 x_{0}^{1} = 2 - 0 = 2

为什么可以直接代值，这道题直接代值本质是算 $F (x) = 2 x$ 的F(0)，代入后发现F(0)存在，且F(x)在x=0处连续（因为可以直接带入x=0,F(x)函数存在），所以极限值等于函数值，等于 $lim_{x \to 0^{+}} 2 x = 0$

\int_{- 1}^{1} \frac{1}{x ^{2}} d x x = 0 为瑕点 \int_{- 1}^{1} \frac{1}{x ^{2}} d x = \int_{- 1}^{0} \frac{1}{x ^{2}} d x + \int_{0}^{1} \frac{1}{x ^{2}} d x - \frac{1}{x}_{- 1}^{0} + (- \frac{1}{x})_{0}^{1}

-1 --- 0（瑕点） --- 1

- \frac{1}{x}_{- 1}^{0}, - \frac{1}{x}_{0}^{1}, - \frac{1}{x}_{- 1}^{0} + (- \frac{1}{x})_{0}^{1} = - x \in (- 1, 0) x \in (0, 1) = x \to 0^{-} lim (- \frac{1}{x}) - (- \frac{1}{- 1}) + (- \frac{1}{1} - x \to 0^{+} lim - (\frac{1}{x})) (- \infty) - 1 - 1 - (- \infty) = + \infty - 2 + \infty = \infty ∴ \int_{- 1}^{1} \frac{1}{x ^{2}} d x 发散

或者只算一个就行了

(- \frac{1}{x})_{0}^{1} = (- \frac{1}{1}) - x \to 0^{+} lim (- \frac{1}{x}) - 1 - (- \infty) = + \infty

\int_{0}^{1} \frac{ln x}{x} d x x = 0 为瑕点 \int_{0}^{1} ln x d (ln x) = \frac{1}{2} ln^{2} x_{0}^{1} = \frac{1}{2} (ln^{2} 1 - x \to 0^{+} lim ln^{2} x) x \to 0^{+}, ln x \to - \infty, ln^{2} x \to + \infty = 0 - \frac{1}{2} \cdot (+ \infty) = - \infty ∴ \int_{0}^{1} \frac{ln x}{x} d x 发散

这道题求完原函数，代入瑕点原函数的极限不存在，所以要取极限。

\int_{0}^{2} \frac{1}{3 ( x - 1 ) ^{2}} d x x = 1 为瑕点 \int_{0}^{1} \frac{1}{3 ( x - 1 ) ^{2}} d x + \int_{1}^{2} \frac{1}{3 ( x - 1 ) ^{2}} d x \int_{0}^{1} (x - 1)^{- \frac{2}{3}} d (x - 1) + \int_{1}^{2} (x - 1)^{- \frac{2}{3}} d (x - 1) 3 (x - 1)^{\frac{1}{3}}_{0}^{1} + 3 (x - 1)^{\frac{1}{3}}_{1}^{2} 0 - 3 (- 1) + 3 - 0 = 6

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x

x=2为瑕点
第一种方法凑 $arcsin x$

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x = \int_{0}^{2} \frac{1}{4 ( 1 - \frac{x ^{2}}{4} )} d x 2 \int_{0}^{2} \frac{1}{1 - \frac{x ^{2}}{4}} d x 2 \int_{0}^{2} \frac{1}{1 - ( \frac{x}{2} ) ^{2}} d x \int_{0}^{2} \frac{1}{1 - ( \frac{x}{2} ) ^{2}} d (\frac{x}{2}) arcsin \frac{x}{2}_{0}^{2} = arcsin 1 - arcsin 0 = \frac{π}{2}

也可以用三角换元来做

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x 令 x = 2 sin t, d x = d (2 sin t) = 2 cos t d t x = 2, 2 = 2 sin t, t = \frac{π}{2} x = 0, t = 0 4 - x^{2} = 4 - 4 sin^{2} t = 4 (1 - sin^{2} t) = 2 cos t \int_{0}^{\frac{π}{2}} \frac{1}{2 cos t} 2 cos t d t = \int_{0}^{\frac{π}{2}} 1 d t = t_{0}^{\frac{π}{2}} = \frac{π}{2}

\int_{0}^{1} \frac{1}{1 - x ^{2}} d x

x	0	1
	取得到	瑕点，取不到

arcsin x_{0}^{1} = x \to 1^{-} lim arcsin x - arcsin 0 = \frac{π}{2} arcsin x_{0}^{1} = x \to 1 lim arcsin x - arcsin 0 = \frac{π}{2} arcsin x_{0}^{1} = arcsin x - arcsin 0 = \frac{π}{2} 这三种写法都可以

另一种左右极限的理解方式，理解即可，可以不看

由于函数 $f (x) = arcsin x$ 的定义域是 $[- 1, 1]$ ，在 $x = 1$ 的右侧（即 $1^{+}$ ）函数是没有定义的。

教材逻辑： 当一个函数在某点的某侧没有定义时，在该点处的极限 $lim_{x \to 1}$ 默认指的就是其有定义的那一侧（即左极限）。

\int_{- 1}^{1} \frac{1}{x ^{4}} d x x \to 0 lim \frac{1}{x ^{4}} = \infty, x = 0 为瑕点 \int_{- 1}^{1} \frac{1}{x ^{4}} d x = \int_{- 1}^{0} \frac{1}{x ^{4}} d x + \int_{0}^{1} \frac{1}{x ^{4}} d x \int_{- 1}^{0} x^{- 4} d x = - \frac{1}{3} x^{- 3}_{- 1}^{0} = - \frac{1}{3} \cdot \frac{1}{x ^{3}}_{- 1}^{0} = - \frac{1}{3} \cdot (x \to 0^{-} lim \frac{1}{x ^{3}} - 1) = \infty ∴ \int_{- 1}^{1} \frac{1}{x ^{4}} d x 发散

\int_{0}^{1} ln (1 - x) d x 定义域： 1 - x > 0, x < 1 x = 1 为瑕点 x \cdot ln (1 - x)_{0}^{1} + \int_{0}^{1} x \cdot \frac{1}{1 - x} d x 先求一下积分原函数 \int_{0}^{1} \frac{x}{1 - x} d x = - \int_{0}^{1} \frac{( 1 - x ) - 1}{1 - x} d x = - [\int_{0}^{1} d x - \int_{0}^{1} \frac{1}{1 - x} d x] = - (1 + \int_{0}^{1} \frac{1}{1 - x} d (1 - x)) = - 1 - ln ∣1 - x ∣_{0}^{1} = - 1 - ln (1 - x)_{0}^{1} = - 1 - lim_{x \to 1^{-}} ln (1 - x) x \cdot ln (1 - x)_{0}^{1} = lim_{x \to 1^{-}} x \cdot ln (1 - x) lim_{x \to 1^{-}} x \cdot ln (1 - x) - lim_{x \to 1^{-}} ln (1 - x) - 1 合并 lim_{x \to 1^{-}} x \cdot ln (1 - x) - ln (1 - x) - 1 lim_{x \to 1^{-}} (x - 1) ln (1 - x) - 1 = - 1

为啥这里可以合并？

这里极限可以直接合并，因为本质是求 \int ln (1 - x) d x \int ln (1 - x) d x = x ln x + \int \frac{x}{1 - x} d x = x ln (1 - x) - \int \frac{( 1 - x ) - 1}{1 - x} d x = x ln (1 - x) - \int d x + \int \frac{1}{1 - x} d x = x ln (1 - x) - \int d x - \int \frac{1}{1 - x} d (1 - x) = x ln (1 - x) - ln ∣1 - x ∣ - x + C [x ln (1 - x) - ln ∣1 - x ∣ - x]_{0}^{1} [x ln (1 - x) - ln (1 - x) - x]_{0}^{1} [(x - 1) ln (1 - x) - x]_{0}^{1} x \to 1^{-} lim (x - 1) ln (1 - x) - x \to 1^{-} lim x = - 1

另一种解法

\int_{0}^{1} ln (1 - x) d x 令 t = 1 - x, x = 1 - t, d x = d (1 - t) = - d t x = 1, t = 0 x = 0, t = 1 - \int_{1}^{0} ln t d t = \int_{0}^{1} ln t d t = (t ln t_{0}^{1} - \int_{0}^{1} d t) - t ln t_{0}^{1} - 1 - (0 - x \to 0^{+} lim \frac{ln t}{\frac{1}{t}}) - 1 x \to 0^{+} lim \frac{\frac{1}{t}}{- \frac{1}{t ^{2}}} = - x \to 0^{+} lim \frac{1}{t} \cdot t^{2} = x \to 0^{+} lim t = 0 ∴ \int_{0}^{1} ln (1 - x) d x = - 1

常见结论

无穷区间积分

\int_{a}^{+ \infty} \frac{d x}{x ^{α} ln ^{β} x} \int_{a}^{+ \infty} \frac{1}{x ^{p}} d x {p > 1, p \leq 1, 收敛 发散 (a > 0) \int_{a}^{b} \frac{d x}{( x - a ) ^{q}}, \int_{a}^{b} \frac{d x}{( b - x ) ^{q}} ⎩ ⎨ ⎧ q < 1, 收敛 q \geq 1, 发散 = ⎩ ⎨ ⎧ α > 1, 收敛, α < 1, 发散, α = 1, {β > 1, 收敛, β \leq 1, 发散 . 积分下限 (a > 1)

求旋转体体积

平行截面面积已知立体体积的求法（理解）

X型图绕X轴旋转

V = \int_{a}^{b} d V = \int_{a}^{b} π f^{2} (x) d x = π \int_{a}^{b} f^{2} (x) d x

若是两个函数 $f_{上} (x), f_{下} (x)$ 间的体积

上面这张图，两个旋转体之间的体积为

V = π \int_{a}^{b} f_{上}^{2} (x) d x - π \int_{a}^{b} f_{下}^{2} (x) d x

Y型图绕Y轴旋转

V = π \int_{c}^{d} f^{2} (y) d y

如果内部有嵌套

V = π \int_{c}^{d} [f_{右}^{2} (y) - f_{左}^{2} (y)] d y

X型图绕Y轴旋转

V = \int_{a}^{b} 2 π x \cdot f (x) d x = 2 π \int_{a}^{b} x f (x) d x

题目

求由 $y = sin x, 0 \leq x \leq π$ 与x轴所围成区域，绕y轴而成的旋转体体积

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V = \int_{0}^{π} 2 π x \cdot sin x d x = - 2 π \int_{0}^{π} x d (cos x) = - 2 π (x cos_{0}^{π} - \int_{0}^{π} cos x d x) = - 2 π (π cos π - sin π) = - 2 π (- π) = 2 π^{2}

求 $y = x, x = 4$ 及x轴所围成的面积，及绕y轴的旋转体体积

{y = x x = 4 交点 (4, 2)

先求面积

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\int_{0}^{4} x d x = \frac{16}{3} 另一种方式，用 Y 型图 y = x \Rightarrow x = y^{2} {x = y^{2} x = 4 \int_{0}^{2} (4 - y^{2}) d y = 4 y_{0}^{2} - \frac{1}{3} y^{3}_{0}^{2} = 8 - \frac{8}{3} \frac{24}{3} - \frac{8}{3} = \frac{16}{3}

绕y轴的旋转体体积:
第一种方法套X型图绕Y轴旋转的公式

V = \int_{a}^{b} 2 π x \cdot f (x) d x = 2 π \int_{0}^{4} (x \cdot x) d x

第二种方法套Y型图绕Y轴旋转的公式，做两个体积之差。

求y=x-1与x=2所围成的面积与y轴旋转后的体积

欧拉积分

Use the parity of the function to compute the definite integral

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\int_{0}^{100 π} ∣ sin x ∣ d x I = 100 \int_{0}^{π} sin x d x = 100 \cdot (- cos x)_{0}^{π} = 100 \cdot (- cos π - (- cos 0)) = 100 \cdot (- (- 1) - (- 1)) = 100 \cdot (2) = 200

若f(x)的周期为T，

\int_{0}^{n T} f (x) d x = n \int_{0}^{T} f (x) d x

\int_{0}^{2023 π} 1 - cos^{2} x d x \int_{0}^{2023 π} ∣ sin x ∣ d x f (x) 的周期 T = π ∴ I = 2023 \int_{0}^{π} sin x d x = 4046

设f(x)是以T为周期的连续函数，即 $f (x + T) = f (x)$ ，则
$\int_{a}^{a + T} f (x) d x = \int_{0}^{T} f (x) d x$ .

\int_{2}^{2 + π} ∣ sin x ∣ d x = \int_{0}^{π} ∣ sin x ∣ d x

\int_{2022 π}^{2022 π + π} sin x d x = \int_{0}^{π} sin x d x = 2

\int_{2}^{2 + 5 π} sin x d x = \int_{0}^{5 π} sin x d x = \int_{0}^{4 π} sin x d x + \int_{4 π}^{5 π} sin x d x = 2 \int_{0}^{2 π} sin x d x + \int_{0}^{π} sin x d x = 0 + 2 = 2

注，辅助角公式：

A sin α + B cos α = A^{2} + B^{2} sin (α + φ) 其中 tan φ = \frac{B}{A}

Quartz 4

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教材定义（了解）