a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3-b^3=(a-b)(a^2+ab+b^2) a3−b3=(a−b)(a2+ab+b2)
a n − b n = ( a − b ) ( a n − 1 + a n − 2 b + ⋯ + a b n − 2 + b n − 1 ) a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}) an−bn=(a−b)(an−1+an−2b+⋯+abn−2+bn−1)
a b c 3 ≤ a + b + c 3 ≤ a 2 + b 2 + c 2 3 \sqrt[3]{abc}\leq\frac{a+b+c}3\leq\sqrt{\frac{a^2+b^2+c^2}3} 3abc≤3a+b+c≤3a2+b2+c2
( a c + b d ) 2 ≤ ( a 2 + b 2 ) ( c 2 + d 2 ) (ac+bd)^2\leq(a^2+b^2)(c^2+d^2) (ac+bd)2≤(a2+b2)(c2+d2)
a n = a 1 + ( n − 1 ) d a_n=a_1+(n-1)d an=a1+(n−1)d
S n = 1 2 ( a 1 + a n ) S_n=\frac12(a_1+a_n) Sn=21(a1+an)
a n = a 1 q n − 1 a_n=a_1q^{n-1} an=a1qn−1
S n = a 1 ( 1 − q n ) 1 − q = a 1 − a n q 1 − q {{S}_{{n}}}={\frac {a_{1}(1-q^{n})}{1-q}}=\frac{a_1-a_nq}{1-q} Sn=1−qa1(1−qn)=1−qa1−anq
1 2 + 2 2 + 3 2 + ⋯ + n 2 = 1 6 n ( n + 1 ) ( 2 n + 1 ) 1^2+2^2+3^2+\cdots+n^2 = \frac16n(n+1)(2n+1) 12+22+32+⋯+n2=61n(n+1)(2n+1)
sin 2 α = 2 sin α cos α \sin 2\alpha = 2\sin \alpha \cos \alpha sin2α=2sinαcosα
cos 2 α = cos 2 α − sin 2 α = 2 cos 2 α − 1 = 1 − 2 sin 2 α \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 2\cos^2 \alpha - 1 = 1 - 2\sin^2 \alpha cos2α=cos2α−sin2α=2cos2α−1=1−2sin2α
tan 2 α = 2 tan α 1 − tan 2 α \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha} tan2α=1−tan2α2tanα
sin ( α ± β ) = sin α cos β ± cos α sin β \sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta sin(α±β)=sinαcosβ±cosαsinβ
cos ( α ± β ) = cos α cos β ∓ sin α sin β \cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta cos(α±β)=cosαcosβ∓sinαsinβ
tan ( α ± β ) = tan α ± tan β 1 ∓ tan α tan β \tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} tan(α±β)=1∓tanαtanβtanα±tanβ
sin 2 α = 1 2 ( 1 − cos 2 α ) \sin^2 \alpha = \frac12(1-\cos 2\alpha) sin2α=21(1−cos2α)
cos 2 α = 1 2 ( 1 + cos 2 α ) \cos^2 \alpha = \frac12(1+\cos 2\alpha) cos2α=21(1+cos2α)
sin 2 α + cos 2 α = 1 \sin^2 \alpha + \cos^2 \alpha = 1 sin2α+cos2α=1
sec 2 α = tan 2 α + 1 \sec^2 \alpha = \tan^2 \alpha + 1 sec2α=tan2α+1
csc 2 α = cot 2 α + 1 \csc^2 \alpha = \cot^2\alpha + 1 csc2α=cot2α+1
sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] \sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)] sinαcosβ=21[sin(α+β)+sin(α−β)]
cos α sin β = 1 2 [ sin ( α + β ) − sin ( α − β ) ] \cos\alpha\sin\beta=\frac12[\sin(\alpha+\beta)-\sin(\alpha-\beta)] cosαsinβ=21[sin(α+β)−sin(α−β)]
cos α cos β = 1 2 [ cos ( α + β ) + cos ( α − β ) ] \cos\alpha\cos\beta=\frac12[\cos(\alpha+\beta)+\cos(\alpha-\beta)] cosαcosβ=21[cos(α+β)+cos(α−β)]
sin α sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] \sin\alpha\sin\beta=\frac12[\cos(\alpha-\beta)-\cos(\alpha+\beta)] sinαsinβ=21[cos(α−β)−cos(α+β)]
sin α + sin β = 2 sin α + β 2 cos α − β 2 \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} sinα+sinβ=2sin2α+βcos2α−β
sin α − sin β = 2 sin α − β 2 cos α + β 2 \sin\alpha-\sin\beta=2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2} sinα−sinβ=2sin2α−βcos2α+β
cos α + cos β = 2 cos α + β 2 cos α − β 2 \cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} cosα+cosβ=2cos2α+βcos2α−β
cos α − cos β = − 2 sin α + β 2 cos α + β 2 \cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2} cosα−cosβ=−2sin2α+βcos2α+β
扇形面积: S = r 2 θ 2 S=\frac{r^2\theta}{2} S=2r2θ
扇形弧长: l = r θ l=r\theta l=rθ
球体体积: V = 4 3 π R 3 V=\frac43\pi R^3 V=34πR3
球体表面积: S = 4 π R 2 S=4\pi R^2 S=4πR2
u v = e v ln u u^v=e^{v\ln u} uv=evlnu
sin x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! = x − x 3 3 ! + x 5 5 ! + o ( x 5 ) , − ∞ < x < + ∞ \sin x=\sum\limits^\infty_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}+o(x^5), -\infty
cos x = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! = 1 − x 2 2 ! + x 4 4 ! + o ( x 4 ) , − ∞ < x < + ∞ \cos x=\sum\limits^\infty_{n=0}(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4), -\infty
ln ( 1 + x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 x n n = x − x 2 2 + x 3 3 + o ( x 3 ) , − 1 < x ≤ 1 \ln(1+x)=\sum\limits^\infty_{n=1}(-1)^{n-1}\frac{x^n}{n}=x-\frac{x^2}2+\frac{x^3}3+o(x^3), -1
ln ( 1 − x ) = − ∑ n = 1 ∞ x n n = − x − x 2 2 − x 3 3 + o ( x 3 ) , − 1 ≤ x < 1 \ln(1-x)=-\sum\limits^\infty_{n=1}\frac{x^n}{n}=-x-\frac{x^2}2-\frac{x^3}3+o(x^3), -1\leq x<1 ln(1−x)=−n=1∑∞nxn=−x−2x2−3x3+o(x3),−1≤x<1
e x = ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 ! + x 3 3 ! + o ( x 3 ) , − ∞ < x < + ∞ e^x=\sum\limits^\infty_{n=0}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3), -\infty
1 1 + x = ∑ n = 0 ∞ ( − 1 ) n x n = 1 − x + x 2 − x 3 + o ( x 3 ) , − 1 < x < 1 \frac1{1+x}=\sum\limits^\infty_{n=0}(-1)^nx^n=1-x+x^2-x^3+o(x^3), -1
1 1 − x = ∑ n = 0 ∞ x n = 1 + x + x 2 + x 3 + o ( x 3 ) , − 1 < x < 1 \frac1{1-x}=\sum\limits^\infty_{n=0}x^n=1+x+x^2+x^3+o(x^3), -1
x ( 1 − x ) 2 = ∑ n = 1 ∞ n x n , − 1 < x < 1 \frac x{(1-x)^2}=\sum\limits^\infty_{n=1}nx^n,-1
( 1 + x ) α = 1 + α x + α ( α − 1 ) 2 ! x 2 + o ( x 2 ) (1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+o(x^2) (1+x)α=1+αx+2!α(α−1)x2+o(x2)
tan x = x + x 3 3 + o ( x 3 ) \tan x=x+\frac{x^3}3+o({x^3}) tanx=x+3x3+o(x3)
arcsin x = x + x 3 3 ! + o ( x 3 ) \arcsin x=x+\frac{x^3}{3!}+o(x^3) arcsinx=x+3!x3+o(x3)
arctan x = x − x 3 3 + o ( x 3 ) \arctan x=x-\frac{x^3}3+o(x^3) arctanx=x−3x3+o(x3)
lim x → ∞ ( 1 + 1 x ) x = e \lim\limits_{x\to\infty}(1+\frac1x)^x=e x→∞lim(1+x1)x=e
f ′ ( x 0 ) = lim Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x = lim x → x 0 f ( x ) − f ( x 0 ) x − x 0 f'(x_0) = \lim\limits_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x} = \lim\limits_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} f′(x0)=Δx→0limΔxf(x0+Δx)−f(x0)=x→x0limx−x0f(x)−f(x0)
d [ u v ] = u d v + v d u d[uv]=udv+vdu d[uv]=udv+vdu
d ( u v ) = v d u − u d v v 2 d(\frac uv)=\frac{vdu-udv}{v^2} d(vu)=v2vdu−udv
( x a ) ′ = a x a − 1 , a > 0 , a ≠ 1 (x^a)'=ax^{a-1}, a>0, a\not=1 (xa)′=axa−1,a>0,a=1
( a x ) ′ = a x ln a (a^x)'=a^x\ln a (ax)′=axlna
( e x ) ′ = e x (e^x)'=e^x (ex)′=ex
( log a x ) ′ = 1 x ln a (\log_ax)'=\frac1{x\ln a} (logax)′=xlna1
( ln ∣ x ∣ ) ′ = 1 x (\ln |x|)'=\frac1x (ln∣x∣)′=x1
[ ln ( x + x 2 + 1 ) ] ′ = 1 x 2 + 1 [\ln(x+\sqrt{x^2+1})]'=\frac1{\sqrt{x^2+1}} [ln(x+x2+1)]′=x2+11
[ ln ( x + x 2 − 1 ) ] ′ = 1 x 2 − 1 [\ln(x+\sqrt{x^2-1})]'=\frac1{\sqrt{x^2-1}} [ln(x+x2−1)]′=x2−11
( sin x ) ′ = cos x (\sin x)'=\cos x (sinx)′=cosx
( cos x ) ′ = − sin x (\cos x)'=-\sin x (cosx)′=−sinx
( tan x ) ′ = sec 2 x (\tan x)'=\sec^2x (tanx)′=sec2x
( cot x ) ′ = − csc 2 x (\cot x)'=-\csc^2x (cotx)′=−csc2x
( sec x ) ′ = sec x tan x (\sec x)'=\sec x\tan x (secx)′=secxtanx
( csc x ) ′ = − csc x cot x (\csc x)'=-\csc x\cot x (cscx)′=−cscxcotx
( arcsin x ) ′ = 1 1 − x 2 (\arcsin x)'=\frac1{\sqrt{1-x^2}} (arcsinx)′=1−x21
( arccos x ) ′ = − 1 1 − x 2 (\arccos x)'=-\frac1{\sqrt{1-x^2}} (arccosx)′=−1−x21
( arctan x ) ′ = 1 1 + x 2 (\arctan x)'=\frac1{1+x^2} (arctanx)′=1+x21
( arccot x ) ′ = − 1 1 + x 2 (\text{arccot }x)'=-\frac1{1+x^2} (arccot x)′=−1+x21
( u v ) ( n ) = ∑ k = 0 n C n k u k v ( n − k ) (uv)^{(n)} = \sum\limits_{k=0}^n C_n^k u^kv^{(n-k)} (uv)(n)=k=0∑nCnkukv(n−k)
f ( x ) = ∑ n = 0 ∞ f ( n ) ( x 0 ) ( x − x 0 ) n n ! f(x) = \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(x_0)(x-x_0)^n}{n!} f(x)=n=0∑∞n!f(n)(x0)(x−x0)n
f ( x ) = ∑ n = 0 ∞ f ( n ) ( 0 ) ( x ) n n ! f(x) = \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(0)(x)^n}{n!} f(x)=n=0∑∞n!f(n)(0)(x)n
m ≤ μ ≤ M ⇒ f ( ξ ) = μ m\leq\mu\leq M \Rightarrow f(\xi) = \mu m≤μ≤M⇒f(ξ)=μ
f ( a ) ⋅ f ( b ) < 0 ⇒ f ′ ( ξ ) = 0 f(a)\cdot f(b) < 0 \Rightarrow f'(\xi) = 0 f(a)⋅f(b)<0⇒f′(ξ)=0
x = x 0 x=x_0 x=x0处连续可导取极值 ⇒ f ′ ( x 0 ) = 0 \Rightarrow f'(x_0)=0 ⇒f′(x0)=0(充分不必要条件)
f ( a ) = f ( b ) ⇒ f ′ ( ξ ) = 0 f(a)=f(b)\Rightarrow f'(\xi) = 0 f(a)=f(b)⇒f′(ξ)=0
f ( b ) − f ( a ) = f ′ ( ξ ) ( b − a ) f(b)-f(a) = f'(\xi)(b-a) f(b)−f(a)=f′(ξ)(b−a)
f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( ξ ) g ′ ( ξ ) \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)} g(b)−g(a)f(b)−f(a)=g′(ξ)f′(ξ)
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + ⋯ + f ( n ) ( x 0 ) ( x − x 0 ) n n ! + f ( n + 1 ) ( ξ ) ( x − x 0 ) n + 1 ( n + 1 ) ! f(x) = f(x_0) + f'(x_0)(x-x_0) + \cdots + \frac{f^{(n)}(x_0)(x-x_0)^n}{n!} + \frac{f^{(n+1)}(\xi)(x-x_0)^{n+1}}{(n+1)!} f(x)=f(x0)+f′(x0)(x−x0)+⋯+n!f(n)(x0)(x−x0)n+(n+1)!f(n+1)(ξ)(x−x0)n+1
∫ a b f ( x ) d x = f ( ξ ) ( b − a ) \int_a^b f(x)\ dx = f(\xi)(b-a) ∫abf(x) dx=f(ξ)(b−a)
∫ a b f ( x ) g ( x ) d x = f ( ξ ) ∫ a b g ( x ) d x , g ( x ) \int_a^b f(x)g(x)\ dx = f(\xi)\int_a^b g(x)\ dx,\ g(x) ∫abf(x)g(x) dx=f(ξ)∫abg(x) dx, g(x)不变号
凹函数 F ( x 1 ) + F ( x 2 ) 2 ≥ F ( x 1 + x 2 2 ) \frac{F(x_1)+F(x_2)}{2}\geq F(\frac{x_1+x_2}2) 2F(x1)+F(x2)≥F(2x1+x2)
sin x < x < tan x ( 0 < x < π 2 ) \sin x
arctan x ≤ x ≤ arcsin x ( 0 ≤ x ≤ 1 ) \arctan x\leq x \leq \arcsin x(0\leq x\leq 1) arctanx≤x≤arcsinx(0≤x≤1)
e x ≥ x + 1 e^x\geq x+1 ex≥x+1
ln x ≤ x − 1 \ln x\leq x-1 lnx≤x−1
1 1 + x < ln ( 1 + 1 x ) < 1 x ( x > 0 ) \frac1{1+x}<\ln(1+\frac1x)<\frac1x(x>0) 1+x1<ln(1+x1)<x1(x>0)
k = ∣ y ′ ′ ∣ [ 1 + ( y ′ ) 2 ] 3 2 k=\frac{|y''|}{[1+(y')^2]^{\frac32}} k=[1+(y′)2]23∣y′′∣
R = 1 k = [ 1 + ( y ′ ) 2 ] 3 2 ∣ y ′ ′ ∣ R=\frac1k=\frac{[1+(y')^2]^{\frac32}}{|y''|} R=k1=∣y′′∣[1+(y′)2]23
∫ x k d x = 1 k + 1 + C , k ≠ − 1 \int x^k\ dx=\frac{1}{k+1}+C,\ k\not=-1 ∫xk dx=k+11+C, k=−1
∫ 1 x d x = ln ∣ x ∣ + C \int \frac{1}{x}\ dx=\ln|x|+C ∫x1 dx=ln∣x∣+C
∫ e x d x = e x + C \int e^x\ dx = e^x+C ∫ex dx=ex+C
∫ a x d x = 1 ln a ⋅ a x + C \int a^x \ dx= \frac{1}{\ln a}\cdot a^x+C ∫ax dx=lna1⋅ax+C
∫ 1 a 2 + x 2 d x = 1 a arctan x a + C \int \frac1{a^2+x^2}\ dx = \frac1a\arctan\frac xa+C ∫a2+x21 dx=a1arctanax+C
∫ 1 a 2 − x 2 d x = arcsin x a + C \int\frac1{\sqrt{a^2-x^2}}\ dx = \arcsin\frac xa+C ∫a2−x21 dx=arcsinax+C
∫ 1 x 2 − a 2 d x = 1 2 a ln ∣ x − a x + a ∣ + C \int\frac1{x^2-a^2}\ dx = \frac1{2a}\ln|\frac{x-a}{x+a}|+C ∫x2−a21 dx=2a1ln∣x+ax−a∣+C
∫ 1 a 2 − x 2 d x = 1 2 a ln ∣ x + a x − a ∣ + C \int\frac1{a^2-x^2}\ dx = \frac1{2a}\ln|\frac{x+a}{x-a}|+C ∫a2−x21 dx=2a1ln∣x−ax+a∣+C
∫ 1 x 2 − a 2 d x = ln ( x + x 2 − a 2 ) + C \int\frac1{\sqrt{x^2-a^2}}\ dx = \ln(x+\sqrt{x^2-a^2})+C ∫x2−a21 dx=ln(x+x2−a2)+C
∫ 1 x 2 + a 2 d x = ln ( x + x 2 + a 2 ) + C \int\frac1{\sqrt{x^2+a^2}}\ dx = \ln(x+\sqrt{x^2+a^2})+C ∫x2+a21 dx=ln(x+x2+a2)+C
∫ sin x = − cos x + C \int \sin x = -\cos x + C ∫sinx=−cosx+C
∫ cos x = sin x + C \int \cos x = \sin x + C ∫cosx=sinx+C
∫ tan x = − ln ∣ cos x ∣ + C \int\tan x = -\ln|\cos x| + C ∫tanx=−ln∣cosx∣+C
∫ cot x = ln ∣ sin x ∣ + C \int\cot x = \ln|\sin x| + C ∫cotx=ln∣sinx∣+C
∫ sec x = ln ∣ sec x + tan x ∣ + C \int\sec x = \ln|\sec x +\tan x| + C ∫secx=ln∣secx+tanx∣+C
∫ csc x = ln ∣ csc x − cot x ∣ + C \int\csc x = \ln|\csc x -\cot x| + C ∫cscx=ln∣cscx−cotx∣+C
∫ sec 2 x = tan x + C \int \sec^2 x = \tan x + C ∫sec2x=tanx+C
∫ csc 2 x = − cot x + C \int \csc^2 x = -\cot x + C ∫csc2x=−cotx+C
∫ sec x tan x = sec x + C \int\sec x\tan x = \sec x + C ∫secxtanx=secx+C
∫ csc x cot x = − csc x + C \int\csc x\cot x = -\csc x + C ∫cscxcotx=−cscx+C
∫ u d v = u v − ∫ v d u \int u\ dv = uv - \int v\ du ∫u dv=uv−∫v du
lim n → ∞ ∑ i = 1 n f ( 1 n ⋅ i ) ⋅ 1 n = ∫ 0 1 f ( x ) d x \lim\limits_{n\to\infty}\sum\limits_{i=1}^nf(\frac 1n\cdot i)\cdot\frac1n=\int_0^1f(x) dx n→∞limi=1∑nf(n1⋅i)⋅n1=∫01f(x)dx
当 n n n为大于等于2的偶数时, ∫ 0 π 2 s i n n x d x = ∫ 0 π 2 c o s n x d x = n − 1 n ⋅ n − 3 n − 2 ⋯ 1 2 ⋅ π 2 \int_{0}^{\frac{\pi}{2}}{\mathrm{sin}}^nx\ dx=\int_{0}^{\frac{\pi}{2}}{\mathrm{cos}}^nx\ dx=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots\frac{1}{2}\cdot\frac{\pi}{2} ∫02πsinnx dx=∫02πcosnx dx=nn−1⋅n−2n−3⋯21⋅2π
当 n n n为大于等于3的奇数时, ∫ 0 π 2 s i n n x d x = ∫ 0 π 2 c o s n x d x = n − 1 n ⋅ n − 3 n − 2 ⋯ 2 3 \int_{0}^{\frac{\pi}{2}}{\mathrm{sin}}^nx\ dx=\int_{0}^{\frac{\pi}{2}}{\mathrm{cos}}^nx\ dx=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots\frac{2}{3} ∫02πsinnx dx=∫02πcosnx dx=nn−1⋅n−2n−3⋯32
∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx ∫abf(x) dx=∫abf(a+b−x) dx
∫ a b f ( x ) d x = 1 2 ∫ a b [ f ( x ) + f ( a + b − x ) ] d x \int_a^b f(x)\ dx = \frac12\int_a^b [f(x) + f(a+b-x)]\ dx ∫abf(x) dx=21∫ab[f(x)+f(a+b−x)] dx
∫ a b f ( x ) d x = ∫ a a + b 2 [ f ( x ) + f ( a + b − x ) ] d x \int_a^b f(x)\ dx = \int_a^{\frac{a+b}2} [f(x) + f(a+b-x)]\ dx ∫abf(x) dx=∫a2a+b[f(x)+f(a+b−x)] dx
∫ 0 π x f ( sin x ) d x = π 2 ∫ 0 π f ( sin x ) d x \int_0^\pi xf(\sin x)\ dx = \frac\pi2\int_0^\pi f(\sin x)\ dx ∫0πxf(sinx) dx=2π∫0πf(sinx) dx
∫ 0 π x f ( sin x ) d x = π ∫ 0 π 2 f ( sin x ) d x \int_0^\pi xf(\sin x)\ dx = \pi\int_0^{\frac\pi2} f(\sin x)\ dx ∫0πxf(sinx) dx=π∫02πf(sinx) dx
极坐标系: S = ∫ α β 1 2 r 2 ( θ ) d θ S=\int_\alpha^\beta\frac12r^2(\theta)\ d\theta S=∫αβ21r2(θ) dθ
直角坐标方程: s = ∫ a b 1 + [ y ′ ( x ) ] 2 d x s=\int_a^b\sqrt{1+[y'(x)]^2}\ dx s=∫ab1+[y′(x)]2 dx
参数方程: s = ∫ α β [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t s=\int_\alpha^\beta \sqrt{[x'(t)]^2+[y'(t)]^2}\ dt s=∫αβ[x′(t)]2+[y′(t)]2 dt
极坐标方程: s = ∫ α β [ r ( θ ) ] 2 + [ r ′ ( θ ) ] 2 d θ s=\int_\alpha^\beta \sqrt{[r(\theta)]^2+[r'(\theta)]^2}\ d\theta s=∫αβ[r(θ)]2+[r′(θ)]2 dθ
V x = π ∫ a b ∣ y 1 2 ( x ) − y 2 2 ( x ) ∣ d x V_x=\pi\int_a^b|y_1^2(x)-y_2^2(x)|\ dx Vx=π∫ab∣y12(x)−y22(x)∣ dx
V y = 2 π ∫ a b x ∣ y 1 ( x ) − y 2 ( x ) ∣ d x V_y = 2\pi\int_a^b x|y_1(x)-y_2(x)|\ dx Vy=2π∫abx∣y1(x)−y2(x)∣ dx
直角坐标方程绕 x x x轴: S = 2 π ∫ a b ∣ y ( x ) ∣ 1 + [ y ′ ( x ) ] 2 d x S=2\pi\int_a^b|y(x)|\sqrt{1+[y'(x)]^2}\ dx S=2π∫ab∣y(x)∣1+[y′(x)]2 dx
极坐标方程绕 x x x轴: S = 2 π ∫ α β ∣ y ( t ) ∣ [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t S=2\pi\int_\alpha^\beta|y(t)|\sqrt{[x'(t)]^2+[y'(t)]^2}\ dt S=2π∫αβ∣y(t)∣[x′(t)]2+[y′(t)]2 dt
x ˉ = ∫ a b x f ( x ) d x ∫ a b f ( x ) d x \bar{x}=\frac{\int_a^bxf(x)\ dx}{\int_a^bf(x)\ dx} xˉ=∫abf(x) dx∫abxf(x) dx
y ˉ = 1 2 ∫ a b f 2 ( x ) d x ∫ a b f ( x ) d x \bar{y}=\frac{\frac12\int_a^b f^2(x)\ dx}{\int_a^b f(x)\ dx} yˉ=∫abf(x) dx21∫abf2(x) dx
V = ∫ a b A ( x ) d x V=\int_a^bA(x)\ dx V=∫abA(x) dx
压强 p = ρ g h p=\rho gh p=ρgh
静水压力P= ∫ a b ρ g x ⋅ [ f ( x ) − g ( x ) ] d x \int_a^b \rho gx\cdot[f(x)-g(x)]\ dx ∫abρgx⋅[f(x)−g(x)] dx
细杆质心 x ˉ = ∫ a b x ρ ( x ) d x ∫ a b ρ ( x ) d x \bar{x}=\frac{\int_a^bx\rho(x)\ dx}{\int_a^b\rho(x)\ dx} xˉ=∫abρ(x) dx∫abxρ(x) dx
∫ 1 + ∞ 1 x p d x \int_1^{+\infty}\frac1{x^p}\ dx ∫1+∞xp1 dx当 p > 1 p>1 p>1时收敛, p < = 1 p<=1 p<=1时发散(越大越收敛)
∫ 0 1 1 x p d x \int_0^1\frac1{x^p}\ dx ∫01xp1 dx当 p > = 1 p>=1 p>=1时发散, 0 < p < 1 0 0<p<1
F ′ ( x ) = d d x [ ∫ ϕ 1 ( x ) ϕ 2 ( x ) f ( t ) d t ] = f [ ϕ 2 ( x ) ] ϕ 2 ′ ( x ) − f [ ϕ 1 ( x ) ] ϕ 1 ′ ( x ) F'(x)=\frac{d}{dx}[\int_{\phi_1(x)}^{\phi_2(x)}f(t)\ dt]=f[\phi_2(x)]\phi_2'(x)-f[\phi_1(x)]\phi_1'(x) F′(x)=dxd[∫ϕ1(x)ϕ2(x)f(t) dt]=f[ϕ2(x)]ϕ2′(x)−f[ϕ1(x)]ϕ1′(x)
Δ z = lim Δ x → 0 Δ y → 0 f ( x + Δ x , y + Δ x ) − f ( x , y ) = A Δ x + B Δ y + o ( ρ ) \Delta z=\lim\limits_{\Delta x\to0 \atop \Delta y\to 0}f(x+\Delta x, y+\Delta x)-f(x, y) = A\Delta x+B\Delta y + o(\rho) Δz=Δy→0Δx→0limf(x+Δx,y+Δx)−f(x,y)=AΔx+BΔy+o(ρ),其中, A = f x ′ ( x , y ) , B = f y ′ ( x , y ) , ρ = x 2 + y 2 A=f'_x(x, y), B=f'_y(x, y), \rho=\sqrt{x^2+y^2} A=fx′(x,y),B=fy′(x,y),ρ=x2+y2
d z = f x ′ ( x , y ) d x + f y ′ ( x , y ) d y dz = f'_x(x, y)\ dx + f'_y(x, y)\ dy dz=fx′(x,y) dx+fy′(x,y) dy
Δ z x = lim Δ x → 0 f ( x + Δ x , y ) − f ( x , y ) = A Δ x + o ( ρ ) \Delta z_x = \lim\limits_{\Delta x\to0}f(x+\Delta x, y) - f(x, y) = A\Delta x + o(\rho) Δzx=Δx→0limf(x+Δx,y)−f(x,y)=AΔx+o(ρ)
f x ′ ( x 0 , y 0 ) = lim x → x 0 f ( x , y 0 ) − f ( x 0 , y 0 ) x − x 0 f'_x(x_0, y_0)=\lim\limits_{x\to x_0} \frac{f(x, y_0)-f(x_0, y_0)}{x-x_0} fx′(x0,y0)=x→x0limx−x0f(x,y0)−f(x0,y0)
∂ z ∂ x = − F x ′ F y ′ \frac{\partial z}{\partial x}=-\frac{F'_x}{F'_y} ∂x∂z=−Fy′Fx′
∂ z ∂ y = − F y ′ F y ′ \frac{\partial z}{\partial y}=-\frac{F'_y}{F'_y} ∂y∂z=−Fy′Fy′
d y d x = ∂ ( F , G ) ∂ ( x , z ) ∂ ( F , G ) ∂ ( y , z ) \frac{dy}{dx}=\frac{\frac{\partial(F, G)}{\partial(x, z)}}{\frac{\partial(F, G)}{\partial(y, z)}} dxdy=∂(y,z)∂(F,G)∂(x,z)∂(F,G)
d z d x = ∂ ( F , G ) ∂ ( y , x ) ∂ ( F , G ) ∂ ( y , z ) \frac{dz}{dx}=\frac{\frac{\partial(F, G)}{\partial(y, x)}}{\frac{\partial(F, G)}{\partial(y, z)}} dxdz=∂(y,z)∂(F,G)∂(y,x)∂(F,G)
f ( x , y ) = f ( x 0 , y 0 ) + ( f x ′ , f y ′ ) X 0 ( Δ x Δ y ) + 1 2 ( Δ x Δ y ) ( f x x ′ ′ f x y ′ ′ f y x ′ ′ f y y ′ ′ ) X 0 ( Δ x Δ y ) + R 2 f(x, y) = f(x_0, y_0) + (f'_x, f'_y)_{X_0}\left(\begin{array}{cccc} \Delta x\\ \Delta y\end{array}\right) + \frac12\left(\begin{array}{cccc} \Delta x & \Delta y \end{array}\right)\left(\begin{array}{cccc} f''_{xx} & f''_{xy} \\ f''_{yx} & f''_{yy} \end{array}\right)_{X_0}\left(\begin{array}{cccc} \Delta x\\ \Delta y\end{array}\right) + R_2 f(x,y)=f(x0,y0)+(fx′,fy′)X0(ΔxΔy)+21(ΔxΔy)(fxx′′fyx′′fxy′′fyy′′)X0(ΔxΔy)+R2
∬ D f ( x , y ) d σ = lim n → ∞ ∑ i = 1 n ∑ j = 1 n f ( a + b − a n i , c + d − c n j ) ⋅ b − a n ⋅ d − c n = ∫ a b d x ∫ c d f ( x , y ) d y \iint\limits_{D}f(x, y)\ d\sigma=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\sum\limits_{j=1}^nf(a+\frac{b-a}{n}i, c+\frac{d-c}{n}j)\cdot\frac{b-a}{n}\cdot\frac{d-c}{n}=\int_a^b\ dx\int_c^d f(x, y)\ dy D∬f(x,y) dσ=n→∞limi=1∑nj=1∑nf(a+nb−ai,c+nd−cj)⋅nb−a⋅nd−c=∫ab dx∫cdf(x,y) dy
V = ∬ D x y ∣ z ( x , y ) ∣ d σ V=\iint\limits_{Dxy}|z(x, y)|\ d\sigma V=Dxy∬∣z(x,y)∣ dσ
m = ∬ D ρ ( x , y ) σ m=\iint\limits_D \rho(x, y)\sigma m=D∬ρ(x,y)σ
x ˉ = ∬ D x ρ ( x , y ) d σ ∬ D ρ ( x , y ) d σ \bar{x}=\frac{\iint\limits_D x\rho(x, y)\ d\sigma}{\iint\limits_D \rho(x, y)\ d\sigma} xˉ=D∬ρ(x,y) dσD∬xρ(x,y) dσ
y ˉ = ∬ D y ρ ( x , y ) d σ ∬ D ρ ( x , y ) d σ \bar{y}=\frac{\iint\limits_D y\rho(x, y)\ d\sigma}{\iint\limits_D \rho(x, y)\ d\sigma} yˉ=D∬ρ(x,y) dσD∬yρ(x,y) dσ
I x = ∬ D y 2 ρ ( x , y ) d σ I_x=\iint\limits_D y^2\rho(x, y)\ d\sigma Ix=D∬y2ρ(x,y) dσ
I y = ∬ D x 2 ρ ( x , y ) d σ I_y=\iint\limits_D x^2\rho(x, y)\ d\sigma Iy=D∬x2ρ(x,y) dσ
I O = ∬ D ( x 2 + y 2 ) ρ ( x , y ) d σ I_O=\iint\limits_D (x^2+y^2)\rho(x, y)\ d\sigma IO=D∬(x2+y2)ρ(x,y) dσ
能写成 y ′ = f ( x ) ⋅ g ( y ) y'=f(x)\cdot g(y) y′=f(x)⋅g(y),直接分离变量
能写成 y ′ = f ( a x + b y + c ) y'=f(ax+by+c) y′=f(ax+by+c),令 u = a x + b y + c u=ax+by+c u=ax+by+c
能写成 y ′ = f ( y x ) y'=f(\frac yx) y′=f(xy)或 y ′ = f ( x y ) y'=f(\frac xy) y′=f(yx),令 u = y x u=\frac yx u=xy 或 u = x y u=\frac xy u=yx
能写成 y ′ + p ( x ) y = q ( x ) y'+p(x)y=q(x) y′+p(x)y=q(x),用公式法: y = e − ∫ p ( x ) d x [ ∫ e ∫ p ( x ) d x q ( x ) d x + C ] y=e^{-\int p(x)\ dx}[\int e^{\int p(x)\ dx}q(x)\ dx + C] y=e−∫p(x) dx[∫e∫p(x) dxq(x) dx+C]
能写成 y ′ + p ( x ) y = q ( x ) y n ( n ≠ 0 , 1 ) y'+p(x)y=q(x)y^n(n\not=0,1) y′+p(x)y=q(x)yn(n=0,1),令 z = y 1 − n z=y^{1-n} z=y1−n,再用公式法
不显含 y y y,令 y ′ = p y'=p y′=p, y ′ ′ = d p d x y''=\frac{dp}{dx} y′′=dxdp
不显含 x x x,令 y ′ = p y'=p y′=p, y ′ ′ = d p d y p y''=\frac{dp}{dy}p y′′=dydpp
y ′ ′ + p y ′ + q y = f ( x ) y''+py'+qy=f(x) y′′+py′+qy=f(x)
λ 2 + p λ + q = 0 \lambda^2+p\lambda+q=0 λ2+pλ+q=0
若 p 2 − 4 q > 0 p^2-4q>0 p2−4q>0, y = C 1 e λ 1 x + C 2 e λ 2 x y=C_1e^{\lambda_1x}+C_2e^{\lambda_2x} y=C1eλ1x+C2eλ2x
若 p 2 − 4 q = 0 p^2-4q=0 p2−4q=0, y = ( C 1 + C 2 x ) e λ x y=(C_1+C_2x)e^{\lambda_x} y=(C1+C2x)eλx
若 p 2 − 4 q < 0 p^2-4q<0 p2−4q<0, y = e α x ( C 1 cos β x + C 2 sin β x ) y=e^{\alpha x}(C_1\cos\beta x+C_2\sin \beta x) y=eαx(C1cosβx+C2sinβx)
当自由项 f ( x ) = P n ( x ) e α x f(x)=P_n(x)e^{\alpha x} f(x)=Pn(x)eαx时,特解设为 y ∗ = e α x Q n ( x ) x k y^*=e^{\alpha x}Q_n(x)x^k y∗=eαxQn(x)xk, k k k为与 α \alpha α相同的 λ \lambda λ的个数.
当自由项 f ( x ) = e α x [ P m ( x ) cos β x + P n ( x ) sin β x ] f(x)=e^{\alpha x}[P_m(x)\cos\beta x+P_n(x)\sin \beta x] f(x)=eαx[Pm(x)cosβx+Pn(x)sinβx]时,特解设为 y ∗ = e α x [ Q l ( 1 ) ( x ) cos β x + Q l ( 2 ) ( x ) sin β x ] x k y^*=e^{\alpha x}[Q_l^{(1)}(x)\cos\beta x+Q_l^{(2)}(x)\sin\beta x]x^k y∗=eαx[Ql(1)(x)cosβx+Ql(2)(x)sinβx]xk, l = max { m , n } l=\max\{m, n\} l=max{m,n}, k k k取决于 α ± β i \alpha\pm\beta i α±βi是否为特征根
x 2 y ′ ′ + p x y ′ + q y = f ( x ) x^2y''+pxy'+qy=f(x) x2y′′+pxy′+qy=f(x)
x > 0 x>0 x>0,令 x = e t x=e^t x=et
x < 0 x<0 x<0,令 x = − e t x=-e^t x=−et
如 y ′ ′ ′ + p 1 y ′ ′ + p 2 y ′ + p 3 y = 0 y'''+p_1y''+p_2y'+p_3y=0 y′′′+p1y′′+p2y′+p3y=0
λ 3 + p 1 λ 2 + p 2 λ + p 3 = 0 \lambda^3+p_1\lambda^2+p_2\lambda+p_3=0 λ3+p1λ2+p2λ+p3=0
单实根: y = C e λ x y=Ce^{\lambda x} y=Ceλx
重实根: y = ( C 1 + C 2 x + C 3 x 2 + ⋯ + C k x k − 1 e λ x ) y=(C_1+C_2x+C_3x^2+\cdots+C_kx^{k-1}e^{\lambda x}) y=(C1+C2x+C3x2+⋯+Ckxk−1eλx)(有高阶必有低阶)
单复根 α ± β i \alpha\pm\beta i α±βi: y = e a x ( C 1 cos β x + C 2 sin β x ) y=e^{ax}(C_1\cos\beta x+C_2\sin\beta x) y=eax(C1cosβx+C2sinβx)(成对出现)
f ( x ) = [ ∫ f ( x ) d x ] ′ f(x)=[\int f(x)\ dx]' f(x)=[∫f(x) dx]′
f ( x ) = f ( x 0 ) + ∫ x 0 x f ′ ( t ) d t f(x)=f(x_0)+\int_{x_0}^x f'(t)\ dt f(x)=f(x0)+∫x0xf′(t) dt
f ( x ) = 1 2 [ f ( x − ) + f ( x + ) ] f(x)=\frac12[f(x^-)+f(x^+)] f(x)=21[f(x−)+f(x+)]
f ( x ) ∼ a 0 2 + ∑ n = 1 ∞ ( a n cos n π x l + b n sin n π x l ) f(x)\sim\frac{a_0}{2}+\sum\limits^\infty_{n=1}(a_n\cos\frac{n\pi x}{l}+b_n\sin\frac{n\pi x}{l}) f(x)∼2a0+n=1∑∞(ancoslnπx+bnsinlnπx)
a 0 = 1 l ∫ − l l f ( x ) d x a_0=\frac1l\int_{-l}^lf(x)\ dx a0=l1∫−llf(x) dx
a n = 1 l ∫ − l l f ( x ) cos n π x l d x a_n=\frac1l\int_{-l}^lf(x)\cos\frac{n\pi x}l\ dx an=l1∫−llf(x)coslnπx dx
b n = 1 l ∫ − l l f ( x ) sin n π x l d x b_n=\frac1l\int_{-l}^lf(x)\sin\frac{n\pi x}l\ dx bn=l1∫−llf(x)sinlnπx dx
切向量: τ = ( x ′ ( t 0 ) , y ′ ( t 0 ) , z ′ ( t 0 ) ) \boldsymbol\tau=(x'(t_0), y'(t_0), z'(t_0)) τ=(x′(t0),y′(t0),z′(t0))
切线方程: x − x 0 x ′ ( t 0 ) = y − y 0 y ′ ( t 0 ) = z − z 0 z ′ ( t 0 ) \frac{x-x_0}{x'(t_0)}=\frac{y-y_0}{y'(t_0)}=\frac{z-z_0}{z'(t_0)} x′(t0)x−x0=y′(t0)y−y0=z′(t0)z−z0
法平面方程: x ′ ( t 0 ) ( x − x 0 ) + y ′ ( t 0 ) ( y − y 0 ) + z ′ ( t 0 ) ( z − z 0 ) = 0 x'(t_0)(x-x_0)+y'(t_0)(y-y_0)+z'(t_0)(z-z_0)=0 x′(t0)(x−x0)+y′(t0)(y−y0)+z′(t0)(z−z0)=0
切向量: τ = ( 1 , y ′ ( x 0 ) , z ′ ( x 0 ) ) \boldsymbol\tau=(1, y'(x_0), z'(x_0)) τ=(1,y′(x0),z′(x0))
切线方程: x − x 0 1 = y − y 0 y ′ ( x 0 ) = z − z 0 z ′ ( x 0 ) \frac{x-x_0}{1}=\frac{y-y_0}{y'(x_0)}=\frac{z-z_0}{z'(x_0)} 1x−x0=y′(x0)y−y0=z′(x0)z−z0
法平面方程: 1 ⋅ ( x − x 0 ) + y ′ ( x ) ( y − y 0 ) + z ′ ( x ) ( z − z 0 ) = 0 1\cdot(x-x_0)+y'(x)(y-y_0)+z'(x)(z-z_0)=0 1⋅(x−x0)+y′(x)(y−y0)+z′(x)(z−z0)=0
法向量: n = ( F x ′ ∣ P 0 , F y ′ ∣ P 0 , F z ′ ∣ P 0 ) \boldsymbol n=(F'_x|_{P_0}, F'_y|_{P_0}, F'_z|_{P_0}) n=(Fx′∣P0,Fy′∣P0,Fz′∣P0)
法平面方程: F x ′ ∣ P 0 ⋅ ( x − x 0 ) + F y ′ ∣ P 0 ⋅ ( y − y 0 ) + F z ′ ∣ P 0 ( z − z 0 ) F'_x|_{P_0}\cdot(x-x_0)+F'_y|_{P_0}\cdot(y-y_0)+F'_z|_{P_0}(z-z_0) Fx′∣P0⋅(x−x0)+Fy′∣P0⋅(y−y0)+Fz′∣P0(z−z0)
法线方程: x − x 0 F x ′ ∣ P 0 = y − y 0 F y ′ ∣ P 0 = z − z 0 F z ′ ∣ P 0 \frac{x-x_0}{F'_x|_{P_0}}=\frac{y-y_0}{F'_y|_{P_0}}=\frac{z-z_0}{F'_z|_{P_0}} Fx′∣P0x−x0=Fy′∣P0y−y0=Fz′∣P0z−z0
法向量: n = ∣ i j k x u ′ y u ′ z u ′ x v ′ y v ′ z v ′ ∣ \boldsymbol n = \left|\begin{array}{cccc}\boldsymbol i&\boldsymbol j&\boldsymbol k\\x'_u&y'_u&z'_u\\x'_v&y'_v&z'_v\end{array}\right| n=∣∣∣∣∣∣ixu′xv′jyu′yv′kzu′zv′∣∣∣∣∣∣
切平面方程: A ( x − x 0 ) + B ( y − y 0 ) − C ( z − z 0 ) = 0 A(x-x_0)+B(y-y_0)-C(z-z_0)=0 A(x−x0)+B(y−y0)−C(z−z0)=0
法线方程: x − x 0 A = y − y 0 B = z − z 0 C \frac{x-x_0}A=\frac{y-y_0}B=\frac{z-z_0}C Ax−x0=By−y0=Cz−z0
柱面的每一个切面的法向量都与某一个向量垂直(只需找一个 τ \boldsymbol\tau τ向量即可)
在 x O y xOy xOy面上的投影消除 z z z, z = 0 z=0 z=0
设出曲线某一点 ( x 0 , y 0 , z 0 )