a n = a 1 + ( n − 1 ) d a_n=a_1+(n-1)d an=a1+(n−1)d S n = n 2 [ 2 a 1 + ( n − 1 ) d ] = n 2 ( a 1 + a n ) = n a 1 + n ( n − 1 ) 2 d S_n=\frac{n}{2}\left[2a_1+(n-1)d\right]=\frac{n}{2}(a_1+a_n)=na_1+\frac{n(n-1)}{2}d Sn=2n[2a1+(n−1)d]=2n(a1+an)=na1+2n(n−1)d
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 S_n=\frac{a_1(1-q^n)}{1-q} Sn=1−qa1(1−qn)
1 + 2 + ⋯ + n = n ( n + 1 ) 2 1+2+\dots+n=\frac{n(n+1)}{2} 1+2+⋯+n=2n(n+1) 1 2 + 2 2 + ⋯ + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6} 12+22+⋯+n2=6n(n+1)(2n+1) 1 1 × 2 + 1 2 × 3 + ⋯ + 1 n × ( n + 1 ) = n n + 1 \frac{1}{1\times2}+\frac{1}{2\times3}+\dots+\frac{1}{n\times(n+1)}=\frac{n}{n+1} 1×21+2×31+⋯+n×(n+1)1=n+1n
1 + tan 2 α = sec 2 α 1+\tan^2\alpha=\sec^2\alpha 1+tan2α=sec2α 1 + cot 2 α = csc 2 α 1+\cot^2\alpha=\csc^2\alpha 1+cot2α=csc2α a sin x + b sin x = a 2 + b 2 sin ( x + φ ) a\sin x+b\sin x=\sqrt{a^2+b^2}\sin(x+\varphi) asinx+bsinx=a2+b2 sin(x+φ)
sin 3 α = − 4 sin 3 α + 3 sin α \sin3\alpha=-4\sin^3\alpha+3\sin\alpha sin3α=−4sin3α+3sinα cos 3 α = 4 cos 3 α − 3 cos α \cos3\alpha=4\cos^3\alpha-3\cos\alpha cos3α=4cos3α−3cosα tan 2 α = 2 tan α 1 − tan 2 α \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha} tan2α=1−tan2α2tanα cot 2 α = cot 2 α − 1 2 cot α \cot2\alpha=\frac{\cot^2\alpha-1}{2\cot\alpha} cot2α=2cotαcot2α−1
tan α 2 = 1 − cos α sin α = sin α 1 + cos α = ± 1 − cos α 1 + cos α \tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}=\pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}} tan2α=sinα1−cosα=1+cosαsinα=±1+cosα1−cosα cot α 2 = sin α 1 − cos α = 1 + cos α sin α = ± 1 + cos α 1 − cos α \cot\frac{\alpha}{2}=\frac{\sin\alpha}{1-\cos\alpha}=\frac{1+\cos\alpha}{\sin\alpha}=\pm\sqrt{\frac{1+\cos\alpha}{1-\cos\alpha}} cot2α=1−cosαsinα=sinα1+cosα=±1−cosα1+cosα
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 cot ( α ± β ) = cot α cot β ∓ 1 cot β ∓ cot α \cot(\alpha\pm\beta)=\frac{\cot\alpha\cot\beta\mp1}{\cot\beta\mp\cot\alpha} cot(α±β)=cotβ∓cotαcotαcotβ∓1
sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] \sin\alpha\cos\beta=\frac{1}{2}[\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=\frac{1}{2}[\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=\frac{1}{2}[\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=\frac{1}{2}[\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 sin α − β 2 \cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} cosα−cosβ=−2sin2α+βsin2α−β
当 μ = tan x 2 ( − π < x < π ) , 则 sin x = 2 μ 1 + μ 2 当\mu=\tan\frac{x}{2}(-\pi<x<\pi),则\sin x=\frac{2\mu}{1+\mu^2} 当μ=tan2x(−π<x<π),则sinx=1+μ22μ
x 1 + x 2 = − b a x_1+x_2=-\frac{b}{a} x1+x2=−ab x 1 x 2 = c a x_1x_2=\frac{c}{a} x1x2=ac
设 y = a x 2 + b x + c , 则 顶 点 : p ( − b 2 a , c − b 2 4 a ) 设y=ax^2+bx+c,则顶点:p(-\frac{b}{2a},c-\frac{b^2}{4a}) 设y=ax2+bx+c,则顶点:p(−2ab,c−4ab2)
∣ A x 0 + B y 0 + C ∣ A 2 + B 2 \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} A2+B2 ∣Ax0+By0+C∣
( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 (a+b)^3=a^3+3a^2b+3ab^2+b^3 (a+b)3=a3+3a2b+3ab2+b3 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 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) n 为 正 整 数 : a n − b n = ( a − b ) ( a n − 1 + a n − 2 b + ⋯ + a b n − 2 + b n − 1 ) n为正整数:a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}) n为正整数:an−bn=(a−b)(an−1+an−2b+⋯+abn−2+bn−1) n 为 正 偶 数 : a n − b n = ( a + b ) ( a n − 1 − a n − 2 b + ⋯ + a b n − 2 − b n − 1 ) n为正偶数:a^n-b^n=(a+b)(a^{n-1}-a^{n-2}b+\dots+ab^{n-2}-b^{n-1}) n为正偶数:an−bn=(a+b)(an−1−an−2b+⋯+abn−2−bn−1) n 为 正 奇 数 : a n + b n = ( a + b ) ( a n − 1 − a n − 2 b + ⋯ − a b n − 2 + b n − 1 ) n为正奇数:a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\dots-ab^{n-2}+b^{n-1}) n为正奇数:an+bn=(a+b)(an−1−an−2b+⋯−abn−2+bn−1) 二 项 式 定 理 : ( a + b ) n = lim k = 0 n C n k a n − k b k 二项式定理:(a+b)^n=\lim^n_{k=0}C^k_na^{n-k}b^k 二项式定理:(a+b)n=k=0limnCnkan−kbk
( 2 n ) ! ! = 2 ⋅ 4 ⋅ 6 … ( 2 n ) = 2 n ⋅ n ! (2n)!!=2\cdot4\cdot6\dots(2n)=2^n\cdot n! (2n)!!=2⋅4⋅6…(2n)=2n⋅n! ( 2 n − 1 ) ! ! = 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) (2n-1)!!=1\cdot3\cdot5\dots(2n-1) (2n−1)!!=1⋅3⋅5…(2n−1)
∣ a ± b ∣ ≤ ∣ a ∣ + ∣ b ∣ |a\pm b|\le|a|+|b| ∣a±b∣≤∣a∣+∣b∣ ∣ ∣ a ∣ + ∣ b ∣ ∣ ≤ ∣ a − b ∣ ||a|+|b||\le|a-b| ∣∣a∣+∣b∣∣≤∣a−b∣ ∣ ∫ a b f ( x ) d x ∣ ≤ ∫ a b ∣ f ( x ) ∣ d x |\int_a^bf(x)dx|\le\int_a^b|f(x)|dx ∣∫abf(x)dx∣≤∫ab∣f(x)∣dx
a b ≤ a + b 2 ≤ a 2 + b 2 2 ( a , b > 0 ) \sqrt{ab}\le\frac{a+b}{2}\le\sqrt{\frac{a^2+b^2}{2}}(a,b>0) ab ≤2a+b≤2a2+b2 (a,b>0)
设 a > b > 0 , 则 { a n > b n , n > 0 a n < b n , n < 0 设a>b>0,则 \left\{\begin{matrix} a^n>b^n, & n>0 \\ a^n<b^n, & n<0 \end{matrix}\right. 设a>b>0,则{an>bn,an<bn,n>0n<0
若 0 < a < x < b , 0 < c < y < d , 则 c b < y x < d a 若0<a<x<b,0<c<y<d,则\frac{c}{b}<\frac{y}{x}<\frac{d}{a} 若0<a<x<b,0<c<y<d,则bc<xy<ad
sin x < x < tan x ( 0 < x < π 2 ) \sin x<x<\tan x(0<x<\frac{\pi}{2}) sinx<x<tanx(0<x<2π)
sin x < x ( x > 0 ) \sin x<x(x>0) sinx<x(x>0)
arcsin x ≤ x ≤ arcsin x ( 0 ≤ x ≤ 1 ) \arcsin x\le x\le\arcsin x(0\le x\le1) arcsinx≤x≤arcsinx(0≤x≤1)
e x ≥ x + 1 e^x\ge x+1 ex≥x+1
x − 1 ≥ ln x ( x > 0 ) x-1\ge\ln x(x>0) x−1≥lnx(x>0)
1 1 + x < ln ( 1 + 1 x ) < 1 x ( x > 0 ) \frac{1}{1+x}<\ln(1+\frac{1}{x})<\frac{1}{x}(x>0) 1+x1<ln(1+x1)<x1(x>0)
sin x = x − x 3 3 ! + o ( x 3 ) \sin x=x-\frac{x^3}{3!}+o(x^3) sinx=x−3!x3+o(x3) cos x = 1 − x 2 2 ! + x 4 4 ! + o ( x 4 ) \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4) cosx=1−2!x2+4!x4+o(x4) 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) tan x = x + x 3 3 + o ( x 3 ) \tan x=x+\frac{x^3}{3}+o(x^3) tanx=x+3x3+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) ln ( 1 + x ) = x − x 2 2 + x 3 3 + o ( x 3 ) \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3) ln(1+x)=x−2x2+3x3+o(x3) e x = 1 + x + x 2 2 ! + x 3 3 ! + o ( x 3 ) e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3) ex=1+x+2!x2+3!x3+o(x3) ( 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) 1 1 − x = 1 + x + x 2 + x 3 + o ( x 3 ) \frac{1}{1-x}=1+x+x^2+x^3+o(x^3) 1−x1=1+x+x2+x3+o(x3) 1 + x 2 = 1 + x 2 2 − x 4 8 + o ( x 4 ) \sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4) 1+x2 =1+2x2−8x4+o(x4) y = f ( x ) = ∑ n = 0 ∞ f ( n ) ( x 0 ) n ! ( x − x 0 ) n = ∑ n = 0 ∞ f ( n ) ( 0 ) n ! x n y=f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n y=f(x)=n=0∑∞n!f(n)(x0)(x−x0)n=n=0∑∞n!f(n)(0)xn
sin x ∼ x 、 tan x ∼ x 、 arcsin x ∼ x 、 arcsin x ∼ x 、 ln ( 1 + x ) ∼ x 、 e x − 1 ∼ x \sin x\sim x、\tan x\sim x、\arcsin x\sim x、\arcsin x\sim x、\ln(1+x)\sim x、e^x-1\sim x sinx∼x、tanx∼x、arcsinx∼x、arcsinx∼x、ln(1+x)∼x、ex−1∼x
a x − 1 ∼ x ln a 、 1 − cos x ∼ 1 2 x 2 、 ( 1 + x ) α − 1 ∼ α x 、 1 − cos α x ∼ a 2 x 2 a^x-1\sim x\ln a、1-\cos x\sim\frac{1}{2}x^2、(1+x)^\alpha-1\sim\alpha x、1-\cos^\alpha x\sim\frac{a}{2}x^2 ax−1∼xlna、1−cosx∼21x2、(1+x)α−1∼αx、1−cosαx∼2ax2
lim x ⟶ ∞ sin x x = 1 \lim_{x\longrightarrow\infty}\frac{\sin x}{x}=1 x⟶∞limxsinx=1 lim x ⟶ ∞ ( 1 + 1 x ) x = e \lim_{x\longrightarrow\infty}(1+\frac{1}{x})^x=e x⟶∞lim(1+x1)x=e
( log α x ) ′ = 1 x ln α ( α > 0 , α ≠ 0 ) (\log_\alpha x)'=\frac{1}{x\ln\alpha}(\alpha>0,\alpha\ne0) (logαx)′=xlnα1(α>0,α=0) ( arcsin x ) ′ = 1 1 − x 2 (\arcsin x)'=\frac{1}{\sqrt{1-x^2}} (arcsinx)′=1−x2 1 ( tan x ) ′ = sec 2 x (\tan x)'=\sec^2x (tanx)′=sec2x ( arccos x ) ′ = − 1 1 − x 2 (\arccos x)'=-\frac{1}{\sqrt{1-x^2}} (arccosx)′=−1−x2 1 ( cot x ) ′ = − csc 2 x (\cot x)'=-\csc^2x (cotx)′=−csc2x ( arctan x ) ′ = 1 1 + x 2 (\arctan x)'=\frac{1}{1+x^2} (arctanx)′=1+x21 ( a r c c o t x ) ′ = − 1 1 + x 2 (arccotx)'=-\frac{1}{1+x^2} (arccotx)′=−1+x21 ( 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 [ ln ( x + x 2 + 1 ) ] ′ = 1 x 2 + 1 \left[\ln\left(x+\sqrt{x^2+1}\right)\right]'=\frac{1}{\sqrt{x^2+1}} [ln(x+x2+1 )]′=x2+1 1 [ ln ( x + x 2 − 1 ) ] ′ = 1 x 2 − 1 \left[\ln\left(x+\sqrt{x^2-1}\right)\right]'=\frac{1}{\sqrt{x^2-1}} [ln(x+x2−1 )]′=x2−1 1
d y d x = d y d t d x d t \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} dxdy=dtdxdtdy d 2 y d x 2 = d ( d y d t ) d t d x d t = y ′ ′ ( t ) x ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) [ x ′ ( t ) ] 3 \frac{d^2y}{dx^2}=\frac{\frac{d(\frac{dy}{dt})}{dt}}{\frac{dx}{dt}}=\frac{y''(t)x'(t)-x''(t)y'(t)}{\left[x'\left(t\right)\right]^3} dx2d2y=dtdxdtd(dtdy)=[x′(t)]3y′′(t)x′(t)−x′′(t)y′(t)
y x ′ = d y d x = 1 d x d y = 1 x y ′ y'_x=\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{x'_y} yx′=dxdy=dydx1=xy′1 y x x ′ ′ = d 2 y d x 2 = d ( d y d x ) d x = d ( 1 x y ′ ) d x = d ( 1 x y ′ ) d y ⋅ 1 x y ′ = − x y y ′ ′ ( x y ′ ) 3 y''_{xx}=\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dx}=\frac{d(\frac{1}{x'_y})}{dx}=\frac{d(\frac{1}{x'_y})}{dy}\cdot\frac{1}{x'_y}=-\frac{x''_{yy}}{(x'_y)^3} yxx′′=dx2d2y=dxd(dxdy)=dxd(xy′1)=dyd(xy′1)⋅xy′1=−(xy′)3xyy′′
F ( x ) = ∫ φ 1 ( x ) φ 2 ( x ) f ( t ) d t F(x)=\int_{\varphi_1(x)}^{\varphi_2(x)}f(t)dt F(x)=∫φ1(x)φ2(x)f(t)dt F ′ ( x ) = f [ φ 2 ( x ) ] φ 2 ′ ( x ) − f [ φ 1 ( x ) ] φ 1 ′ ( x ) F'(x)=f[\varphi_2(x)]\varphi_2'(x)-f[\varphi_1(x)]\varphi_1'(x) F′(x)=f[φ2(x)]φ2′(x)−f[φ1(x)]φ1′(x)
( u v ) ( n ) = ∑ k = 0 n C n k u ( n − k ) v ( k ) (uv)^{(n)}=\sum_{k=0}^nC_n^ku^{(n-k)}v^{(k)} (uv)(n)=k=0∑nCnku(n−k)v(k)
曲 率 : k = ∣ y ′ ′ ∣ ( 1 + y ′ 2 ) 3 2 曲率:k=\frac{|y''|}{(1+y'^2)^\frac{3}{2}} 曲率:k=(1+y′2)23∣y′′∣ 极 坐 标 : k = r 2 + 2 r ′ 2 − r r ′ ′ ( r 2 + r ′ 2 ) 3 2 极坐标:k=\frac{r^2+2r'^2-rr''}{(r^2+r'^2)^\frac{3}{2}} 极坐标:k=(r2+r′2)23r2+2r′2−rr′′ 曲 率 中 心 : ( x − y ′ ( y ′ 2 + 1 ) y ′ ′ , x + y ′ 2 + 1 y ′ ′ ) 曲率中心:(x-\frac{y'(y'^2+1)}{y''},x+\frac{y'^2+1}{y''}) 曲率中心:(x−y′′y′(y′2+1),x+y′′y′2+1)
∫ 1 x d x = ln ∣ x ∣ + C \int\frac{1}{x}dx=\ln|x|+C ∫x1dx=ln∣x∣+C ∫ a x d x = a x ln a + C ( a > 0 且 a ≠ 0 ) \int a^xdx=\frac{a^x}{\ln a}+C~(a>0且a\ne0) ∫axdx=lnaax+C (a>0且a=0) ∫ tan x d x = − ln ∣ cos x ∣ + C \int\tan xdx=-\ln|\cos x|+C ∫tanxdx=−ln∣cosx∣+C ∫ cot x d x = ln ∣ sin x ∣ + C \int\cot xdx=\ln|\sin x|+C ∫cotxdx=ln∣sinx∣+C ∫ 1 cos x d x = ln ∣ sec x + tan x ∣ + C \int\frac{1}{\cos x}dx=\ln|\sec x+\tan x|+C ∫cosx1dx=ln∣secx+tanx∣+C ∫ d x sin x = ln ∣ csc x − cot x ∣ + C \int\frac{dx}{\sin x}=\ln|\csc x-\cot x|+C ∫sinxdx=ln∣cscx−cotx∣+C ∫ sec 2 x d x = tan x + C \int\sec^2xdx=\tan x+C ∫sec2xdx=tanx+C ∫ csc 2 x d x = − cot x + C \int\csc^2xdx=-\cot x+C ∫csc2xdx=−cotx+C ∫ sec x tan x d x = sec x + C \int\sec x\tan xdx=\sec x+C ∫secxtanxdx=secx+C ∫ csc x cot x d x = − csc x + C \int\csc x\cot xdx=-\csc x+C ∫cscxcotxdx=−cscx+C ∫ 1 1 + x 2 d x = arctan x + C \int\frac{1}{1+x^2}dx=\arctan x+C ∫1+x21dx=arctanx+C ∫ 1 a 2 + x 2 d x = 1 a arctan x a + C ( a > 0 ) \int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C(a>0) ∫a2+x21dx=a1arctanax+C(a>0) ∫ 1 1 − x 2 d x = arcsin x + C \int\frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C ∫1−x2 1dx=arcsinx+C ∫ 1 a 2 − x 2 d x = arcsin x a + C ( a > 0 ) \int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin\frac{x}{a}+C(a>0) ∫a2−x2 1dx=arcsinax+C(a>0) ∫ 1 x 2 + a 2 d x = ln ( x + x 2 + a 2 ) + C \int\frac{1}{\sqrt{x^2+a^2}}dx=\ln(x+\sqrt{x^2+a^2})+C ∫x2+a2 1dx=ln(x+x2+a2 )+C ∫ 1 x 2 − a 2 d x = ln ∣ x + x 2 − a 2 ∣ + C ( ∣ x ∣ > ∣ a ∣ ) \int\frac{1}{\sqrt{x^2-a^2}}dx=\ln|x+\sqrt{x^2-a^2}|+C(|x|>|a|) ∫x2−a2 1dx=ln∣x+x2−a2 ∣+C(∣x∣>∣a∣) ∫ 1 x 2 − a 2 d x = 1 2 a ln ∣ x − a x + a ∣ + C \int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C ∫x2−a21dx=2a1ln∣x+ax−a∣+C ∫ 1 a 2 − x 2 d x = 1 2 a ln ∣ x + a x − a ∣ + C \int\frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln|\frac{x+a}{x-a}|+C ∫a2−x21dx=2a1ln∣x−ax+a∣+C ∫ a 2 − x 2 d x = a 2 2 arcsin x a + x 2 a 2 − x 2 + C ( a > ∣ x ∣ ≥ 0 ) \int\sqrt{a^2-x^2}dx=\frac{a^2}{2}\arcsin\frac{x}{a}+\frac{x}{2}\sqrt{a^2-x^2}+C(a>|x|\ge0) ∫a2−x2 dx=2a2arcsinax+2xa2−x2 +C(a>∣x∣≥0) ∫ sin 2 x d x = x 2 − sin 2 x 4 + C \int\sin^2xdx=\frac{x}{2}-\frac{\sin2x}{4}+C ∫sin2xdx=2x−4sin2x+C ∫ cos 2 x d x = x 2 + sin 2 x 4 + C \int\cos^2xdx=\frac{x}{2}+\frac{\sin2x}{4}+C ∫cos2xdx=2x+4sin2x+C ∫ tan 2 x d x = tan x − x + C \int\tan^2xdx=\tan x-x+C ∫tan2xdx=tanx−x+C ∫ cot 2 x d x = − cot x − x + C \int\cot^2xdx=-\cot x-x+C ∫cot2xdx=−cotx−x+C
偶 函 数 : ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x 偶函数:\int_{-a}^{a}f(x)dx=2\int_0^af(x)dx 偶函数:∫−aaf(x)dx=2∫0af(x)dx 奇 函 数 : ∫ − a a = 0 奇函数:\int_{-a}^a=0 奇函数:∫−aa=0 ∫ 0 π x f ( sin x ) d x = π 2 ∫ 0 π f ( sin x ) d x = π ∫ 0 π 2 f ( sin x ) d x \int_0^\pi xf(\sin x)dx=\frac{\pi}{2}\int_0^\pi f(\sin x)dx=\pi\int_0^\frac{\pi}{2}f(\sin x)dx ∫0πxf(sinx)dx=2π∫0πf(sinx)dx=π∫02πf(sinx)dx ∫ 0 π 2 f ( sin x ) d x = ∫ 0 π 2 f ( cos x ) d x \int_0^\frac{\pi}{2}f(\sin x)dx=\int_0^\frac{\pi}{2}f(\cos x)dx ∫02πf(sinx)dx=∫02πf(cosx)dx ∫ 0 π 2 f ( sin x , cos x ) d x = ∫ 0 π 2 f ( cos x , sin x ) d x \int_0^\frac{\pi}{2}f(\sin x,\cos x)dx=\int_0^\frac{\pi}{2}f(\cos x,\sin x)dx ∫02πf(sinx,cosx)dx=∫02πf(cosx,sinx)dx ∫ a b f ( x ) d x = ∫ − π 2 π 2 f ( a + b 2 + b − a 2 sin t ) ⋅ b − a 2 cos t d t \int_a^bf(x)dx=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}f(\frac{a+b}{2}+\frac{b-a}{2}\sin t)\cdot\frac{b-a}{2}\cos tdt ∫abf(x)dx=∫−2π2πf(2a+b+2b−asint)⋅2b−acostdt ∫ 1 x y f ( t ) d t = x ∫ 1 y f ( t ) d t + y ∫ 1 x f ( t ) d t \int_1^{xy}f(t)dt=x\int_1^yf(t)dt+y\int_1^xf(t)dt ∫1xyf(t)dt=x∫1yf(t)dt+y∫1xf(t)dt
∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_a^bf(x)dx=\int_a^bf(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^bf(x)dx=\frac{1}{2}\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^bf(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 2 sin n x d x = ∫ 0 π 2 cos n x d x = { n − 1 n ⋅ n − 3 n − 2 … 2 3 ⋅ 1 n 为 大 于 1 的 奇 数 n − 1 n ⋅ n − 3 n − 2 … 1 2 ⋅ π 2 n 为 正 偶 数 \int_0^\frac{\pi}{2}\sin^nxdx=\int_0^\frac{\pi}{2}\cos^nxdx= \left\{\begin{matrix} \frac{n-1}{n}\cdot\frac{n-3}{n-2}\dots\frac{2}{3}\cdot1 & n为大于1的奇数 \\ \frac{n-1}{n}\cdot\frac{n-3}{n-2}\dots\frac{1}{2}\cdot\frac{\pi}{2} & n为正偶数 \end{matrix}\right. ∫02πsinnxdx=∫02πcosnxdx={nn−1⋅n−2n−3…32⋅1nn−1⋅n−2n−3…21⋅2πn为大于1的奇数n为正偶数 ∫ 0 π sin n x d x = { 2 ⋅ n − 1 n ⋅ n − 3 n − 2 … 2 3 ⋅ 1 n 为 大 于 1 的 奇 数 2 ⋅ n − 1 n ⋅ n − 3 n − 2 … 1 2 ⋅ π 2 n 为 正 偶 数 \int_0^\pi\sin^nxdx= \left\{\begin{matrix} 2\cdot\frac{n-1}{n}\cdot\frac{n-3}{n-2}\dots\frac{2}{3}\cdot1 & n为大于1的奇数 \\ 2\cdot\frac{n-1}{n}\cdot\frac{n-3}{n-2}\dots\frac{1}{2}\cdot\frac{\pi}{2} & n为正偶数 \end{matrix}\right. ∫0πsinnxdx={2⋅nn−1⋅n−2n−3…32⋅12⋅nn−1⋅n−2n−3…21⋅2πn为大于1的奇数n为正偶数 ∫ 0 π cos n x d x = { 0 n 为 正 奇 数 2 ⋅ n − 1 n ⋅ n − 3 n − 2 … 1 2 ⋅ π 2 n 为 正 偶 数 \int_0^\pi\cos^nxdx= \left\{\begin{matrix} 0 & n为正奇数 \\ 2\cdot\frac{n-1}{n}\cdot\frac{n-3}{n-2}\dots\frac{1}{2}\cdot\frac{\pi}{2} & n为正偶数 \end{matrix}\right. ∫0πcosnxdx={02⋅nn−1⋅n−2n−3…21⋅2πn为正奇数n为正偶数 { ∫ 0 2 π sin n x d x ∫ 0 2 π cos n x d x = { 0 n 为 正 奇 数 4 ⋅ n − 1 n ⋅ n − 3 n − 2 … 1 2 ⋅ π 2 n 为 正 偶 数 \left\{\begin{matrix} \int_0^{2\pi}\sin^nxdx \\ \int_0^{2\pi}\cos^nxdx \end{matrix}\right.= \left\{\begin{matrix} 0 & n为正奇数 \\ 4\cdot\frac{n-1}{n}\cdot\frac{n-3}{n-2}\dots\frac{1}{2}\cdot\frac{\pi}{2} & n为正偶数 \end{matrix}\right. {∫02πsinnxdx∫02πcosnxdx={04⋅nn−1⋅n−2n−3…21⋅2πn为正奇数n为正偶数
∫ a b f ( x ) d x = ∫ 0 1 ( b − a ) f [ a + ( b − a ) t ] d t \int_a^bf(x)dx=\int_0^1(b-a)f\left[a+(b-a)t\right]dt ∫abf(x)dx=∫01(b−a)f[a+(b−a)t]dt ∫ − a a f ( x ) d x = ∫ 0 a [ f ( x ) + f ( − x ) ] d x ( a > 0 ) \int_{-a}^af(x)dx=\int_0^a\left[f(x)+f(-x)\right]dx(a>0) ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx(a>0)
S = ∫ a b ∣ y 1 ( x ) − y 2 ( x ) ∣ d x S=\int_a^b|y_1(x)-y_2(x)|dx S=∫ab∣y1(x)−y2(x)∣dx S = 1 2 ∫ α β ∣ r 1 2 ( θ ) − r 2 2 θ ∣ d θ S=\frac{1}{2}\int_\alpha^\beta|r_1^2(\theta)-r_2^2{\theta}|d\theta S=21∫αβ∣r12(θ)−r22θ∣dθ S = ∫ a b y ( t ) d x ( t ) = ∫ α β ∣ y ( t ) ⋅ x ′ ( t ) ∣ d t S=\int_a^by(t)dx(t)=\int_\alpha^\beta|y(t)\cdot x'(t)|dt S=∫aby(t)dx(t)=∫αβ∣y(t)⋅x′(t)∣dt 椭 圆 面 积 公 式 : S = π a b 椭圆面积公式:S=\pi ab 椭圆面积公式:S=πab
绕 x 轴 : V = ∫ a b π y 2 ( x ) d x 绕x轴:V=\int_a^b\pi y^2(x)dx 绕x轴:V=∫abπy2(x)dx 绕 y 轴 : V = 2 π ∫ a b x ∣ y ( x ) ∣ d x 绕y轴:V=2\pi\int_a^bx|y(x)|dx 绕y轴:V=2π∫abx∣y(x)∣dx
y ˉ = 1 b − a ∫ a b y ( x ) d x , 注 : 平 均 值 是 积 分 中 值 定 理 中 的 ξ \bar{y}=\frac{1}{b-a}\int_a^by(x)dx,注:平均值是积分中值定理中的\xi yˉ=b−a1∫aby(x)dx,注:平均值是积分中值定理中的ξ
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 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 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
y ( x ) 绕 x 轴 : S = 2 π ∫ a b ∣ y ( x ) ∣ 1 + [ y ′ ( x ) ] 2 d x y(x)绕x轴:S=2\pi\int_a^b|y(x)|\sqrt{1+[y'(x)]^2}dx y(x)绕x轴:S=2π∫ab∣y(x)∣1+[y′(x)]2 dx y ( t ) 绕 x 轴 : S = 2 π ∫ a b ∣ y ( t ) ∣ [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d x y(t)绕x轴:S=2\pi\int_a^b|y(t)|\sqrt{[x'(t)]^2+[y'(t)]^2}dx y(t)绕x轴:S=2π∫ab∣y(t)∣[x′(t)]2+[y′(t)]2 dx
x ˉ = ∬ D x d σ ∬ D d σ = ∫ a b d x ∫ 0 f ( x ) x d y ∫ a b d x ∫ 0 f ( x ) d y = ∫ a b x f ( x ) d x ∫ a b f ( x ) d x \bar{x}=\frac{\iint\limits_{D}xd\sigma}{\iint\limits_{D}d\sigma}=\frac{\int_a^bdx\int_0^{f(x)}xdy}{\int_a^bdx\int_0^{f(x)}dy}=\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx} xˉ=D∬dσD∬xdσ=∫abdx∫0f(x)dy∫abdx∫0f(x)xdy=∫abf(x)dx∫abxf(x)dx y ˉ = ∬ D y d σ ∬ D d σ = ∫ a b d x ∫ 0 f ( x ) y d y ∫ a b d x ∫ 0 f ( x ) d y = 1 2 ∫ a b f 2 ( x ) d x ∫ a b f ( x ) d x \bar{y}=\frac{\iint\limits_{D}yd\sigma}{\iint\limits_{D}d\sigma}=\frac{\int_a^bdx\int_0^{f(x)}ydy}{\int_a^bdx\int_0^{f(x)}dy}=\frac{\frac{1}{2}\int_a^bf^2(x)dx}{\int_a^bf(x)dx} yˉ=D∬dσD∬ydσ=∫abdx∫0f(x)dy∫abdx∫0f(x)ydy=∫abf(x)dx21∫abf2(x)dx
V = ∫ b a S ( x ) d x V=\int_b^aS(x)dx V=∫baS(x)dx
∬ D f ( x , y ) d σ = f ( μ , ξ ) ⋅ σ , ( μ , ξ ) ∈ D \iint\limits_{D}^{}f(x,y)d\sigma=f(\mu,\xi)\cdot\sigma,~~(\mu,\xi)\in D D∬f(x,y)dσ=f(μ,ξ)⋅σ, (μ,ξ)∈D
r = a ( 1 − c o s θ ) ( a > 0 ) r=a(1-cosθ)(a>0) r=a(1−cosθ)(a>0) r = a ( 1 + c o s θ ) ( a > 0 ) r=a(1+cosθ)(a>0) r=a(1+cosθ)(a>0)
r = a s i n 3 θ ( a > 0 ) r=asin3θ(a>0) r=asin3θ(a>0)
r = a θ ( a > 0 , θ > = 0 ) r=aθ(a>0,θ>=0) r=aθ(a>0,θ>=0)
r ² = 2 a ² c o s 2 θ r²=2a²cos2θ r²=2a²cos2θ r ² = a ² c o s 2 θ r²=a²cos2θ r²=a²cos2θ r ² = a ² s i n 2 θ r²=a²sin2θ r²=a²sin2θ
{ x = r ( t − s i n t ) y = r ( 1 − c o s t ) \left\{\begin{matrix} x=r(t-sint) \\ y=r(1-cost) \end{matrix}\right. {x=r(t−sint)y=r(1−cost)
x ⅔ + y ⅔ = r ⅔ x⅔+y⅔=r⅔ x⅔+y⅔=r⅔ { x = r c o s ³ t y = r s i n ³ t \left\{\begin{matrix} x=rcos³t \\ y=rsin³t \end{matrix}\right. {x=rcos³ty=rsin³t