高等数学 A(上)公式

BUPT 高等数学 A(上)课程涉及的公式,自制

填空题

卡片正面 卡片背面
\[\lim_{x\rightarrow\infty}\frac{a_0x^n+a_1x^{n-1}+\cdots+a_n}{b_0x^m+b_1x^{m-1}+\cdots+b_m}=\left\{ \begin{align*}&[...] & n=m \\& [...] & n<m \\& [...] & n>m \end{align*} \right.\] \[\lim_{x\rightarrow\infty}\frac{a_0x^n+a_1x^{n-1}+\cdots+a_n}{b_0x^m+b_1x^{m-1}+\cdots+b_m}=\left\{ \begin{align*}& \frac {a_0}{b_0} & n=m \\& 0 & n<m \\ & \infty & n>m \end{align*} \right.\]
上凹\[f\left(\frac{x_1+x_2}2\right)[...] \frac{f(x_1)+f(x_2)}2\] 上凹\[f\left(\frac{x_1+x_2}2\right)< \frac{f(x_1)+f(x_2)}2\]
上凸\[f\left(\frac{x_1+x_2}2\right)[...] \frac{f(x_1)+f(x_2)}2\] 上凸\[f\left(\frac{x_1+x_2}2\right)> \frac{f(x_1)+f(x_2)}2\]
斜渐近线\[\begin{align*}&a=[...]\\&b=[...]\end{align*}\] 斜渐近线\[\begin{align*}&a= \lim_{x\rightarrow\infty}\frac{f(x)}x \\&b= \lim_{x\rightarrow\infty}[f(x)-ax]\end{align*}\]
万能公式\[\begin{align*}&u= [...] \\& \sin x= [...] \\& \cos x=[...] \\& \mathop{}\!\mathrm{d}x = [...]\end{align*} \] 万能公式\[\begin{align*}&u= \tan\frac x2 \\& \sin x= \frac{2u}{1+u^2} \\& \cos x= \frac{1-u^2}{1+u^2} \\& \mathop{}\!\mathrm{d}x = \frac{2}{1+u^2}\mathop{}\!\mathrm{d}u\end{align*}\]
\[\left| \int_a^bf(x)\mathop{}\!\mathrm{d}x\right|[...] \int_a^b|f(x)|\mathop{}\!\mathrm{d}x\] \[\left| \int_a^bf(x)\mathop{}\!\mathrm{d}x\right| \leq \int_a^b|f(x)|\mathop{}\!\mathrm{d}x\]
第二类换元法\(\sqrt{a^2-x^2},x=[...]\) 第二类换元法\(\sqrt{a^2-x^2},x= a\sin t\)
第二类换元法\(\sqrt{a^2+x^2},x=[...]\) 第二类换元法\(\sqrt{a^2+x^2},x= a\tan t\)
第二类换元法\(\sqrt{x^2-a^2},x=[...]\) 第二类换元法\(\sqrt{x^2-a^2},x= a\sec t\)
\[I_n=\int_0^{\frac\pi2}\sin^nx\mathop{}\!\mathrm{d}x=\int_0^{\frac\pi2}\cos^nx\mathop{}\!\mathrm{d}x=\left\{ \begin{align*}& [...] & n\text{为正偶数} \\& [...] & n\text{为大于 1 的正奇数} \end{align*}\right.\] \[I_n=\int_0^{\frac\pi2}\sin^nx\mathop{}\!\mathrm{d}x=\int_0^{\frac\pi2}\cos^nx\mathop{}\!\mathrm{d}x=\left\{ \begin{align*} & \frac{n-1}n\cdot\frac{n-3}{n-2}\cdots\frac34\cdot\frac12\cdot\frac\pi2 & n\text{为正偶数} \\ & \frac{n-1}n\cdot\frac{n-3}{n-2}\cdots\frac45\cdot\frac23 & n\text{为大于 1 的正奇数} \end{align*}\right.\]
\[\int_a^{+\infty}\frac{\mathop{}\!\mathrm{d}x}{x^p}(a>0)\left\{\begin{align*}&[...] & p>1 \\& [...] & p\leq 1\end{align*}\right.\] \[\int_a^{+\infty}\frac{\mathop{}\!\mathrm{d}x}{x^p}(a>0)\left\{\begin{align*}&\text{收敛} & p>1 \\& \text{发散} & p\leq 1\end{align*}\right.\]
\[\int_a^b\frac{\mathop{}\!\mathrm{d}x}{(x-a)^p}(a>0)\left\{\begin{align*}&[...] & p<1 \\& [...] & p\geq 1\end{align*}\right.\] \[\int_a^b\frac{\mathop{}\!\mathrm{d}x}{(x-a)^p}(a>0)\left\{\begin{align*}&\text{收敛} & p<1 \\ & \text{发散} & p\geq 1\end{align*}\right.\]
\[\sec^2 x=\tan^2 x[...]\] \[\sec^2 x=\tan^2 x+1\]
\[\csc^2 x=\cot^2 x[...]\] \[\csc^2 x=\cot^2 x+1\]

基础

卡片正面 卡片背面
\[\lim_{x\rightarrow 0}(1+x)^{\frac1x}=\] \[e\]
\[\lim_{n\rightarrow \infty}\sqrt[n]a(a>0)=\] \[1\]
\[\lim_{n\rightarrow \infty}\sqrt[n]n=\] \[1\]
\[\lim_{x\rightarrow 0^+}x^x=\] \[1\]
\[\tan x \sim \] \[x\]
\[\sin x \sim \] \[x\]
\[\arctan x \sim \] \[x\]
\[\arcsin x \sim \] \[x\]
\[1-\cos x \sim \] \[\frac 12 x^2\]
\[\ln(1+x) \sim\] \[x\]
\[e^x -1 \sim\] \[x\]
\[a^x -1 \sim\] \[x\ln a\]
\[(1+x)^a-1 \sim\] \[ax\]
\[(u\pm v)'=\] \[u'\pm v'\]
\[(uv)'=\] \[u'v+uv'\]
\[\left( \frac uv \right)'\] \[\frac{u'v-uv'}{v^2}\]
\[(C)'=\] \[0\]
\[(x^\mu)'=\] \[\mu x^{\mu-1}\]
\[(\sin x)'=\] \[\cos x\]
\[(\cos x)'=\] \[-\sin x\]
\[(\tan x)'=\] \[\sec^2x\]
\[(\cot x)'=\] \[-\csc^2x\]
\[(e^x)'=\] \[e^x\]
\[(a^x)'=\] \[a^x\ln a\]
\[(\sec x)'=\] \[\sec x\cdot\tan x\]
\[(\csc x)'=\] \[-\csc x\cdot\cot x\]
\[(\ln x)'=\] \[\frac 1x\]
\[(\log_ax)'=\] \[\frac 1{x\ln a}\]
\[(\arcsin x)'=\] \[\frac 1{\sqrt{1-x^2}}\]
\[(\arccos x)'=\] \[-\frac 1{\sqrt{1-x^2}}\]
\[(\arctan x)'=\] \[\frac 1{1+x^2}\]
\[(\operatorname {arccot} x)'=\] \[-\frac 1{1+x^2}\]
\[(\sqrt x)'=\] \[\frac 1{2\sqrt x}\]
\[[u(x)\pm v(x)]^{(n)}=\] \[u(x)^{(n)}\pm v(x)^{(n)}\]
\[[Cu(x)]^{(n)}=\] \[Cu^{(n)}(x)\]
\[[u(ax+b)]^{(n)}=\] \[a^nu^{(n)}(ax+b)\]
\[[u(x)\cdot v(x)]^{(n)}=\] \[\sum_{k=0}^n C_n^ku^{(n-k)}(x)v^{(k)}(x)\]
\[\left(x^n\right)^{(n)}=\] \[n!\]
\[\left(e^{ax+b}\right)^{(n)}=\] \[a^ne^{ax+b}\]
\[\left(a^x\right)^{(n)}=\] \[a^x\ln^na\]
\[\sin (ax+b)^{(n)}=\] \[a^n\sin(ax+b+n\cdot\frac\pi2)\]
\[\cos (ax+b)^{(n)}=\] \[a^n\cos(ax+b+n\cdot\frac\pi2)\]
\[\left(\frac1{ax+b}\right)^{(n)}=\] \[\frac{(-1)^na^nn!}{(ax+b)^{n+1}}\]
\[\mathop{}\!\mathrm{d}(u\pm v)=\] \[\mathop{}\!\mathrm{d}u\pm \mathop{}\!\mathrm{d}v\]
\[\mathop{}\!\mathrm{d}(Cu)=\] \[C\mathop{}\!\mathrm{d}u\]
\[\mathop{}\!\mathrm{d}(uv)=\] \[v\mathop{}\!\mathrm{d}u+u\mathop{}\!\mathrm{d}v\]
\[\mathop{}\!\mathrm{d}\left(\frac uv\right)=\] \[\frac {v\mathop{}\!\mathrm{d}u-u\mathop{}\!\mathrm{d}v}{v^2}\]
\[\mathop{}\!\mathrm{d}(C)=\] \[0\]
\[\mathop{}\!\mathrm{d}(x^\mu)=\] \[\mu x^{\mu-1}\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(\sin x)=\] \[\cos x\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(\cos x)=\] \[-\sin x\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(\tan x)=\] \[\sec^2x\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(\cot x)=\] \[-\csc^2x\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(\sec x)=\] \[\sec x\cdot\tan x\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(\csc x)=\] \[-\csc x\cdot\cot x\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(e^x)=\] \[e^x\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(a^x)=\] \[a^x\ln a\mathop{}\!\mathrm{d}x\]
\[\mathop{}\!\mathrm{d}(\ln x)=\] \[\frac {\mathop{}\!\mathrm{d}x}x\]
\[\mathop{}\!\mathrm{d}(\log_ax)=\] \[\frac {\mathop{}\!\mathrm{d}x}{x\ln a}\]
\[\mathop{}\!\mathrm{d}(\arcsin x)=\] \[\frac {\mathop{}\!\mathrm{d}x}{\sqrt{1-x^2}}\]
\[\mathop{}\!\mathrm{d}(\arccos x)=\] \[-\frac {\mathop{}\!\mathrm{d}x}{\sqrt{1-x^2}}\]
\[\mathop{}\!\mathrm{d}(\arctan x)=\] \[\frac{\mathop{}\!\mathrm{d}x}{1+x^2}\]
\[\mathop{}\!\mathrm{d}(\operatorname {arccot} x)=\] \[-\frac{\mathop{}\!\mathrm{d}x}{1+x^2}\]
\[\int k\mathop{}\!\mathrm{d}x=\] \[kx+C\]
\[\int x^\mu \mathop{}\!\mathrm{d}x=\] \[\frac 1{\mu+1}x^{\mu+1}+C\]
\[\int\frac{\mathop{}\!\mathrm{d}x}x=\] \[\ln |x|+C\]
\[\int\frac1{1+x^2}\mathop{}\!\mathrm{d}x=\] \[\arctan x+C=-\operatorname {arccot} x+C\]
\[\int\frac1{\sqrt{1-x^2}}\mathop{}\!\mathrm{d}x=\] \[\arcsin x+C=-\arccos x+C\]
\[\int\cos x\mathop{}\!\mathrm{d}x=\] \[\sin x+C\]
\[\int\sin x\mathop{}\!\mathrm{d}x=\] \[-\cos x+C\]
\[\int\frac{\mathop{}\!\mathrm{d}x}{\cos^2 x}=\int\sec^2 x\mathop{}\!\mathrm{d}x=\] \[\tan x+C\]
\[\int\frac{\mathop{}\!\mathrm{d}x}{\sin^2 x}=\int\csc^2 x\mathop{}\!\mathrm{d}x=\] \[-\cot x+C\]
\[\int\sec x\tan x\mathop{}\!\mathrm{d}x=\] \[\sec x+C\]
\[\int\csc x\cot x\mathop{}\!\mathrm{d}x=\] \[-\csc x+C\]
\[\int e^x\mathop{}\!\mathrm{d}x=\] \[e^x+C\]
\[\int a^x\mathop{}\!\mathrm{d}x=\] \[\frac{a^x}{\ln a}+C\]
\[\int f(ax+b)\mathop{}\!\mathrm{d}x=\] \[\frac 1a\int f(ax+b)\mathop{}\!\mathrm{d}(ax+b)\]
\[\int f(x^n)x^{n-1}\mathop{}\!\mathrm{d}x=\] \[\frac 1n\int f(x^n)\mathop{}\!\mathrm{d}x^n\]
\[\int f(x^n)\frac 1x\mathop{}\!\mathrm{d}x=\] \[\frac 1n\int f(x^n)\frac 1{x^n}\mathop{}\!\mathrm{d}x^n\]
\[\int f(\sin^kx)\cos x\mathop{}\!\mathrm{d}x=\] \[\int f(\sin^kx)\mathop{}\!\mathrm{d}\sin x\]
\[\int f(\cos^kx)\sin x\mathop{}\!\mathrm{d}x=\] \[-\int f(\cos^kx)\mathop{}\!\mathrm{d}\cos x\]
\[\int f(\tan x)\sec^2x\mathop{}\!\mathrm{d}x=\int \frac{f(\tan x)}{\cos^2x}\mathop{}\!\mathrm{d}x=\] \[\int f(\tan x)\mathop{}\!\mathrm{d}\tan x\]
\[\int f(\cot x)\csc^2x\mathop{}\!\mathrm{d}x=\int \frac{f(\cot x)}{\sin^2x}\mathop{}\!\mathrm{d}x=\] \[-\int f(\cot x)\mathop{}\!\mathrm{d}\cot x\]
\[\int f(a^x)a^x\mathop{}\!\mathrm{d}x=\] \[\frac 1{\ln a}\int f(a^x)\mathop{}\!\mathrm{d}a^x\]
\[\int f(e^x)e^x\mathop{}\!\mathrm{d}x=\] \[\int f(e^x)\mathop{}\!\mathrm{d}e^x\]
\[\int f(\ln x)\frac 1x\mathop{}\!\mathrm{d}x=\] \[\int f(\ln x)\mathop{}\!\mathrm{d}\ln x\]
\[\int f(\sqrt x)\frac 1{\sqrt x}\mathop{}\!\mathrm{d}x=\] \[2\int f(\sqrt x)\mathop{}\!\mathrm{d}\sqrt x\]
\[\int \frac {f(\frac 1x)}{x^2}\mathop{}\!\mathrm{d}x=\] \[-\int f(\frac 1x)\mathop{}\!\mathrm{d}\frac 1x\]
\[\int \frac {f(\arctan x)}{1+x^2}\mathop{}\!\mathrm{d}x=\] \[\int f(\arctan x)\mathop{}\!\mathrm{d}\arctan x=-\int f(\operatorname{arccot}x)\mathop{}\!\mathrm{d}\operatorname{arccot}x\]
\[\int \frac {f(\arcsin x)}{\sqrt{1-x^2}}\mathop{}\!\mathrm{d}x=\] \[\int f(\arcsin x)\mathop{}\!\mathrm{d}\arcsin x=-\int f(\arccos x)\mathop{}\!\mathrm{d}\arccos x\]
\[\int \frac {f(\sqrt{1+x^2})x}{\sqrt{1+x^2}}\mathop{}\!\mathrm{d}x=\] \[\int f(\sqrt{1+x^2})\mathop{}\!\mathrm{d}\sqrt{1+x^2}\]
\[\int\tan x\mathop{}\!\mathrm{d}x=\] \[-\ln |\cos x|+C\]
\[\int\cot x\mathop{}\!\mathrm{d}x=\] \[\ln |\sin x|+C\]
\[\int\sec x\mathop{}\!\mathrm{d}x=\] \[\ln |\sec x+\tan x|+C\]
\[\int\csc x\mathop{}\!\mathrm{d}x=\] \[\ln |\csc x-\cot x|+C\]
\[\int\frac 1{a^2+x^2}\mathop{}\!\mathrm{d}x=\] \[\frac 1a\arctan\frac xa+C\]
\[\int\frac 1{x^2-a^2}\mathop{}\!\mathrm{d}x=\] \[\frac 1{2a}\ln\left|\frac {x-a}{x+a}\right|+C\]
\[\int\frac 1{\sqrt{a^2-x^2}}\mathop{}\!\mathrm{d}x=\] \[\arcsin\frac xa+C\]
\[\int\frac 1{\sqrt{x^2+a^2}}\mathop{}\!\mathrm{d}x=\] \[\ln|x+\sqrt{x^2+a^2}|+C\]
\[\int\frac 1{\sqrt{x^2-a^2}}\mathop{}\!\mathrm{d}x=\] \[\ln|x+\sqrt{x^2-a^2}|+C\]
\[\int_0^{\frac{\pi}2}f(\sin x)\mathop{}\!\mathrm{d}x=\] \[\int_0^{\frac{\pi}2}f(\cos x)\mathop{}\!\mathrm{d}x\]
\[\int_0^{\frac{\pi}2}f(\cos x)\mathop{}\!\mathrm{d}x=\] \[\int_0^{\frac{\pi}2}f(\sin x)\mathop{}\!\mathrm{d}x=\]
\[\int_0^{\pi}xf(\sin x)\mathop{}\!\mathrm{d}x=\] \[\frac \pi2\int_0^{\pi}f(\sin x)\mathop{}\!\mathrm{d}x\]
\[\int_a^{a+T}f(x)\mathop{}\!\mathrm{d}x=\] \[\int_0^Tf(x)\mathop{}\!\mathrm{d}x\]
\[\int_a^{a+nT}f(x)\mathop{}\!\mathrm{d}x=\] \[n\int_0^Tf(x)\mathop{}\!\mathrm{d}x\]
\[g(y)\mathop{}\!\mathrm{d}y=f(x)\mathop{}\!\mathrm{d}x\] \[\int g(y)\mathop{}\!\mathrm{d}y=\int f(x)\mathop{}\!\mathrm{d}x\\G(y)=F(x)+C\]
\[\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}=\varphi\left(\frac yx\right)\] \[\begin{align*}u=\frac yx\\u+x\frac{\mathop{}\!\mathrm{d}u}{\mathop{}\!\mathrm{d}x}=\varphi(u)\\\int\frac{\mathop{}\!\mathrm{d}u}{\varphi(u)-u}=\int\frac{\mathop{}\!\mathrm{d}x}{x}\\\Phi(\frac yx)=\ln|x|+C\end{align*}\]
\[\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}+P(x)y=0\] \[y=Ce^{-\int P(x)\mathop{}\!\mathrm{d}x}\]
\[\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}+P(x)y=Q(x)\] \[y=e^{-\int P(x)\mathop{}\!\mathrm{d}x}\left(\int Q(x)e^{\int P(x)\mathop{}\!\mathrm{d}x}\mathop{}\!\mathrm{d}x+C\right)\]
\[\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}+P(x)y=Q(x)y^\alpha\] \[\begin{align*}&y^{-\alpha}\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}+P(x)y^{1-\alpha}=Q(x)\\&z=y^{1-\alpha}\\&\frac{\mathop{}\!\mathrm{d}z}{\mathop{}\!\mathrm{d}x}+(1-\alpha)P(x)z=(1-\alpha)Q(x)\end{align*}\]
\[\sin a+\sin b=\] \[2\sin\frac{a+b}2\cdot\cos\frac{a-b}2\]
\[\sin a-\sin b=\] \[2\cos\frac{a+b}2\cdot\sin\frac{a-b}2\]
\[\cos a+\cos b=\] \[2\cos\frac{a+b}2\cdot\cos\frac{a-b}2\]
\[\cos a-\cos b=\] \[-2\sin\frac{a+b}2\cdot\sin\frac{a-b}2\]
\[\sin a\cdot\sin b=\] \[-\frac 12[\cos(a+b)-\cos(a-b)]\]
\[\cos a\cdot\cos b=\] \[\frac 12[\cos(a+b)+\cos(a-b)]\]
\[\sin a\cdot\cos b=\] \[\frac 12[\sin(a+b)+\sin(a-b)]\]
\[\cos a\cdot\sin b=\] \[\frac 12[\sin(a+b)-\sin(a-b)]\]
\[\arcsin x+\arccos x=\] \[\frac\pi2\]
\[\arctan x+\operatorname{arccot} x=\] \[\frac\pi2\]
罗尔定理:若\(f(x)\)满足:在区间\([a,b]\)上连续,在区间\((a,b)\)内可导,\(f(a)=f(b)\) 则至少存在一点\(\xi\in (a,b)\),使得\(f'(\xi)=0\)
拉格朗日中值定理:若\(f(x)\)满足:在区间\([a,b]\)上连续,在区间\((a,b)\)内可导 则至少存在一点\(\xi\in (a,b)\),使得等式\(f(b)-f(a)=f'(\xi)(b-a)\)成立
柯西中值定理:若\(f(x)\)\(g(x)\)满足:在区间\([a,b]\)上连续,在区间\((a,b)\)内可导,且在区间\((a,b)\)\(g'(x)\neq0\) 则至少存在一点\(\xi\in (a,b)\),使得等式\(\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}\)成立
泰勒公式\(f(x)=\) \[\begin{align*}&f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x)\\&R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}=o[(x-x_0)^n]\end{align*}\]
麦克劳林公式\(f(x)=\) \[\begin{align*}&f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots +\frac{f^{(n)}(0)}{n!}x^n+R_n(x)\\&R_n(x)=\frac{f^{(n+1)}(\theta x)}{(n+1)!}x^{n+1}=o(x^n),\theta\in(0,1)\end{align*}\]
麦克劳林公式\(e^x=\) \[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}+o(x^n)\]
麦克劳林公式\(\sin x=\) \[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+o(x^{2n+1})\]
麦克劳林公式\(\cos x=\) \[1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n})\]
麦克劳林公式\(\ln(1+x)=\) \[x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+(-1)^n\frac{x^{n+1}}{(n+1)}+o(x^{n+1})\]
麦克劳林公式\((1+x)^\alpha=\) \[1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\cdots+\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n+o(x^n)\]
积分中值定理:若\(f(x)\in C[a,b]\) 则至少存在一点\(\xi\in [a,b]\),使得\[\int_a^bf(x)\mathop{}\!\mathrm{d}x=f(\xi)(b-a)\]
\(f(-x)=f(x)\),则\[\int_{-a}^a f(x)\mathop{}\!\mathrm{d}x=\] \[2\int_0^af(x)\mathop{}\!\mathrm{d}x\]
\(f(-x)=-f(x)\),则\[\int_{-a}^a f(x)\mathop{}\!\mathrm{d}x=\] \[0\]
定积分介值定理:设\(M\)\(m\)分别是函数\(f(x)\)在区间\([a,b]\)上的最大值和最小值,则: \[m(b-a)\leq\int_a^bf(x)\mathop{}\!\mathrm{d}x\leq M(b-a)\]
由曲线\(y=f(x),y=g(x)\)\(f(x)\leq g(x)\),均在\([a,b]\)上连续)以及直线\(x=a,x=b\)所围成的图形面积为: \[\int_a^b[g(x)-f(x)]\mathop{}\!\mathrm{d}x\]
由曲线\(\rho=\rho(\theta)\)\(\rho(\theta)\geq0\))以及射线\(\theta=\alpha,\theta=\beta\)\(0< \beta-\alpha\leq2\pi\))所围成的图形(曲边扇形)面积为: \[\frac 12\int_\alpha^\beta\rho^2(\theta)\mathop{}\!\mathrm{d}\theta\]
由连续曲线\(y=f(x)\)、直线\(x=a,x=b\)\(a<b\))以及\(x\)轴所围成的曲边梯形绕\(x\)轴一周而成的旋转体的体积为: \[\pi\int_a^bf^2(x)\mathop{}\!\mathrm{d}x\]
由连续曲线\(y=f(x)\)、直线\(x=a,x=b\)\(a<b\))以及\(x\)轴所围成的曲边梯形绕 y 轴一周而成的旋转体的体积为: \[2\pi\int_a^bxf(x)\mathop{}\!\mathrm{d}x\]
如果一立体在过\(x\)轴上点\(x=a,x=b\)\(a<b\))且垂直于\(x\)的两个平面之间,以\(A(x)\)表示过点\(x\)且垂直于\(x\)轴的截面面积,那么该立体体积为: \[\int_a^bA(x)\mathop{}\!\mathrm{d}x\]
如果曲线弧由\(y=f(x)\)\(a\leq x\leq b\))给出,其中\(f(x)\)\([a,b]\)上具有连续导数,那么该曲线的弧长为: \[\int_a^b\sqrt{1+y'^2}\mathop{}\!\mathrm{d}x\]
如果曲线弧由参数方程\(\left\{\begin{align*}x=\varphi(t)\\y=\psi(t)\end{align*}\right.\)\(\alpha\leq t\leq\beta\))给出,其中\(\varphi(t),\psi(t)\)\([\alpha,\beta]\)上具有连续导数,那么该曲线的弧长为: \[\int_\alpha^\beta\sqrt{\varphi'^2(t)+\psi'^2(t)}\mathop{}\!\mathrm{d}t\]
如果曲线弧由极坐标方程\(\rho=\rho(\theta)\)\(\alpha\leq\theta\leq\beta\))给出,其中\(\rho(\theta)\)\([\alpha,\beta]\)上具有连续导数,那么该曲线的弧长为: \[\int_\alpha^\beta\sqrt{\rho^2(\theta)+\rho'^2(\theta)}\mathop{}\!\mathrm{d}\theta\]
\[y''+py'+qy=e^{\lambda x}P_m(x)\] \[y^*=x^ke^{\lambda x}R_m(x)\]\(\lambda\)不是特征方程的根时,\(k=0\);当\(\lambda\)是特征方程的单根时,\(k=1\);当\(\lambda\)是特征方程的二重根时,\(k=2\)
\[y''+py'+qy=e^{\lambda x}[P_l(x)\cos\omega x+Q_n(x)\sin\omega x]\] \[y^*=x^ke^{\lambda x}[R_m^{(1)}(x)\cos\omega x+R_m^{(2)}(x)\sin\omega x]\]\(\lambda\pm\omega\mathrm{i}\)不是特征方程的根时,\(k=0\);当\(\lambda\pm\omega\mathrm{i}\)是特征方程的根时,\(k=1\)
\[y''=f(x,y')\] \[\begin{align*}&y'=p,y''=p'\\&p'=f(x,p)\\&p=y'=\varphi(x,C_1)\\&y=\int\varphi(x,C_1)\mathop{}\!\mathrm{d}x+C_2\end{align*}\]
\[y''=f(y,y')\] \[\begin{align*}&y'=p,y''=\frac{\mathop{}\!\mathrm{d}p}{\mathop{}\!\mathrm{d}x}=\frac{\mathop{}\!\mathrm{d}p}{\mathop{}\!\mathrm{d}y}\cdot\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}=p\frac{\mathop{}\!\mathrm{d}p}{\mathop{}\!\mathrm{d}y}\\&p\frac{\mathop{}\!\mathrm{d}p}{\mathop{}\!\mathrm{d}y}=f(y,p)\\&p=\frac{\mathop{}\!\mathrm{d}y}{\mathop{}\!\mathrm{d}x}=\varphi(y,C_1)\\&\int\frac{\mathop{}\!\mathrm{d}y}{\varphi(y,C_1)}=x+C_2\end{align*}\]
\[y''+py'+qy=0\] \[\begin{array} &r^2+pr+q=0\\r_{1,2}=\frac{-p\pm\sqrt{p^2-4q}}2\\y=\left\{\begin{align*}&C_1e^{r_1x}+C_2e^{r_2x} & p^2-4q>0,r_1\neq r_2 \\ &C_1e^{r_1x}+C_2xe^{r_1x} & p^2-4q=0,r_1=r_2 \\ &e^{\alpha x}(C_1\cos \beta x+C_2\sin\beta x) & p^2-4q<0,r_{1,2}=\alpha\pm\beta\mathrm{i} \\ \end{align*}\right.\end{array}\]
作者

xqmmcqs

发布于

2020-03-28

更新于

2023-03-29

许可协议

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