1.微积分第一基本定理
注释:
如果F′(x)=f(x)F'(x)=f(x)F′(x)=f(x),则∫abf(x)dx=F(a)−F(b)=F(x)∣ab\int_a^bf(x)dx=F(a)-F(b)=F(x)|_a^b∫abf(x)dx=F(a)−F(b)=F(x)∣ab
Ex:
F(x)=x33F(x)=\frac{x^3}{3}F(x)=3x3
F′(x)=x2=f(x),∫abf(x)dx=F(a)−F(b)=b33−a33F'(x)=x^2=f(x),\int_a^bf(x)dx=F(a)-F(b)=\frac{b^3}{3}-\frac{a^3}{3}F′(x)=x2=f(x),∫abf(x)dx=F(a)−F(b)=3b3−3a3
当a=0时,∫abx2dx=x33∣0b=b33当a= 0时,\int_a^bx^2dx=\frac{x^3}{3}|_0^b=\frac{b^3}{3}当a=0时,∫abx2dx=3x3∣0b=3b3
Ex:
求sinx函数与x轴的阴影面积。(0−>π)sinx函数与x轴的阴影面积。(0->\pi)sinx函数与x轴的阴影面积。(0−>π)
∫0πsinxdx=(−cosπ)∣0π=1+1=2\int_0^\pi sinxdx=(-cos \pi)|_0^\pi = 1+1=2∫0πsinxdx=(−cosπ)∣0π=1+1=2
2.定积分的几何意义:
x轴上方面积−-−x轴下方面积。
3.性质
- ∫ab(f(x)+g(x))dx=∫abf(x)dx+∫abg(x)dx\int_a^b(f(x) +g(x))dx=\int_a^bf(x)dx+\int_a^bg(x)dx∫ab(f(x)+g(x))dx=∫abf(x)dx+∫abg(x)dx
- ∫abCf(x)dx=C∫abf(x)dx\int_a^bCf(x)dx=C\int_a^bf(x)dx∫abCf(x)dx=C∫abf(x)dx
- ∫abf(x)dx+∫bcf(x)dx=∫acf(x)dx\int_a^bf(x)dx+\int_b^cf(x)dx=\int_a^cf(x)dx∫abf(x)dx+∫bcf(x)dx=∫acf(x)dx
- ∫aaf(x)dx=0\int_a^af(x)dx=0∫aaf(x)dx=0
- ∫abf(x)dx=−∫baf(x)dx\int_a^bf(x)dx=-\int_b^af(x)dx∫abf(x)dx=−∫baf(x)dx
- 积分估计:
如果f(x)>g(x)f(x)>g(x)f(x)>g(x),则∫abf(x)dx≤∫abg(x)dx(a<b)\int_a^bf(x)dx\leq\int_a^bg(x)dx(a<b)∫abf(x)dx≤∫abg(x)dx(a<b)
Ex:
ex≥1,x≥0e^x\geq 1,x \geq 0ex≥1,x≥0.
∫abexdx≥∫ab1dx\int_a^be^xdx \geq \int_a^b1dx∫abexdx≥∫ab1dx
eb−1≥be^b-1 \geq beb−1≥b
eb≥1+be^b \geq 1+beb≥1+b
Ex:
ex≥1+x,x≥0e^x \geq1+x,x \geq0ex≥1+x,x≥0
∫0bexdx≥∫0b(1+x)dx\int_0^be^xdx \geq \int_0^b(1+x)dx∫0bexdx≥∫0b(1+x)dx
eb−1≥b+b22e^b-1 \geq b+\frac{b^2}{2}eb−1≥b+2b2
eb≥b+1+b22e^b\geq b+1+\frac{b^2}{2}eb≥b+1+2b2
- 变量替换:
∫u1u2g(u)=∫x1x2g(u(x))u′(x)dx\int_{u_1}^{u_2}g(u)=\int_{x_1}^{x_2}g(u(x))u'(x)dx∫u1u2g(u)=∫x1x2g(u(x))u′(x)dx
Ex:
∫12(x3+2)5x2dx\int_1^2(x^3+2)^5x^2dx∫12(x3+2)5x2dx
令u=x2+2,du=3x2u=x^2+2,du=3x^2u=x2+2,du=3x2
原式=∫310u513du=\int_3^{10}u^5\frac{1}{3}du=∫310u531du
=118u6∣310=10618−3618=\frac{1}{18}u^6|_3^{10}=\frac{10^6}{18}-\frac{3^6}{18}=181u6∣310=18106−1836
Wanning:
∫−11x2dx=≠∫11u12udu=0u=x2,du=2xdx.dx=du2x\int_{-1}^{1}x^2 dx =\neq \int_1^1u\frac{1}{2\sqrt{u}}du=0\\
u=x^2,du=2xdx.\\
dx=\frac{du}{2x}∫−11x2dx≠=∫11u2u1du=0u=x2,du=2xdx.dx=2xdu
u=x2,u′(x)=2x{>0,(x>0)<0,(x<0)u=x^2,u'(x)=2x \left\{\begin{matrix} & >0, &(x>0) \\ & <0, &(x<0) \end{matrix}\right. u=x2,u′(x)=2x{>0,<0,(x>0)(x<0)