sinx ~ x； tanx ~ x； arcs inx ~ x； arctanx ~ x；

ln( 1+x)~ x； e x- 1 ~ x； 1-cosx ~ 1/2 . x 2； ( 1+x) a - 1 ~ ax(a ≠ 0)。

Note: Let f(x) and g(x) be the equivalent infinitesimal of x→x0, and then perform the following operations:

lim x→x0 f(x)u(x)/v(x)= lim x→x0[f(x)u(x)/v(x]。 g(x)/f(x)] = lim x→x0 g(x)u(x)/v(x)

(c)' = 0； (x a )' = a.x a- 1 (a is a constant real number and a power function);

(sinx)' = cosx； (cosx)' =-sinx；

(tanx)' =( 1/cosx)2 = sec 2 x； (cotx)' =-( 1/sinx)2 =-CSC 2 x；

(secx)' = secx tanx； (cscx)' =-cscx cotx；

(log a x)' = 1/(x . lna)(a & gt； 0，a≠0)； (lnx)' = 1/x；

(a x)' = a x . lna(a & gt； 0，a≠0)； (e x)' = e x；

(arcsinx)' = 1/√( 1-x ^ 2)； (arc cosx)' =- 1/√( 1-x ^ 2)；

(arctanx)' = 1/( 1+x ^ 2)； (arccotx)' =- 1/( 1+x ^ 2)。

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∫[f(x)g(x)]dx =∫f(x)dx∫g(x)dx； ∫KF(x)dx = k∫f(x)dx； (You can add, subtract, multiply)

(∫f(x)dx)' = f(x)； d(∵f(x)dx)= f(x)dx； ∫df(x)= f(x)+C； (Reciprocity of Derivative and Differential Operations)

∫x μ dx = x μ+ 1 /(μ+ 1)，μ≦- 1，x & gt0; ∫ 1/xdx = ln|x|+C，x≠0；

∫e x dx = e x+C； ∫a x dx = a x /lna+C, a≠0 and a >；; 0;

∫cosxdx = sinx+C； ∫sinxdx =-cosx+C；

∫sec 2 xdx = tanx+C； ∫CSC 2 xdx =-cotx+C；

∫ 1/( 1+x^2)dx = arctanx+c； ∫ 1/√( 1-x^2)dx = arcs inx+c

Area of a plane figure: S = ∫ a b [f 2 (x)-f 1 (x)]dx For example, find the area solution of a closed plane figure with parabola y 1 = x 2 and y2 = 2-x^2. Two lines y2 are in y 1.

Find the volume by parallel cross-sectional area: ω is a three-dimensional geometry, and the plane perpendicular to the X axis is at x∈[a, b]. The triangular area or rectangular area of the cross-sectional plane is a function of X, which can be expressed as A(x), so the X-axis solid volume V = ∫ ABBA (X) DX is calculated by two cylindrical surfaces. Solution: determine the closed part of the two volumes 1/8, and the vertical X-axis cross-sectional plane is z = y = √ (A 2-x

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Volume of the body of revolution: ω is defined by the plane figure 0 ≤ |y| ≤ |f(x)|, x∈[a, b]. Taking the rotation around the X axis as an example, the cross-sectional area function of ω is a (x) = π [f (x)] 2, x∈[a, b], so it rotates. For example 1, try to derive the volume formula of the cone with the above formula. Solution: Let the height of the cone be H and the radius of the bottom circle be R (that is, Y). The drawing shows that the X-axis rotator is an approximate flat cylinder with the base radius of 0 ≤ | y | ≤ x.r/h, the height of x∈[0, h] dx, and v = π ∫ 0h [x.r/h] 2dx.

e x =∑n = 1∞x n = 1+x+x ^ 2/2！ +...+ x n / n！ +...，x∈(-∞，+∞)；

sinx =∑n = 1∞x 2n+ 1 = x-x 3/3！ +...+ .x 2n+ 1 +...；

cosx =∑n = 1∞x 2n = 1-x 2/2！ +...+ .x 2n +...；

ln(x+ 1)=∑n = 1∞x n = x-x ^ 2/2！ + x 3 / 3！ +...+ x n +...；

( 1+x)- 1 = 1/( 1+x)=∑n = 0∞(- 1)n . x n，x∈(- 1， 1)。