n →∞ 2sin x 2 (sin x + ···+sinnx)=cos x 2 − cos 2n +1 2 x f [a, b] lim n→∞ b − a n n  k=1 f(a + k(b −a) n )=  b a f(x)dx lim n→∞ 1 n (( 1 n ) 2 +( 2 n ) 2 + ···+( n n ) 2 ) lim n→∞ 1 n (e 3 n + e 3.2 n + ···+ e 3n n ) lim n→∞ ( 1 n +1 + 1 n +2 + ···+ 1 2n ) lim n→∞ 1 p +2 p + ···+ n p n p+1 S n = 1 n  (1 + 1 n )sin 2π n +(1+ 2 n )sin 4π n + ···+(1+ n n )sin 2nπ n  lim n→+∞ S n lim n→+∞ S n f [0, 1]  1 0 f − 1 n n  k=1 f( k n )=O( 1 n )  b a f(x)dx =  b a f(t)dt =  b a f(u)du  2 0 (3x 2 − 5)dx =3  2 0 x 2 dx − 5  2 0 dx =3( 2 3 3 − 0) − 5(2 − 0). |f| f [0, 1] {0, 1 10 , 2 10 , 3 10 , ··· , 1} f(x)=sin 1 x ,f(0) = 7 f(x)= 1 n x = 1 n ,n∈ N f(x)=0 D(x)=0 x D(x)=1 x f [a, b] f(x)=g(x) g f [a, b] f(x)=g(x) g f [a, b] F (x)=  x a f [a, b] |F (x) −F(y)|≤max t∈[a,b] |f(t)||x −y| 0 ≤ x ≤ 1 x 2 √ 2 ≤ x 2 √ 1+x ≤ x 2 1 3 √ 2 ≤  1 0 x 2 √ 1+x dx ≤ 2 3 2π 2 9 ≤  π 2 π 6 2x sin x dx ≤ 4 9 π 2 f [a, b] f ≥ 0 c f(c) > 0  b a f>0  b a f =0 f ≡ 0 f,g [a, b]   b a f(x)g(x)dx  2 ≤  b a f 2 (x)dx  b a g 2 (x)dx n =1, 2, 3, ···  n 1 ln xdx < ln n! <  n+1 1 ln xdx e  n e  n <n! <e  n +1 e  n+1 n!=  n e  n O(n) f :[1, +∞) → R S n = n  k=1 f(k) I n =  n 1 f(x)dx f(k) <  k k−1 f(x)dx < f(k − 1) (k =2, 3, ···) (S n − I n ) n∈N [0,f(1)] 1+ 1 2 + ···+ 1 n − ln n f(x)=x + n  k=1 (a k cos kx + b k sin kx) (−π, π) f [a, b] d dx   x a f(t)dt  = f(x) f(x)=  x 0  t + t 6 dt df dx df dt f F (x)=  x 2 0 f F  (x) ϕ [a, b] f ϕ([a, b]) d dx (  ϕ(x) ϕ(a) f(t)dt)=f(ϕ(x))ϕ  (x). ◦  a 0 x 2  a 2 − x 2 dx  a 1 √ a 2 − x 2 x dx ◦  1 0 xe x dx  π/2 0 x cos xdx  π/2 0 e x cos xdx ◦  1 0 dx x 2 − 5x +6  1 0 xdx (1 + x) 2  1 0 x 5 dx 1+x 2  1 0 dx x 4 +4x 2 +3 ◦  √ 2 √ 2/3 dx x √ x 2 − 1  7 2 dx √ 2+x +1 ◦  π 0 sin xdx cos 2 x − 3  π 0 sin 4 xdx  π/4 0 tan 6 xdx  π/4 0 dx cos 4 x f [−a, a] f  a −a f =2  a 0 f f  a −a f =0 f T [0,T] a ∈ R  a+T a f =  T 0 f f [0, 1]  π 0 f(sin x)dx =2  π/2 0 f(sin x)dx  π 0 xf(sin x)dx = π 2  π 0 f(sin x)dx  π 0 x sin x 1+cos 2 x dx  π 0 x 3 sin x 1+cos 2 x dx f(x)= 1 (1 + x) 2 f(x)= d dx ( −1 1+x )  +∞ −∞ f(x)dx = lim a→+∞  a −a f(x)dx = lim a→+∞ ( −1 1+a − 1 1+a )=0.  +∞ 1 dx x 2/3  +∞ 1 dx x 4/3  1 0 dx x 2/3  1 0 dx x 4/3  +∞ 0 x 2 +1 x 4 +1 dx  +∞ 0 x cos xdx  +∞ 0 x ln xdx  +∞ 1 xdx √ x − 1  +∞ 0 x 2 dx x 4 + x 2 +1  +∞ 1 x n dx 1+x m n ≥ 0  +∞ 1 ln(1 + x)dx x n  +∞ 0 cos xdx 2+x n  1 0 dx √ 1 − x 2  1 0 √ xdx √ 1 − x 4  1 0 √ xdx e sin x − 1  1 0 ln xdx 1 − x 2  +∞ 0 sin x x 3/2 dx p, q, p 1 , ··· ,p n  b a dx (x − a) p  +∞ 0 x p 1+x dx  +∞ 1 dx x p ln q x Γ(p)=  +∞ 0 e −x x p−1 dx B(p, q)=  1 0 x p (1 − x) q dx  π/2 0 dx cos p x sin q x  +∞ −∞ dx |x − a 1 | p 1 |x − a 2 | p 2 ···|x −a n | p n  +∞ 0 cos x x dx f [0, 1]  1 0 f(x) √ 1 − x 2 dx  1 0 f(x) √ 1 − x 2 dx =  π/2 0 f(sin u)du F (x)=  x 0 sin t t 3/2 dt (0 <x<∞) max x = π  1 0 ln xdx lim n→∞ n √ n! n R 2 y = x 2 +4,y= x +4 x 2 a 2 + y 2 b 2 =1,y= x 2 y =ln( k x ),y=0,x=1,x= e (k>0) k ∈ N <e− 2 x = a(t − sin t),y= a(1 −cos t) y =0 r 2 = a 2 cos 2ϕ r = a(1 + cos ϕ) I(t)=  1 0 |e x − t|dx R 3 y = b a  a 2 − x 2 , −a ≤ x ≤ a y 2 =4− x, 0 ≤ x ≤ 4 R 2 y 2 = 1 9 x 3 , 0 ≤ x ≤ 1,y >0 y 2 =2px, a ≤ x ≤ b x = a cos 3 t, y = a sin 3 t r =sin 3 ϕ 3 , 0 ≤ ϕ ≤ π/2 0, 61111 ··· 1, 33333 ··· −2, 343434 ··· e π ln 2 S =1+2+4+8+16+··· 2S =2+4+8+···= S − 1 S = −1 a 1 + a 2 + a 3 + ··· S a 2 + a 3 + ··· S − a 1 1 1.2 + 1 2.3 + 1 3.4 + 1 4.5 + 1 5.6 + ··· 1 1.4 + 1 4.7 + 1 7.10 + 1 10.13 + ··· 1 1.3 + 1 4.6 + 1 7.9 + 1 10.12 + 1 13.15 + ··· 1 2 − 1 4 + 1 8 − 1 16 + 1 32 + ··· ∞  k=0 2 k +3 k 6 k ∞  k=0 ( 1 − x 1+x ) k ∞  k=0 ( √ k +2− 2 √ k +1+ √ k) ∞  k=1 1 k(k +1)(k +2) ∞  k=1 1 k(k + m) (m ∈ N) ∞  k=0 k 4 k! ∞  k=0 1+k 1+k 2 ∞  k=0 3 4+2 k ∞  k=0 k ln k k 2 +2k +3 ∞  k=1 k!3 k k k ∞  k=0 (k!) 2 (2k)! ∞  k=2 1 (ln k) k ∞  k=0 (1 + 1 k ) 2k e k ∞  k=1 1 k p ∞  k=2 1 k ln k ∞  k=2 1 k p ln q k ∞  k=0 sin kx a k = 1 √ k + (−1) k k ∞  k=1 (−1) k a k a k > 0 a k → 0 ( 1 2 ) 0 +( 1 4 ) 1 +( 1 2 ) 2 +( 1 4 ) 3 +( 1 2 ) 4 + ··· S = ∞  k=1 1 k p n S n p>0 1 (k +1) p <  k+1 k 1 x p dx < 1 k p (k =1, 2, ···) p>1 p>1 S n−1 +  +∞ n 1 x p dx < S < S n +  +∞ n 1 x p dx 1 (p − 1)(n +1) p−1 <S− S n < 1 (p − 1)n p−1 a k ,b k > 0 ∞  k=0 a k ∞  k=0 b k ∞  k=0 a k b k , ∞  k=0 a 2 k , ∞  k=0 (a k + b k ) 2 , ∞  k=0 √ a k k 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + ··· =1+( 1 2 − 1) + 1 3 +( 1 4 − 1 2 )+ 1 5 +( 1 6 − 1 3 )+··· =(1+ 1 2 + 1 3 + 1 4 + ···) − 1 − 1 2 − 1 3 − 1 4 −··· =(1+ 1 2 + 1 3 + 1 4 + ···) − (1 + 1 2 + 1 3 + 1 4 + ···) =0. 1+x 2 + x + x 4 + x 6 + x 3 + x 8 + x 10 + x 5 + ···= 1 1 − x |x| < 1 . ··· 1 1.4 + 1 4.7 + 1 7 .10 + 1 10 .13 + ··· 1 1.3 + 1 4.6 + 1 7.9 + 1 10 .12 + 1 13 .15 + ··· 1 2 − 1 4 + 1 8 − 1 16 + 1 32 + ··· ∞  k=0 2 k +3 k 6 k ∞  k=0 ( 1 − x 1+ x ) k ∞  k=0 ( √ k +2− 2 √ k +1+ √ k) ∞  k =1 1 k(k. + 1 3 +( 1 4 − 1 2 )+ 1 5 +( 1 6 − 1 3 )+··· = (1+ 1 2 + 1 3 + 1 4 + ···) − 1 − 1 2 − 1 3 − 1 4 −··· = (1+ 1 2 + 1 3 + 1 4 + ···) − (1 + 1 2 + 1 3 + 1 4 + ···) =0. 1+ x 2 + x + x 4 + x 6 + x 3 + x 8 + x 10 + x 5 +. π/2 0, 611 11 ··· 1, 33333 ··· −2, 343434 ··· e π ln 2 S =1+ 2+4+8 +16 +··· 2S =2+4+8+···= S − 1 S = 1 a 1 + a 2 + a 3 + ··· S a 2 + a 3 + ··· S − a 1 1 1. 2 + 1 2.3 + 1 3.4 + 1 4.5 + 1 5.6 + ··· 1 1.4 + 1 4.7 + 1 7 .10 + 1 10 .13 +