Int_2 ^ 3 (2x + 1) / (x ^ 3 - 5x ^ 2 + 4x) dx?

Jawapan:

#-1.11164#

Penjelasan:

# "Ini adalah integral dari fungsi rasional." #

# "Prosedur piawai dipecah dalam pecahan separa." #

# "Mula-mula, kita mencari nol penyebut:" #

# x ^ 3 - 5 x ^ 2 + 4 x = 0 #

# => x (x - 1) (x - 4) = 0 #

# => x = 0, 1, atau 4 #

# "Jadi kami berpecah dalam pecahan separa:" #

# (2x + 1) / (x ^ 3-5x ^ 2 + 4x) = A / x + B / (x-1) + C / (x-4) #

# => 2x + 1 = A (x-1) (x-4) + B x (x-4) + C x (x-1) #

# => A + B + C = 0, -5 A - 4 B - C = 2, 4A = 1 #

# => A = 1/4, B = -1, C = 3/4 #

# "Jadi kami ada" #

# (1/4) int {dx} / x - int {dx} / (x-1) + (3/4) int {dx} / (x-4)

# = (1/4) ln (| x |) - ln (| x-1 |) + (3/4) ln (| x-4 |) + C #

# "Sekarang kita menilai antara 2 dan 3:" #

(3) - ln (2) + membatalkan ((3/4) ln (1)) - (1/4) ln (2) + membatalkan (ln (1) / 4) ln (2) #

# = (1/4) ln (3) - 2 ln (2) #

#= -1.11164#

Apakah semua nilai untuk k yang mana int_2 ^ kx ^ 5dx = 0?

Lihat di bawah. int_2 ^ kx ^ 5 dx = 1/6 (k ^ 6-2 ^ 6) dan k ^ 6-2 ^ 6 = (k ^ 3 + 2 ^ 3) (k ^ 3-2 ^ 3) tetapi k ^ + 2 ^ 3 = (k +2) (k ^ 2-2k + 2 ^ 2) dan k ^ 3-2 ^ 3 = (k-2) (k ^ 2 + 2k + 2 ^ 2) -2 ^ 6 = (k +2) (k ^ 2-2k + 2 ^ 2) (k-2) (k ^ 2 + 2k + 2 ^ 2) atau {(k + 2 = 0) 2-2k + 2 ^ 2 = 0), (k-2 = 0), (k ^ 2 + 2k + 2 ^ 2 = 0):} maka akhirnya nilai sebenar k = {-2,2} {-1pm saya sqrt3,1pm i sqrt3}

Bagaimana untuk menyelesaikannya? Int_2 ^ 85-xdx =?

= 9 int_2 ^ 8 | 5-x | dx = int_2 ^ 5 (5-x) dx + int_5 ^ 8 (x-5) dx = [5x - x ^ 2/2 + C1] _2 ^ 5 + [x ^ 2/2 - 5x + C2] 8 = 12.5 + C1 - 8 - C1 - 8 + C2 + 12.5 - C2 = 9 "Pada langkah pertama kami hanya menggunakan takrif | ... |:" | x | = {(-x, "," x <= 0), (x, "," x> = 0):} "Jadi" | 5 - x | (5 - x <= 0), (5 - x, "," 5 - x> = 0):} = {(x - 5, "," x> = 5) , (5 - x, "," x <= 5):} "Jadi kes had x = 5 membahagikan selang integrasi dalam dua bahagian: [2, 5] dan [5, 8]."