Question

N=9
Ajutooooooor
Urgent

Answer 1

Răspuns:

Ex  2

∫(4x-9)./(2x-4)dx=∫(4x-8-1)/(2x-4)dx=∫(4x-8)dx/(2x-4)-∫dx/(2x-4)=

∫2(2x-4)dx/(2x-4)-1/2∫dx/(x-2)=2∫dx-1/2ln(x-2)=2x-1/2ln(x-2)+c

Ex3

A=∫(6x²-x-9)dx=

∫6x²dx-∫xdx-9∫dx=

6x³/3-x²/2+9x=

(2x³-x²/2+9)║₂⁴=

(2*4³-4²/2+9)-(2*2³-2²/2+9)=

(2*64-16/2+9)-(2*8-4/2+9)=

128-8+9-(16-2+9)=

129-23=106u²

Ex 5

limitele  de  integrare

3x²+1=2x+1

3x²-2x=0

x(3x-2)=0

x1=0

x2=2/3

V=π∫(f(x)-g(x)]²dx   x∈[0,2/3]

V=π∫(3x²+1)-(2x+1)]²dx=

π∫[3x²+1-2x-1)²dx=

π∫(3x²-2x)²dx=

π∫(9x⁴-12x³+4x²)dx=

π[9x⁵/5-12*4x⁴/4+4x³/3]║₀²⁾³=

π   calculele   le   faci   tu

Ex5

x²-8x+7=0

x1=1    x2=7

x²-8x+7=(x-1)(x-7)

(x+9)/((x-1)(x-7)=A/(x-1)+B/(x-7)=

(AX-7A+Bx-B)/(x-1)(x-7)=>

x+9=x(A+B)-7A-B

Sistem

{a+B=1

{-7A-B=9

le  adui

A-7A+B-b=1+9

-6A=10

A=-10/6= -5/3

-5/3+B=1

B=1+5/3=8/3

=>∫(x+9)dx/(x-1)(x-7)=

∫-5/3*dx/(x-1)+∫8/3dx/(x-7)=

[-5/3ln(x-1)+8/3  ln  (x-7)]║₂³=

-5/3ln(2

Explicație pas cu pas: