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:
