Indefinite integrals ~ geometrical applications

(d^2)y/d(x^2) = 6
dy/dx = 6x + C (slope)

sub (0,4)
2 = 6(0) + C
C = 2

dy/dx = 6x + 2
y = 3x^2 + 2x + D

sub(0,4)
4 = 0 + 0 + D
D = 4

therefore the equation is y = 3x^2 + 2x + 4

at every point on a curve , (d^2)y/d(x^2) = 6

咁in左佢先
dy/ dx = 6x + C
咁佢話有一點Slope 係2
咁即係

Slope = 6 (x個點) + C
2 =6(0) + C
C =2

再IN多次
dy/ dx = 6x + 2
y = 6x^2 + 2x +C

咁佢話本新有curve有(0,4)哩點
咁代晒入去 y = 6x^2 + 2x +C
4= 6(0)^2 + 2(0) +C
C=4

所以
y = 6x^2 + 2x + 4

2007-11-29 01:54:59 補充：
打左錯第二次個in~再IN多次dy/ dx = 6x 2y = 3x^2 2x C咁佢話本新有curve有(0,4)哩點咁代晒入去 y = 3x^2 2x C4= 3(0)^2 2(0) CC=4所以y = 3x^2 2x 4

(d^2)y/(dx^2) = 6 , [d/dx*d/dx*v=d^2v/(dx)^2]
dv/dx=6x+d ,d is constant
dv/dx∣(0,4)=6(0)+d=2
d=2
∴dv/dx=6x+2

v=3x^2+2x+c ,c is constant
sub (0.4) into curve
4=0-0+c
c=4

∴the curve is v=3x^2+2x+4