A cubic polynomial M(x) is such that M(2)=0, and has relative extremes at x=1 and x=1/3.
Determine M(x) given that ∫_{1}^{1}M(x)dx= 14/3.
Let the equation be ax^2 + bx^2 + cx + d.
Substituting x=2,
8a + 4b  2c + d = 0
The derivative is
3ax^2 + 2bx + c
and substituting x=1 and x=1/3 to find the extrema, where the derivative vanishes, gives
3a  2b + c = 0
a + 2b + 3c = 0
leading to c = a and b = a
Integrating the original cubic:
ax^4/4 + bx^3/3 + cx^2/2 + dx
and evaluating from 1 to 1 and equating to 14/3, give
b + 3d = 7 or
3d = 7  a
so
b = a = 7  3d
c = 3d  7
Going back to the original cubic evaluated at 2:
4b  2c + d = 8a
28  12d  6d + 14 + d = 56  24d
7d = 14
d = 2
So going back to where b, a and c were defined in terms of d:
c = 1; a = 1; b = 1
So M(x) = x^3 + x^2  x + 2

Posted by Charlie
on 20070117 10:52:41 