Find all the polynomials f(t) of degree less than or equal to 3, such that f(0) = 3, f(1) = 2, f(3) = 0 and the integral from 0 to 2 of f ' (t) dt = 4
- Help with this problem would be appreciated a lot. Thanks.
You have to solve a 4x4 system of linear equations.
The polynomial is of the form $\displaystyle ax^{3}+bx^{2}+cx+d.$ The first three data give you three of the equations you need in order to determine a, b, c, and d.
Since the integral of the derivative of f between 0 and 2 is equal to f(2)-f(0), you have a fourth equation here. So, have a happy Gauss-elimination (or whatever way you call it) practice!
No you plug in your values into your 3rd degree polynomial:
$\displaystyle
ax^{3}+bx^{2}+cx+d.
$
so f(0)=3:
$\displaystyle a(0)^{3}+b(0)^{2}+c(0)+d=3$
f(1)=2
$\displaystyle a(1)^{3}+b(1)^{2}+c(1)+d=2$
f(3)=0
$\displaystyle a(3)^{3}+b(3)^{2}+c(3)+d=0$
f(2)-f(0)
$\displaystyle f(2)-f(0) = a(2)^{3}+b(2)^{2}+c(2)+d-(a(0)^{3}+b(0)^{2}+c(0)+d)=8a+4b+2c = 4$
$\displaystyle \left [\begin{array}{cccc}
0 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 \\
8 & 4 & 2 & 0 \\
27 & 9 & 3 & 1 \end{array} \right]*\left [\begin{array}{c}
a \\
b \\
c \\
d \end{array} \right]=\left [\begin{array}{c}
3 \\
2 \\
4 \\
0 \end{array} \right]$
I put the f(3) equation in the last row btw.