Evaluate $a+2b+3c$ If $$\lim_{h \to 0}\frac { \int _{ 0 }^{ h }{ f(x)dx-h(af(0)+bf(h/3)+cf(h)) }  }{ { h }^{ 4 } } $$ is a finite non-zero number,how to evaluate $a+2b+3c$?
Ok only hints at first.I want to solve this myself.But can't understand how to start.Any suggestion appreciated.Thanks.
 A: HINTS:
If $f$ is sufficiently smooth, then we can write
$$f(\alpha h)=f(0)+f'(0)\alpha h+\frac1{2!} f''(0)(\alpha h)^2+\frac1{3!} f'''(0)(\alpha h)^3+\frac{1}{4!}f''''(0)(\alpha h)^4+O(h^5)$$
and
$$\int_o^hf(x)\,dx=f(0)h+\frac1{2!}f'(0)h^2+\frac1{3!}f''(0)h^3+\frac{1}{4!}f'''(0)h^4+(h^5)$$
A: Set $f(x) = 1, x, x^2, x^3$.
You will get a series of equations
for $a, b, c$.
Solve them and here is what you get.
$\frac { \int _{ 0 }^{ h } f(x)dx-h(af(0)+bf(h/3)+cf(h))   }{ { h }^{ 4 } }
$
For $f(x) = 1$:
$ \int _{ 0 }^{ h } dx-h(a+b+c) 
=h-h(a+b+c) 
$,
so
$a+b+c = 1$.
For $f(x) = x$:
$ \int _{ 0 }^{ h } xdx-h(a\cdot 0+b(h/3)+ch) 
=h^2/2-h^2(b/3+c) 
$,
so
$b/3+c = 1/2$
or
$b+3c = 3/2$.
For $f(x) = x^2$:
$ \int _{ 0 }^{ h } x^2dx-h(a\cdot 0+b(h/3)^2+ch^2)
=h^3/3-h^3(b/9+c) 
$,
so
$b/9+c = 1/3$
or
$b+9c = 3$.
We get
$a+b+c = 1$,
$b+3c = 3/2$,
$b+9c = 3$.
Subtracting the 2nd and 3rd,
$9c = 3/2$
or
$c = 1/6$.
From the last,
$b = 3-9/6 = 3/2$.
From the first,
$a = 1-3/2-1/6
=-2/3
$.
So
$a+2b+3c
=-2/3+3+1/2
=17/6
$.
Note that,
for $f(x) = x^3$:
$ \int _{ 0 }^{ h } x^3dx-h(a\cdot 0+b(h/3)^3+ch^3) 
=h^4/4-h^4(b/27+c) 
=h^4(1/4-3/(2\cdot 27)+1/6) 
=(h^4/(4\cdot 27)(27-4-18) 
=5h^4/(4\cdot 27) 
$,
