1. Definition of Integral

Could somebody please explain how, from the definition of integral, we can deduce the following:

$\displaystyle \int_a^{a+h} f(x) \, dx = f(a)h + \frac{f'(a)}{2} h^2 + \frac{f''(\xi)}{6}h^3$

Obviously some form of the Taylor expansion with Lagrangian Remainders is going on, but where does the multiplying by h on the RHS come into it, why are the numerators moved 'back' one term than they usually are in a Taylor expansion, and how do we arrive at this result 'from the definition of integral'?

AnonymitySquared

2. Originally Posted by AnonymitySquared
Could somebody please explain how, from the definition of integral, we can deduce the following:

$\displaystyle \int_a^{a+h} f(x) \, dx = f(a)h + \frac{f'(a)}{2} h^2 + \frac{f''(\xi)}{6}h^3$

Obviously some form of the Taylor expansion with Lagrangian Remainders is going on, but where does the multiplying by h on the RHS come into it, why are the numerators moved 'back' one term than they usually are in a Taylor expansion, and how do we arrive at this result 'from the definition of integral'?

AnonymitySquared
Substitute $f(x) = f(a) + f'(a) (x - a) + \frac{f''(a)}{2} (x - a)^2 + \frac{f^{(3)} (a)}{6} (x - a)^3 + ....$ and integrate term-by-term.

3. Originally Posted by mr fantastic
Substitute $f(x) = f(a) + f'(a) (x - a) + \frac{f''(a)}{2} (x - a)^2 + \frac{f^{(3)} (a)}{6} (x - a)^3 + ....$ and integrate term-by-term.
Actually, I got it, in a manner that is a bit less laborious than that!

$\int_{a}^{a+h} f(x)\,dx = g(a+h)-g(a)$

$g(a+h) = g(a) + g'(a)h + \frac{g''(a)h^2}{2} + \frac{f''(\xi)h^3}{6}$

$\xi \in [x_0,x]$

$\therefore g(a+h) - g(a) = g'(a)h + \frac{g''(a)h^2}{2} + \frac{g'''(\xi)h^3}{6}$

$\therefore \int_{a}^{a+h} f(x)\,dx = g'(a)h + \frac{g''(a)h^2}{2} + \frac{g'''(\xi)h^3}{6}$

By the fundamental theorem of calculus:

$g'(a) = f(a) \, \, \, g''(a) = f'(a) \, \, \, g'''(a) = f''(a)$, hence:

$\int_{a}^{a+h} f(x)\,dx = f(a)h + \frac{f'(a)h^2}{2} + \frac{f''(\xi)h^3}{6}$

(Y)