So im trying to show that $E_2(x)$ = $\int_{a}^{x} \frac{f'''(t)(x-t)^2}{2}dt$. In the problem they want this to be done by using the equation
$E_2(x)$ = $E_1(x)$ - $\frac{f''(a)(x-a)^2}{2}$
where we know
$E_1(x) = \int_{a}^{x} f''(t)(x-t)dt$
my attempt to solve it:
want to show that $E_2(x)$ = $\int_{a}^{x} \frac{f'''(t)(x-t)^2}{2}dt$.
We have:
$E_2(x)$ = $E_1(x)$ - $\frac{f''(a)(x-a)^2}{2}$
$E_1(x) = \int_{a}^{x} f''(t)(x-t)dt$
$E_2(x)$ = $\int_{a}^{x} f''(t)(x-t)dt$ - $\frac{f''(a)(x-a)^2}{2}$
doing integration by parts differentiating $f''(t)$, and integrating $(x-t)$ we get:
$E_2(x)$ = $ \tfrac{f''(t)(x-t)^2}{2}\Big|_a^x - \int_{a}^{x} \frac{f'''(t)(x-t)^2}{2}dt$ - $\frac{f''(a)(x-a)^2}{2}$
$E_2(x) = $ - $\int_{a}^{x} \frac{f'''(t)(x-t)^2}{2}dt$ - ${f''(a)(x-a)^2}$
However at that point i dont see how i should progress. thanks in advance