The book I am studying, has this question, show $F(x+h)=F(x)+hF'(x)+\frac{h^2}{2}F''(x)+h^2\phi (h)$ where $\phi(h) \to 0$ as $h\to 0$. It is noted that this is a Taylor expansion, but the method suggested was to use $F(x+h)-F(x)=\int\_x^{x+h}F'(y)dy$ and $F'(y)=F'(x)+(y-x)F''(x)+(y-x)\psi(y-x)$ where $\psi(h) \to 0$ as $h\to 0$. I tried it and am having trouble at the last step.\begin{align\*} F(x+h)-F(x)&=\int\_x^{x+h}F'(y)dy\\ &= \int\_x^{x+h}F'(x)+(y-x)F''(x)+(y-x)\psi(y-x)dy\\ &=hF'(x)+\frac{h^2}{2}F''(x)+\int\_x^{x+h}(y-x)\psi(y-x)dy\\&=hF'(x)+\frac{h^2}{2}F''(x)+\int\_0^{h}k\psi(k)dk \end{align\*}

How do I finish it off, i.e. how do I show $\int\_0^{h}k\psi(k)dk=h^2\phi (h)$. Thanks