Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 11E (Analysis I)

For continuous $f$, $F(x)={\int }_a^{x}f(t)dt$ is differentiable with $F\prime =f$. Conversely, if $F\prime$ is continuous, ${\int }_a^{b}F\prime =F(b)-F(a)$.

Put $h(x)=|f(0)|+{\int }_0^{x}g(t)|f(t)|dt$. Then $|f|\le h$ and $h\prime \le gh$. If $h>0$, $(\log h)\prime \le g$; integration gives $h(x)\le h(0)e^{{\int }_0^{x}g}$. For $f(0)=0$, apply the same argument after adding $\epsilon$ and let $\epsilon \downarrow 0$.

For $h=f_1-f_2$, $h\prime =\sqrt{1+x}(f_1+f_2)h$ and $h(0)=0$. On every compact subinterval the coefficient is continuous, so Gronwall inequality gives $h=0$; hence equality holds throughout $[0,a)$.