Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 12E (Analysis I)

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

The fundamental theorem giveswhich is the desired inequality.

The assumptions imply $f\ge 0$. Since $(f^2)\prime =2ff\prime \le 2f$ and both sides vanish at zero, integration gives $f(x{)}^2\le 2F(x)$ where $F(x)={\int }_0^{x}f$. NowThe bracket vanishes at zero, proving ${\int }_0^x{f}^3\le ({\int }_0^{x}f{)}^2$.

No. Take the constant function $f(t)=c>0$, which has $f\prime =0$. The claimed inequality becomes $c^{3}x\le c^2{x}^2$, false whenever $0<x<c$.