Paper 1, Section I, 4F4 \mathbf{F}

Analysis I | Part IA, 2012

Let f,g:[0,1]Rf, g:[0,1] \rightarrow \mathbb{R} be continuous functions with g(x)0g(x) \geqslant 0 for x[0,1]x \in[0,1]. Show that

01f(x)g(x)dxM01g(x)dx\int_{0}^{1} f(x) g(x) d x \leqslant M \int_{0}^{1} g(x) d x

where M=sup{f(x):x[0,1]}M=\sup \{|f(x)|: x \in[0,1]\}.

Prove there exists α[0,1]\alpha \in[0,1] such that

01f(x)g(x)dx=f(α)01g(x)dx\int_{0}^{1} f(x) g(x) d x=f(\alpha) \int_{0}^{1} g(x) d x

[Standard results about continuous functions and their integrals may be used without proof, if clearly stated.]

Typos? Please submit corrections to this page on GitHub.