Question

Iff(t)is a continuous function, what is the relationship between

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}{\mathbf{\left(}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right)}}^{\mathbf{2}}\mathit{d}\mathit{t}\mathbf{}\mathbf{and}\mathbf{}{\left({\int }_0^{1}f\left(t\right)\mathrm{dt}\right)}^{\mathbf{2}}$

Hint: Use the Cauchy–Schwarz inequality.

Short answer

${\left({\int }_0^{1}f\left(t\right)dt\right)}^2\le {\int }_0^1{\left(f\left(t\right)\right)}^{2}dt$

Step-by-step solution

Cauchy–Schwarz inequality.

If$\overrightarrow{\mathbf{x}}$and$\overrightarrow{\mathbf{y}}$are vectors in${\mathbf{R}}^{\mathbf{n}}$then$\left|\overrightarrow{x},\overrightarrow{y}\right|\mathbf{\le }||\overrightarrow{x}||\mathbf{.}||\overrightarrow{y}||$.

This statement is an equality if (and only if) and are parallel.

It will apply theCauchy–Schwarz inequality.

By taking the functions f and 1.

Observe that following,

Hence, the relationship between ${\int }_0^1{\left(f\left(t\right)\right)}^{2}dt\mathrm{and}{\left({\int }_0^{1}f\left(t\right)dt\right)}^2\mathrm{is}{\left({\int }_0^{1}f\mathit{\left(}\mathit{t}\mathit{\right)}dt\right)}^2\le {\int }_0^1{\left(f\left(t\right)\right)}^{2}dt$.