From the fact you know, find in your head the average value of

a) ${\mathbf{\text{x}}}^{\mathbf{\text{3}}}\mathbf{-}\mathbf{\text{3sinh2x}}\mathbf{+}{\mathbf{\text{sin}}}^{\mathbf{\text{2}}}\mathbf{\text{πx}}\mathbf{+}\mathbf{\text{cos3πx}}$on $\mathbf{\text{(}}\mathbf{-}\mathbf{\text{5,5)}}$

b)${\mathbf{\text{2sin}}}^{\mathbf{\text{2}}}\mathbf{\text{3x}}\mathbf{-}\mathbf{\text{4cosx}}\mathbf{+}\mathbf{\text{5xcosh2x}}\mathbf{-}{\mathbf{\text{xcos}}}^{\mathbf{\text{2}}}\mathbf{\text{x}}$on $\mathbf{(}\mathbf{-}\mathbf{\text{π,π}}\mathbf{)}$

In mathematics, a function is an expression, rule, or law that describes the relationship between the independent variable and the dependent variable.In mathematics, functions are everywhere, and they're crucial for establishing interconnection in the sciences.

Given the expressions

a) ${\text{x}}^{\text{3}}-\text{3sinh2x}+{\text{sin}}^{\text{2}}\text{πx}+\text{cos3πx}$

b)${\text{2sin}}^{\text{2}}\text{(3x)}-\text{4cosx}+\text{5xcosh(2x)}-{\text{xcos}}^{\text{2}}\text{x}$

A mean of a sum will equal the sum of the means.

It implies that ${\text{<x}}^{\text{3}}\text{>}=\text{<}-\text{3sinhx>}=\text{0}$

Since they are odd over the symmetrical interval.

Therefore, 2 and $\frac{2}{3}$will be the base periods of the other two functions where both are integral divisions of $\left(-5,5\right)$.

As a result, the average value of $\text{cos3πx}$will be zero and the average value of ${\text{sin}}^{\text{2}}\text{πx}$is$\frac{1}{2}$.

This implies that,

${\text{<x}}^{\text{3}}-\text{3sinhx}+{\text{sin}}^{\text{2}}\text{(πx)}+\text{cos(3πx)>}=\frac{\text{1}}{\text{2}}$

Since the functions ${\text{xcos}}^{\text{2}}\text{x}$and $\text{xcosh(2x)}$are odd.

Thus, their average will be zero.

And as a result, the average value of will be zero over the interval$(-\text{π,π})$.

Then Finally,

$<{\text{2sin}}^{\text{2}}\text{(3x)}-\text{4cosx}+\text{5xcosh(2x)}-{\text{xcos}}^{\text{2}}\text{x}>=<2{\sin }^2\left(3x\right)>=1$

So, the average value is 1.