Question

Use the properties of integrals to verify that

Short answer

The given function is verified.

Step-by-step solution

Using the property

Apply the property which says if \(f(x) \le g(x)\) for \(x \in (a,b)\),

Then

\(\int_a^b f (x)dx \le \int_a^b g (x)dx\)

Evaluating to get the final result

First we define \(f(x): = {x^4}\cos x\).

On the interval \((0,1)\), notice that functions \(x \mapsto {x^4}\)and \(x \mapsto \cos x\)are both non- negative

Also cos \(x \le 1\), so we conclude

\(\begin{aligned}{c}g(x): &= 0 \le {x^4}\cos x \le {x^4} \cdot 1\\ &= {x^4}\\ &= :h(x) \Rightarrow g(x) \le f(x) \le h(x)\end{aligned}\)

Now apply the property (1) twice

\(\begin{aligned}0 &= \int_0^1 0 dx \le \int_0^1 {{x^4}} \cos xdx \le \int_0^1 {{x^4}} dx\\ &= \left. {\frac{{{x^5}}}{5}} \right|_0^1\\ &= \frac{1}{5}\\ \Rightarrow &0 \le \int_0^1 {{x^4}} \cos xdx \le 0.2\end{aligned}\)

By making use of property \(f(x) \le g(x)\)for \(x \in (a,b)\), then \(\int_a^b f (x)dx \le \int_a^b g (x)dx\).