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Use the weighted mean value theorem to establish some inequalities

Note that

    \[ \sqrt{1-x^2} = \frac{1-x^2}{\sqrt{1-x^2}}, \]

and use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove:

    \[ \frac{11}{24} \leq \int_0^{\frac{1}{2}} \sqrt{1-x^2} \, dx \leq \frac{11}{24} \sqrt{\frac{4}{3}}. \]

Recall the weighted mean value theorem:

For functions f and g continuous on [a,b], if g never changes sign in [a,b] then there exists c \in [a,b] such that

    \[ \int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx. \]


Proof. Let

    \[ f(x) = \frac{1}{\sqrt{1-x^2}}, \qquad g(x) = 1-x^2. \]

Then,

    \begin{align*}   \int_0^{\frac{1}{2}} \sqrt{1-x^2} \, dx &= \int_0^{\frac{1}{2}} \frac{1-x^2}{\sqrt{1-x^2}} \, dx \\  &= \int_0^{\frac{1}{2}} f(x) g(x) \, dx \\  &= f(c) \int_0^{\frac{1}{2}} g(x) \, dx \end{align*}

for some c \in \left[ 0, \frac{1}{2} \right]. Since f is strictly increasing on \left[ 0, \frac{1}{2} \right], we have f(0) \leq f(c) \leq f\left( \frac{1}{2} \right) implies 1 \leq f(c) \leq \sqrt{\frac{4}{3}}. Thus,

    \[ f(0) \int_0^{\frac{1}{2}} g(x) \, dx \leq f(c)  \int_0^{\frac{1}{2}} g(x) \, dx \leq f(1) \int_0^{\frac{1}{2}} g(x) \, dx. \]

We also know

    \[ \int_0^{\frac{1}{2}} g(x) \, dx = \int_0^{\frac{1}{2}} (1-x^2) \, dx = \frac{1}{2} - \frac{1}{24} = \frac{11}{24}. \]

So, putting these together,

    \[ \frac{11}{24} \leq \int_0^{\frac{1}{2}} \sqrt{1-x^2} \, dx \leq \frac{11}{24} \sqrt{\frac{4}{3}}. \]

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