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Use the weighted mean value theorem to establish an inequality

by RoRi

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

    \[ \frac{1}{10 \sqrt{2}} \leq \int_0^1 \frac{x^9}{\sqrt{1+x}} \, dx \leq \frac{1}{10}. \]

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

    \[ f(x) = \frac{1}{\sqrt{1+x}}, \qquad g(x) = x^9. \]

Then substituting our definitions of f and g,

    \[ \int_0^1 \frac{x^9}{\sqrt{1+x}} \, dx = \int_0^1 f(x) g(x) \, dx. \]

Since f and g are continuous and g does not change sign on [0,1] we may apply Theorem 3.16,

    \[ \int_0^1 f(x) g(x) \, dx = f(c) \int_0^1 g(x) \, dx \qquad \text{for some } c \in [0,1]. \]

Since f is strictly decreasing on [0,1], we have

    \[ f(0) \geq f(c) \geq f(1) \quad \implies \quad 1 \geq f(c) \geq \frac{1}{\sqrt{2}}. \]

Then,

    \begin{align*} &&f(1) \int_0^1 g(x) \, dx &\leq f(c) \int_0^1 g(x) \, dx \leq f(0) \int_0^1 g(x) \, dx \\[8pt] \implies && \frac{1}{\sqrt{2}} \int_0^1 x^9 \, dx &\leq f(c) \int_0^1 x^9 \, dx \leq \int_0^1 x^9 \, dx \\[8pt] \implies && \frac{1}{10 \sqrt{2}} &\leq f(c) \int_0^1 \frac{x^9}{\sqrt{1+x}} \, dx \leq \frac{1}{10}. \qquad \blacksquare \end{align*}

4 comments

  1. Mihajlo says:

    It seems safer to do integration on the product of g and f, as in the proof of 3.16 (and not like done here).

    • Mihajlo says:

      Sorry, the way you proved it seems valid (apart from the mentioned error in the last row) and probably the intended way, because it uses the required theorem.

    • jon says:

      remember that f(c)(integral of g(x)) = (integral of f(x)g(x)) which is the original equation. In other words both the function he ended up with and the one being asked for are equivalent.

      *edit although he did make a typo and rewrote f(c) in the final line when he had already factored it back in.

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