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Use the weighted mean value theorem to approximate an integral

Note that

    \[ 1+x^6 = (1+x^2)(1-x^2 + x^4), \]

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

    \[ \frac{1}{1+a^6} \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right) \leq \int_0^a \frac{dx}{1+x^2} \leq a - \frac{a^3}{3} + \frac{a^5}{5}.  \]

Using a = \frac{1}{10} compute the integral to six decimal places.

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

    \[ f(x) = \frac{1}{1+x^6}, \qquad g(x) = 1 - x^2 + x^4. \]

Then,

    \begin{align*}   \int_0^a \frac{dx}{1+x^2} &= \int_0^a \frac{1-x^2+x^4}{1+x^6} \, dx \\  &= \int_0^a f(x) g(x) \, dx \\  &= \frac{1}{c} \int_0^a g(x) \, dx  \end{align*}

for some c \in [0,a]. Since f is strictly decreasing on [0,a], we know f(a) \leq f(c) \leq f(0); thus,

    \[ \frac{1}{1+a^6} \leq f(c) \leq 1. \]

Furthermore,

    \[ \int_0^a g(x) \, dx = \int_0^a (1-x^2 + x^4) \, dx = a - \frac{a^3}{3} + \frac{a^5}{5}. \]

Thus,

    \[ \left( \frac{1}{1+a^6} \right) \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right) \leq \int_0^a \frac{dx}{1+x^2} \leq \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right). \qquad \blacksquare\]

Now, taking a = \frac{1}{10} we compute,

    \begin{align*}  && (.9999999)(.0996687) \leq \int_0^a \frac{dx}{1+x^2} \leq .0996687 \\ \implies && .0996686 \leq \int_0^a \frac{dx}{1+x^2} \leq .0996687 \\ \implies && \int_0^a \frac{dx}{1+x^2} \approx .099669. \end{align*}

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