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,

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

Stumbling Robot

Use the weighted mean value theorem to approximate an integral

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