what-is-the-mean-value-of-1-1-4x-2-for-0-x-1-2-

Question Number 76721 by Rio Michael last updated on 29/Dec/19

what is the mean value of \frac{1}{1 + {4 x}^{2}} for 0 ⩽ x ⩽ \frac{1}{2}

Answered by MJS last updated on 30/Dec/19

mean (f (x)) with a ⩽ x ⩽ b

\frac{1}{b - a} \underset{a}{\int^{b}} f (x) d x

{we}^{'} re calculating the area first and then

the height of a rectangle of the width of

our interval

\frac{1}{\frac{1}{2} - 0} \underset{0}{\int^{\frac{1}{2}}} \frac{d x}{1 + 4 x^{2}} = 2 \underset{0}{\int^{\frac{1}{2}}} \frac{d x}{1 + 4 x^{2}} =

[t = 2 x \to d x = \frac{1}{2} d t]

= \underset{0}{\int^{1}} \frac{d t}{t^{2} + 1} = {[\arctan t]}_{0}^{1} = \frac{π}{4}

Answered by john santu last updated on 30/Dec/19

= \frac{1}{\frac{1}{2} - 0} \int \underset{0}{\overset{\frac{1}{2}}{}} \frac{d x}{1 + 4 x^{2}}

= 2 \int \underset{0}{\overset{\frac{1}{2}}{}} \frac{d x}{1 + 4 x^{2}} = \int \underset{0}{\overset{\frac{1}{2}}{}} \frac{d (2 x)}{1 + {(2 x)}^{2}}

= a r c \tan (2 x) ∣ \underset{0}{\overset{\frac{1}{2}}{}} = \frac{π}{4}

Commented by Rio Michael last updated on 30/Dec/19

thanks