Question

Find the average value of \(f\left( x \right) = \frac{{\sqrt {{x^2} - 1} }}{x},1 \le x \le 7\)

Short answer

The average value of \(f\left( x \right) = \frac{{\sqrt {{x^2} - 1} }}{x},\) for the interval \(1 \le x \le 7\) is \(0.917\). Using the average formula, \({f_{avg}} = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right).dx} \).

Step-by-step solution

Applying the average formula,

\({f_{avg}} = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right).dx} \)over \(\left( {a,b} \right)\) internal.

Here, we have \(a = 1\) and \(b = 7\)

On substituting in the formula,

\({f_{avg}} = \frac{1}{{7 - 1}}\int\limits_1^7 {\frac{{\sqrt {{x^2} - 1} }}{x}.dx} \)

Using Substitution method,

Consider \(u = \sqrt {{x^2} - 1} \)

\(\begin{aligned}{l} \Rightarrow x &= \sqrt {{u^2} + 1} \\ \Rightarrow dx &= \frac{u}{{\sqrt {{u^2} + 1} }}du\end{aligned}\)

Changing the limits,

Upper limit when \(x = 7,u = \sqrt {48} \)

Lower limit when \(x = 1,u = 0\)

Applying to integral all the assumptions obtained in the above step.

\(\begin{aligned}{l}{f_{avg}} &= \frac{1}{6}\int\limits_0^{\sqrt {48} } {\frac{u}{{\sqrt {{u^2} + 1} }}.\frac{u}{{\sqrt {{u^2} + 1} }}} .du\\ &= \frac{1}{6}\int\limits_0^{\sqrt {48} } {\frac{{{u^2}}}{{{{\sqrt {\left( {{u^2} + 1} \right)} }^2}}}.du} \\ &= \frac{1}{6}\int\limits_0^{\sqrt {48} } {\frac{{{u^2}}}{{{u^2} + 1}}.du} \end{aligned}\)

Adding and subtracting \(1\) from the numerator,

\({f_{avg}} = \frac{1}{6}\int\limits_0^{\sqrt {48} } {\left( {\frac{{{u^2} + 1}}{{{u^2} + 1}} - \frac{1}{{{u^2} + 1}}} \right)} du\)

Simplifying further and applying limits.

\(\begin{aligned}{l}{f_{avg}} &= \frac{1}{6}\int\limits_0^{\sqrt {48} } {1.du - } \frac{1}{6}\int\limits_0^{\sqrt {48} } {\frac{1}{{{u^2} + 1}}.} du\\ &= \frac{1}{6}\left( u \right)_0^{\sqrt {48} } - \frac{1}{6}\left( {ta{n^{ - 1}}\,u} \right)_0^{\sqrt {48} }\\ &= \frac{1}{6}\left( {\sqrt {48} - 0} \right) - \frac{1}{6}\left( {ta{n^{ - 1}}\left( {\sqrt {48} } \right) - ta{n^{ - 1}}\left( 0 \right)} \right)\\ &= 1.154 - \frac{1}{6}\left( {1.427 - 0} \right)\\ &= 1.154 - 0.237\\{f_{avg}} &= 0.917\end{aligned}\)

Therefore, The average value of \(f\left( x \right) = \frac{{\sqrt {{x^2} - 1} }}{x},\) at \(1 \le x \le 7\) is \(0.917\)