Question

Find the average value of the function f(x)=e^2x+1/e^x on the interval [0,ln⁡2].

Answer 1

Given function,

$\sf f(x) = {e}^{2x} + \dfrac{1}{ {e}^{x} }$

We have to find the average value of the function from [0, In(2)].

Average of a function can be calculated by :

$\star \: \boxed{ \boxed{ \displaystyle \sf \overline{f(x)} = \dfrac{1}{b - a} \int_{a}^{b} f(x)dx}}$

Now,

$\longrightarrow \displaystyle \sf \overline{f(x)} = \dfrac{1}{ ln(2) - 0} \int_{0}^{ ln(2) } \bigg( {e}^{2x} + \dfrac{1}{ {e}^{x} } \bigg)dx \\ \\ \longrightarrow \displaystyle \sf \overline{f(x)} = \dfrac{1}{ ln(2)} \bigg( \dfrac{ {e}^{2x} }{2} - \dfrac{1}{ {e}^{x} } \bigg) \bigg|_{0}^{ ln(2) } \\ \\ \longrightarrow \displaystyle \sf \overline{f(x)} = \dfrac{1}{ ln(2)} \bigg( \dfrac{ {e}^{2 ln(2) } }{2} - \dfrac{1}{ {e}^{ ln(2) } } - \dfrac{ {e}^{0} }{2} + \dfrac{1}{ {e}^{0} } \bigg) \\ \\ \longrightarrow \displaystyle \sf \overline{f(x)} = \dfrac{1}{ ln(2)} \{2 - \dfrac{1}{2} + 1 - 1 \} \\ \\ \longrightarrow \displaystyle \sf \boxed{ \boxed{ \sf\overline{f(x)} = \dfrac{3}{2ln(2)}}}$

The average value of the above function is 1.03

Answer 2

The average value will be 1.03

Go through the attachment for the explanation.