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Sagot :
The average value have of the function h on the given interval is [tex]\frac{1}{2} \ln^2 (2)[/tex].
How to find the Average Value of function ?
The given function is [tex]h (u) = \frac{\ln (u)}{u}[/tex] on the interval (1, 5).
The average value of function (a, b) is found by using the formula:
[tex]\frac{1}{(b -a)} \displaystyle \int^{b}_{a} f(x) dx[/tex]
Here the function is [tex]h (u) = \frac{\ln (u)}{u}[/tex] and a = 1, b = 5
Now,
[tex]\dfrac{1}{(5 - 1)} \displaystyle \int^{5}_{1} \frac{\ln (u)}{u}\ du[/tex]
[tex]= \dfrac{1}{4} \displaystyle \int^{5}_{1} \frac{\ln (u)}{u}\ du[/tex]
Substitute [tex]\ln (u) = t, dt = \frac{1}{u}\ du[/tex]
[tex]\dfrac{1}{4} \displaystyle \int^{4}_{1} \frac{\ln (u)}{u} du = \frac{1}{4} \int^{4}_{1} t (dt)[/tex]
[tex]= \frac{1}{4} \left [\frac{t^2}{2} \right ]^{4}_{1}[/tex]
[tex]= \frac{1}{8} \left [\{\ln (u)\}^2 \right ]^{4}_{1}[/tex]
[tex]= \frac{1}{8} [\{\ln (4) \}^2 - \{\ln (1) \}^2 ][/tex]
[tex]= \frac{1}{8} [\{\ln (4)\}^2 - 0][/tex]
[tex]= \frac{1}{8} \ln^2 (4)[/tex]
[tex]= \frac{2}{4} [2 \ln (2)]^2[/tex]
[tex]= \frac{1}{2} \ln^2 (2)[/tex]
Thus from the above conclusion we can say that The average value have of the function h on the given interval is [tex]\frac{1}{2} \ln^2 (2)[/tex].
Learn more about the Integrational Average value please click here : https://brainly.com/question/27419605
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