Question

Teorema do Valor Medio, alguem poderia me ajudar ?

Com t meses de experiência, um funcionário do correio é capaz de separar
Q(t) = 700 - 400e^(-0,5t) cartas por hora. Qual é a velocidade média com que um funcionário consegue separar as correspondências durante os 3 primeiros meses de trabalho?

Answer 1

Queremos calcular o valor médio da função velocidade $Q(t)\,,$ com $0\le t\le 3:$
$\overline{Q}_{t_1\to t_2}=\dfrac{1}{t_2-t_1}\displaystyle\int_{t_1}^{t_2}Q(t)\,dt\\\\\\\\ \overline{Q}_{0\to 3}=\dfrac{1}{3-0}\int_0^3 Q(t)\,dt\\\\\\ \boxed{\begin{array}{c} \overline{Q}_{0\to 3}=\dfrac{1}{3}\displaystyle\int_0^3 (700-400e^{-0,5t})\,dt \end{array}}~~~~~~\mathbf{(i)}$

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Famos encontrar uma primitiva para $Q(t):$

$\displaystyle\int Q(t)\,dt\\\\\\ =\int (700-400e^{-0,5t})\,dt\\\\\\ =\int \left(700-400\cdot \dfrac{0,5}{0,5}\,e^{-0,5t}\right)\,dt\\\\\\ =\int \left(700-\dfrac{400}{0,5}\cdot 0,5\,e^{-0,5t}\right)\,dt\\\\\\ =\int \left(700-800\cdot 0,5\,e^{-0,5t}\right)\,dt\\\\\\ =\int \left(700+800\cdot (-0,5)\,e^{-0,5t}\right)\,dt\\\\\\ =\int 700\,dt+800\int e^{-0,5t}(-0,5)\,dt$
$=\displaystyle 700 t+800\int e^{u}\,du~~~~~~(u=-0,5t)\\\\\\ =700 t+800e^{u}+C\\\\ =700 t+800e^{-0,5t}+C\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int Q(t)\,dt=700t+800e^{-0,5t}+C \end{array}}$

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Para calcular o valor médio de $Q$ dado por $\mathbf{(i)}\,,$ aplicamos o Teorema Fundamental do Cálculo usando a primitiva encontrada acima:

$\overline{Q}_{0\to 3}=\dfrac{1}{3}\displaystyle\int_0^3 (700-400e^{-0,5t})\,dt\\\\\\ =\dfrac{1}{3}\cdot \left.\left(700t+800e^{-0,5t} \right )\right|_0^3\\\\\\ =\dfrac{1}{3}\cdot\left[\left(700\cdot 3+800e^{-0,5\cdot 3} \right )-\left(700\cdot 0+800e^{-0,5\cdot 0} \right ) \right ]\\\\\\ =\dfrac{1}{3}\cdot\left[2\,100+800e^{-1,5}-800e^{0} \right ]\\\\\\ =\dfrac{1}{3}\cdot\left[2\,100-800+800e^{-1,5} \right ]\\\\\\ =\dfrac{1}{3}\cdot\left(1\,300+800e^{-1,5} \right)\approx \boxed{\begin{array}{c} 493\mathrm{~cartas/h}\end{array}}$