Question

De acuerdo al método de integración por cambio de variable. Resuelve la siguiente integral indefinida​

Answer 1

Recordemos que:

                                         $\begin{array}{c}\red{\sf{\dfrac{d(X(t))}{dt}=V(t)}}\\\\\Rightarrow\quad \boxed{\sf{X(t) =\displaystyle \int V(t) \ dt}}\end{array}$

Siendo X(t) la posición y V(t) la velocidad enfunción del tiempo, entonces

                                      $\begin{array}{c}\\\sf{X(t) =\displaystyle \int V(t) \ dt}\\\\\sf{X(t) =\displaystyle \int 60 - e^{-t/10} \ dt}\\\\\sf{X(t) =\displaystyle \int 60\ dt - \displaystyle \int e^{-t/10} \ dt}\\\end{array}$

El cambio de variable que realizaremos será u = -t/10

         $\sf{u = -t/10 \qquad \Rightarrow\qquad du = -1/10 dt \qquad\Rightarrow\qquad \boldsymbol{\sf{dt = -10du}}}$

Reemplazamos en nuestra integral

                                  $\begin{array}{c}\sf{X(t) =\displaystyle \int 60\ dt - \displaystyle \int e^{\boldsymbol{\sf{\green{-t/10}}}} \ dt}\\\\\sf{X(t) =\displaystyle \int 60\ dt - \displaystyle \int e^{u} \ \boldsymbol{\sf{dt}}}\\\\\sf{X(t) =\displaystyle \int 60\ dt - \displaystyle \int e^{u} \ (-10du)}\\\\\sf{X(t) =\displaystyle \int 60\ dt +10 \displaystyle \int e^{u} \ du}\\\\\sf{Conocemos\ que: \int e^x \ dx = e^x, entonces }\\\\\sf{X(t) =60t +10e^{u}}\\\\\boxed{\boxed{\blue{\sf{X(t) =60t +10e^{-t/10}}}}}\end{array}$

Nos pide la distancia  en t = 6 h

                                            $\begin{array}{c}\sf{X(t) =60t +10e^{-t/10}\\\\\sf{X(6) =60(6) +10e^{-(6)/10}\\\\\boxed{\sf{X(6) =365.488 \ millas}}\end{array}$

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