Question

Qual a antiderivada mais geral de h( t ) = 2t - e^t?

Answer 1

$\displaystyle \underline{\text{Antiderivada ( integral ) de }}: \\\\ \text{h(x)} = 2\text t-\text e^{\displaystyle \text t} \\\\ \int {\text h(\text t)}\text{dt} =\int [\ 2\text t -\text e^{\displaystyle \text t}\ ]\text{dt} \\\\\\ \text{H(t)}=\int2\text {t dt } -\int \text e^{\displaystyle\text t}}\text{dt}\\\\\\ \text{H(t)}=2\int \text{t dt} -\text{e}^{\displaystyle \text t} \\\\\\ \text{H(t)} = 2.\frac{\text t^2}{2}-\text{e}^{\displaystyle \text t}+\text C$

$\huge\boxed{\ \text{H(t)}= \text t^2-\text e^{\text t}+\text C \ }\checkmark \\\\\\ \text{letra d}$

Answer 2

✅ Após resolver os cálculos, concluímos que a antiderivada - primitiva ou integral indefinida -  da referida função é:

$\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \int 2t - e^{t}\,dt= t^{2} - e^{t} + c\:\:\:}}\end{gathered} }$

Portanto, a opção correta é:

                  $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf Letra\:D\:\:\:}}\end{gathered} }$

Seja a função:

             $\Large\displaystyle\begin{gathered}\tt h(t) = 2t - e^{t}\end{gathered}$

Se estamos querendo calcular a antiderivada de "h(t)", então estamos querendo saber o seguinte:

            $\Large\displaystyle\begin{gathered}\tt \int 2t - e^{t}\,dt = \:?\end{gathered}$

Então, fazemos:

  $\Large\displaystyle\begin{gathered}\tt \int 2t - e^{t}\,dt = \int 2t\,dt - \int e^{t}\,dt\end{gathered}$

                                $\Large\displaystyle\begin{gathered}\tt = 2\cdot\int t\,dt - \int e^{t}\,dt\end{gathered}$

                                 $\Large\displaystyle\text{ \begin{gathered}\tt = 2\cdot\frac{t^{1 + 1}}{1 + 1} - e^{t} + c\end{gathered} }$

                                 $\Large\displaystyle\begin{gathered}\tt = \frac{2}{2}t^{2} - e^{t} + c\end{gathered}$

                                 $\Large\displaystyle\begin{gathered}\tt = t^{2} - e^{t} + c\end{gathered}$

✅ Portanto, a antiderivada procurada é:

           $\Large\displaystyle\begin{gathered}\tt \int 2t - e^{t}\,dt= t^{2} - e^{t} + c\end{gathered}$    

 

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