Assinale a alternativa verdadeira
a) \int_{0}^{\infty} t e^{-3 t} \operatorname{sen} 2 t d t=\frac{2}{169}
b) \int_{0}^{\infty} t e^{-3 t} \operatorname{sen} 3 t d t=\frac{1}{36}
c) \int_{0}^{\infty} t e^{-3 t} \cos 2 t d t=\frac{12}{169}
d) \int_{0}^{\infty} t e^{-3 t} \cos 2 t d t=\frac{5}{13}
e) \int_{0}^{\infty} t e^{-3 t} \operatorname{sen} 2 t d t=\frac{12}{169}

Passo 1

O enunciado pede para identificar qual igualdade envolvendo \displaystyle \int_{0}^{\infty} t e^{-3t}\sin(2t)\,dt, \displaystyle \int_{0}^{\infty} t e^{-3t}\sin(3t)\,dt ou \displaystyle \int_{0}^{\infty} t e^{-3t}\cos(2t)\,dt
é verdadeira. Vamos usar transformada de Laplace para t\mapsto t\,f(t).

Passo 2

Recordando que \displaystyle L\{f\}(s)=\int_{0}^{\infty}e^{-s t}f(t)\,dt
e que
L\{t\,f(t)\}(s) = -\frac{d}{ds}\bigl(L\{f\}(s)\bigr),
para f(t)=\sin(\omega t) temos
L\{\sin(\omega t)\}(s)=\frac{\omega}{s^{2}+\omega^{2}},
logo
L\{t\sin(\omega t)\}(s) =-\frac{d}{ds}\!\Bigl(\frac{\omega}{s^{2}+\omega^{2}}\Bigr) =\frac{2\,\omega\,s}{\bigl(s^{2}+\omega^{2}\bigr)^{2}}.

Passo 3

No caso \omega=2 e s=3,
\int_{0}^{\infty}t\,e^{-3t}\sin(2t)\,dt =L\{t\sin(2t)\}(3) =\frac{2\cdot2\cdot3}{\bigl(3^{2}+2^{2}\bigr)^{2}} =\frac{12}{13^{2}} =\frac{12}{169}.

Passo 4

Comparando com as alternativas, vemos que \displaystyle\int_{0}^{\infty}t e^{-3t}\sin(2t)\,dt=\tfrac{12}{169}
corresponde à alternativa (e).

Resposta

(e)

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