avaliando cal 3

\[\mathcal{L}(f(x))=\int_0^\infty {{e^{ - st}}f\left( t \right)dt}\]

Assim, temos, por exemplo:

\[\mathcal{L}(e^{kt})=\int_0^\infty {{e^{ - st}}e^{kt}dt}=\int_0^\infty {e^{(k-s)t}dt}=\mathop {\lim }\limits_{A \to \infty } \int_0^A {e^{(k-s)t}dt}=\mathop {\lim }\limits_{A \to \infty }(-\dfrac{e^{(k-s)A}}{k-s}-\dfrac{1}{k-s}) =\dfrac{1}{s-k}\]

Por ser uma operação que envolve integração, uma de suas propriedades é a linearidade. Sejam
\(f(t)\)
e
\(g(t)\)
duas funções, e
\(\alpha, \beta \in \mathbb R$. Pela propriedade da linearidade sabe-se que: <br><span class="ql-formula" data-value="\mathcal{L}(\alpha f(x)+\beta g(x))=\alpha \mathcal{L}(f(x))+\beta \mathcal{L}(g(x))"><span class="katex">\[\mathcal{L}(\alpha f(x)+\beta g(x))=\alpha \mathcal{L}(f(x))+\beta \mathcal{L}(g(x))\]</span></span><br> O seno hiperbólico de\)
kt
\(é definido como: <br><span class="ql-formula" data-value="\sinh(kt)=\dfrac{e^{kt}-e^{-kt}}{2}"><span class="katex">\[\sinh(kt)=\dfrac{e^{kt}-e^{-kt}}{2}\]</span></span><br> Como\)
\mathcal{L}(e^{kt})=\dfrac{1}{s-k}
\(, se\)
k>0
\(, logo temos que: <br><span class="ql-formula" data-value="\mathcal{L}(\sinh(kt))=\mathcal{L}(\dfrac{e^{kt}-e^{-kt}}{2})"><span class="katex">\[\mathcal{L}(\sinh(kt))=\mathcal{L}(\dfrac{e^{kt}-e^{-kt}}{2})\]</span></span><br> <br><span class="ql-formula" data-value="\mathcal{L}(\sinh(kt))=\dfrac{1}{2}\mathcal{L}(e^{kt})-\dfrac{1}{2}\mathcal{L}(e^{-kt})"><span class="katex">\[\mathcal{L}(\sinh(kt))=\dfrac{1}{2}\mathcal{L}(e^{kt})-\dfrac{1}{2}\mathcal{L}(e^{-kt})\]</span></span><br> <br><span class="ql-formula" data-value="\mathcal{L}(\sinh(kt))=\dfrac{1}{s-k}-\dfrac{1}{s+k}"><span class="katex">\[\mathcal{L}(\sinh(kt))=\dfrac{1}{s-k}-\dfrac{1}{s+k}\]</span></span><br> Assim:\)
\boxed{\mathcal{L}(\sinh(kt))=\dfrac{1}{s^2-k2}}
\(\)