Calcule a integral das funções vetoriais.
\left(1\right. ponto) \int_{-\pi / 4}^{\pi / 4}\left(\operatorname{sen} t, 1+\cos t, \operatorname{sec} ^{2} t\right) d t.

Passo 1

O enunciado pede a integral definida do vetor
\displaystyle\int_{-\pi/4}^{\pi/4}(\sin t,\;1+\cos t,\;\operatorname{sec}^2 t)\,dt.
Para funções vetoriais usamos a propriedade de que
\displaystyle\int (f,g,h)\,dt=(\!\int f\,dt,\;\int g\,dt,\;\int h\,dt).
Vamos, portanto, integrar cada componente de -\pi/4 a \pi/4.

Passo 2

Primeira componente:
\int_{-\pi/4}^{\pi/4}\sin t\,dt =\bigl[-\cos t\bigr]_{-\pi/4}^{\pi/4} =-\cos\bigl(\tfrac{\pi}{4}\bigr)-\bigl(-\cos\bigl(-\tfrac{\pi}{4}\bigr)\bigr) =-\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2} =0.

Passo 3

Segunda componente:
\int(1+\cos t)\,dt =\int1\,dt+\int\cos t\,dt =t+\sin t.
Avaliaremos de -\pi/4 a \pi/4:
\bigl(t+\sin t\bigr)\Big|_{-\pi/4}^{\pi/4} =\Bigl(\tfrac{\pi}{4}+\tfrac{\sqrt2}{2}\Bigr) -\Bigl(-\tfrac{\pi}{4}-\tfrac{\sqrt2}{2}\Bigr) =\tfrac{\pi}{2}+\sqrt2.

Passo 4

Terceira componente:
\int_{-\pi/4}^{\pi/4}\operatorname{sec}^2 t\,dt =\bigl[\tan t\bigr]_{-\pi/4}^{\pi/4} =\tan\bigl(\tfrac{\pi}{4}\bigr)-\tan\bigl(-\tfrac{\pi}{4}\bigr) =1-(-1) =2.

Resposta

\int_{-\pi/4}^{\pi/4}(\sin t,\;1+\cos t,\;\operatorname{sec}^2 t)\,dt =(0,\;\tfrac{\pi}{2}+\sqrt2,\;2).

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