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Hola Yani!
La integral de una función vectorial es la integral de cada componente:
1)
Te integro cada componente por separado.
La segunda con un cambio de variable t-1=u
Y la tercera por partes:
$$\begin{align}&\int_1^2t^2i= \frac{t^3}3 \Bigg|_1^2i=( \frac 8 3- \frac 1 3)i= \frac 7 3 i\\&\\&\int_1^2t \sqrt {t-1}\ j \ dt=\\&\\&t-1=u\ ==>t=u+1\\&dt=du\\&t=1 ==>u=0\\&t=2==> u=1\\&\\&= \int_0^1(u+1) u^\frac 1 2du \ j= \int_0^1 u^\frac 3 2 + u^\frac 12 )du\ j=\\&\\&\frac{u^\frac 5 2}{\frac 5 2} + \frac{u^\frac 3 2}{\frac 3 2}\Bigg|_0^1j=( \frac 2 5+ \frac 2 3)j=\frac{16}{15}j\\&\\&\\&\int_1^2 t \sin \pi t dt \ k=\\&\\&t=u==>u'=1\\&\sin \pi t =v'==> v= \frac{-\cos \pi t}{\pi}\\&\\&=\frac{-tcos \pi t}{\pi}+ \frac 1 \pi \int \cos \pi t=\\&\\&\frac{-tcos \pi t}{\pi} + \frac 1 {\pi^2} \sin \pi t \Bigg |_1^2 k=\\&\\&- \frac 2 \pi +0 - (- \frac {-1}\pi+0)=- \frac 3 \pi k\\&\\&\end{align}$$
2.-
$$\begin{align}&\int(e^t i + 2t j + ln t \ k)dt=\\&\\&e^t\ i+ t^2 j+(tlnt\ -t)k\end{align}$$
Saludos
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