Answer
$(\dfrac{e-1}{2})i+(\dfrac{e-1}{e})j+k$
Work Step by Step
We need to integrate the given integral as follows:
$[\int_{0}^{1} t e^{t^2} dt]i+[\int_{0}^{1} e^{-t} dt]j+[\int_{0}^{1} dt]k=[(1/2) e^{t^2}|_{0}^{1}i+[-e^{-t}]_{0}^{1}j+[t]_{0}^{1}k$
or,
$[(1/2) e^{t^2}|_{0}^{1}i+[-e^{-t}]_{0}^{1}j+[t]_{0}^{1}k=[(1/2) (e-1)]i+[1-e^{-1}]j+[1-0]k$
Thus,
$[(1/2) (e-1)]i+[1-e^{-1}]j+[1-0]k=(\dfrac{e-1}{2})i+(\dfrac{e-1}{e})j+k$
