Answer
$\dfrac{1}{4}i+7j+\dfrac{3}{2} k$
Work Step by Step
We need to integrate the given integral as follows:
$\int_0^1 [t^3i+7j +(t+1) k] dt= \dfrac{t^4}{4}i+7tj+(\frac{t^2}{t}+t) k]_0^1$
or,
$\dfrac{t^4}{4}i+7tj+(\frac{t^2}{t}+t) k]_0^1=[\dfrac{1}{4}i+7(1)j+(\frac{1}{1}+1) k]]-\dfrac{0}{4}i+7(0)j+(\frac{0}{0}+0) k]]$
Thus,
$[\dfrac{1}{4}i+7(1)j+(\frac{1}{1}+1) k]]-\dfrac{0}{4}i+7(0)j+(\frac{0}{0}+0) k]]=\dfrac{1}{4}i+7j+\dfrac{3}{2} k$
