A integral definida da função vetorial r(t) = (t² - 1)i + (2t +1)j + (t³)k para t pertencente ao intervalo [0,2] é:

a integral fica assim:

\(integral (t^2-1)i+(2t+1)j+(t^3)k=> (t^3/3-t)i+(t^2+t)j+(t^4/4)k=> [(2^3/3-2)-(0^3/3-0)]i+[(2^2+2)-(0)]j+[(2^4/2)-(0)]k=>> (2/3i+6j+8k) = (2/3,6,8) \)

\(\Longrightarrow \int \limits_0^2 r(t) \, dt = \int \limits_0^2 (t^2 - 1) \, dt \, ( \mbox{i} ) + \int \limits_0^2 (2t+1) \, dt \, ( \mbox{j} ) + \int \limits_0^2 (t^3) \, dt \, ( \mbox{k} ) \)

\(\Longrightarrow \int \limits_0^2 r(t) \, dt = ({1 \over 3}t^3 - t) \bigg |_0^2 ( \mbox{i} ) + (t^2+t) \bigg |_0^2 \, ( \mbox{j} ) + ({1 \over 4}t^4) \bigg |_0^2 \, ( \mbox{k} ) \)

\(\Longrightarrow \int \limits_0^2 r(t) \, dt = \Big [ ({1 \over 3}\cdot 2^3 - 2) - ({1 \over 3}\cdot 0^3 - 0) \Big ] ( \mbox{i} ) + \Big [ (2^2+2) - (0^2+0) \Big ] \, ( \mbox{j} ) + {1 \over 4}(2^4-0^4) \, ( \mbox{k} ) \)

\(\Longrightarrow \int \limits_0^2 r(t) \, dt = \Big [ ({8 \over 3} - 2) - 0 \Big ] ( \mbox{i} ) + \Big [ (4+2) - 0 \Big ] \, ( \mbox{j} ) + {1 \over 4}16 \, ( \mbox{k} ) \)

\(\Longrightarrow \fbox {$ \int \limits_0^2 r(t) \, dt = {2 \over 3} ( \mbox{i} ) +6 \, ( \mbox{j} ) + 4 \, ( \mbox{k} ) $}\)