Calcule a integral definida: ∫01 [t3i + 7j + (t + 1)k]dt.

(t^4/4 i + 7t j + (t² + t)k)|10 = 1/4 i + 7j + 2k

Para calcularmos a integral da função dada devemos realizar os seguintes cálculos abaixo:

\(\begin{align} & I=\int_{{}}^{{}}{{{t}^{3}}i+7j+(t+1)k}dt \\ & \int_{{}}^{{}}{{{t}^{3}}i+7j+(t+1)k}dt=\int_{{}}^{{}}{{{t}^{3}}i+\int_{{}}^{{}}{7j}+\int_{{}}^{{}}{(t+1)k}} \\ & \int_{{}}^{{}}{{{t}^{3}}i+7j+(t+1)k}dt=\frac{{{t}^{3+1}}}{3+1}+\frac{7{{t}^{0+1}}}{0+1}+\frac{{{t}^{1+1}}}{1+1}+\frac{1{{t}^{0+1}}}{0+1} \\ & \int_{{}}^{{}}{{{t}^{3}}i+7j+(t+1)k}dt=\frac{{{t}^{4}}}{4}+7t+\frac{{{t}^{2}}}{2}+t \\ \end{align} \)

Portanto, a integral da função dada será \(\begin{align} & \int_{{}}^{{}}{{{t}^{3}}i+7j+(t+1)k}dt=\frac{{{t}^{4}}}{4}+7t+\frac{{{t}^{2}}}{2}+t \\ \end{align} \).