Evaluate {eq}\int_{0}^{\pi} \sin 2t i + \cos 4t j + t^2 k {/eq}.

Question:

Integration of Vector Function:

Consider the vector function {eq}I\left( x \right) = f\left( x \right){\bf{i}} + g\left( x \right){\bf{j}} + h\left( x \right){\bf{k}} {/eq}, its indefinite integral is presented as {eq}\int {\left[ {f\left( x \right){\bf{i}} + g\left( x \right){\bf{j}}} \right]} dx = \left[ {\int {f\left( x \right)dx} } \right]{\bf{i}} + \left[ {\int {g\left( x \right)dx} } \right]{\bf{j}} + \left[ {\int {h\left( x \right)dx} } \right]{\bf{k}} {/eq}. Similarly, for the same function, {eq}I\left( x \right) = f\left( x \right){\bf{i}} + g\left( x \right){\bf{j}} + h\left( x \right){\bf{k}} {/eq}, the definite integral is, {eq}\int\limits_a^b {\left[ {f\left( x \right){\bf{i}} + g\left( x \right){\bf{j}}} \right]} dx = \left[ {\int\limits_a^b {f\left( x \right)dx} } \right]{\bf{i}} + \left[ {\int\limits_a^b {g\left( x \right)dx} } \right]{\bf{j}} + \left[ {\int\limits_a^b {h\left( x \right)dx} } \right]{\bf{k}} {/eq}, it implies that the integration of vector functions is accomplished by independently integrating each element.

Answer and Explanation: 1