Question

Evaluate the integral.

\(\int_0^2 {\left( {ti - {t^3}j + 3{t^5}k} \right)} dt\)

Short answer

Therefore, the solution is \(\int_0^2 {\left( {t{\bf{i}} - {t^3}{\bf{j}} + 3{t^5}{\bf{k}}} \right)} dt = 2{\bf{i}} - 4{\bf{j}} + 32{\bf{k}}\)

Step-by-step solution

Step 1: Given information

The given value is:

\(\int_0^2 {\left( {ti - {t^3}j + 3{t^5}k} \right)} dt\)

Step 2: Evaluate the provided integral.

Let us evaluate the given integral component-wise as shown below,\(\int_0^2 {\left( {t{\bf{i}} - {t^3}{\bf{j}} + 3{t^5}{\bf{k}}} \right)} dt = \int_0^2 t \cdot dt \cdot {\bf{i}} - \int_0^2 {{t^3}} \cdot dt \cdot {\bf{j}} + 3\int_0^2 {{t^5}} \cdot dt \cdot {\bf{k}}\)

\( = \left( {\frac{{{t^2}}}{2}} \right)_0^2{\bf{i}} - \left( {\frac{{{t^4}}}{4}} \right)_0^2{\bf{j}} + 3\left( {\frac{{{t^6}}}{6}} \right)_0^2{\bf{k}}\)

\( = \frac{4}{2}{\bf{i}} - \frac{{16}}{4}{\bf{j}} + 3 \cdot \frac{{64}}{6}{\bf{k}}\)

\( = 2{\bf{i}} - 4{\bf{j}} + 32{\bf{k}}\)

\(\int_0^2 {\left( {t{\bf{i}} - {t^3}{\bf{j}} + 3{t^5}{\bf{k}}} \right)} dt = 2{\bf{i}} - 4{\bf{j}} + 32{\bf{k}}\)