Question

Evaluate the integral.

\(\int_1^2 {\left( {{t^2}i + t\sqrt {t - 1} j + tsin\pi tk} \right)} dt\)

Short answer

Therefore, the solution is \(\int_1^2 {\left( {{t^2}{\bf{i}} + t\sqrt {t - 1} {\bf{j}} + t\sin \pi t{\bf{k}}} \right)} dt = \frac{7}{3}{\bf{i}} + \frac{{16}}{{15}}{\bf{j}} - \frac{3}{\pi }{\bf{k}}\)

Step-by-step solution

Step 1: Given information

The given value is:

\(\int_1^2 {\left( {{t^2}i + t\sqrt {t - 1} j + tsin\pi tk} \right)} dt\)

For clarity we can integrate the components separately and reassemble everything at the end.

First the\({\bf{i}}:\)

\(\int_1^2 {{t^2}} dt = \left( {\frac{1}{3}{t^3}} \right)_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\)

Step 2: Parts integration is now required for the j component

\(u = t\;\;\;du = dt\)

\(dv = \sqrt {t - 1} dt\;\;\;v = \frac{2}{3}{(t - 1)^{3/2}}\)

\(\int_1^2 t \sqrt {t - 1} dt = \left( {t \cdot \frac{2}{3}{{(t - 1)}^{3/2}}} \right)_1^2 - \int_1^2 {\frac{2}{3}} {(t - 1)^{3/2}}dt\)

\( = \left( {\frac{4}{3} - 0} \right) - \left( {\frac{2}{3} \cdot \frac{2}{5}{{(t - 1)}^{5/2}}} \right)_1^2\)

\( = \frac{4}{3} - \frac{4}{{15}}(1 - 0)\)

\( = \frac{{20}}{{15}} - \frac{4}{{15}}\)

\( = \frac{{16}}{{15}}\)

Parts will also need to integrate the k component

\(u = t\;\;\;du = dt\)

\(dv = \sin \pi tdt\;\;\;v = - \frac{1}{\pi }\cos \pi t\)

\(\int_1^2 t \sin \pi tdt = \left( {t \cdot - \frac{1}{\pi }\cos \pi t} \right)_1^2 - \int_1^2 - \frac{1}{\pi }\cos \pi tdt\)

\( = \left( { - \frac{2}{\pi }\cos (2\pi ) - \left( { - \frac{1}{\pi }\cos \pi } \right)} \right) + \frac{1}{\pi }\left( {\frac{1}{\pi }\sin \pi t} \right)_1^2\)

\( = \left( { - \frac{2}{\pi }(1) - \left( { - \frac{1}{\pi }( - 1)} \right)} \right) + \frac{1}{\pi }(0 - 0)\)

\( = \left( { - \frac{2}{\pi } - \frac{1}{\pi }} \right) + 0\)

\( = - \frac{3}{\pi }\)

Reassemble:

\(\int_1^2 {\left( {{t^2}{\bf{i}} + t\sqrt {t - 1} {\bf{j}} + t\sin \pi t{\bf{k}}} \right)} dt = \frac{7}{3}{\bf{i}} + \frac{{16}}{{15}}{\bf{j}} - \frac{3}{\pi }{\bf{k}}\)