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Section 16.3 : Line Integrals - Part II

4. Evaluate \( \displaystyle \int\limits_{C}{{1 + {x^3}\,dx}}\) where \(C\) is the right half of the circle of radius 2 with counter clockwise rotation followed by the line segment from \(\left( {0,2} \right)\) to \(\left( { - 3, - 4} \right)\). See the sketch below for the direction.

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To help with the problem let’s label each of the curves as follows,

Now let’s parameterize each of these curves.

\({C_1}:\,\,\vec r\left( t \right) = \left\langle {2\cos \left( t \right),2\sin \left( t \right)} \right\rangle \hspace{0.25in} - \frac{1}{2}\pi \le t \le \frac{1}{2}\pi \)
\({C_2}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,2} \right\rangle + t\left\langle { - 3, - 4} \right\rangle = \left\langle { - 3t,2 - 6t} \right\rangle \hspace{0.25in}0 \le t \le 1\)

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Now we need to compute the line integral for each of the curves.

\[\begin{align*}\int\limits_{{{C_1}}}{{1 + {x^3}\,dx}} & = \int_{{ - \,\,\frac{1}{2}\pi }}^{{\frac{1}{2}\pi }}{{\left[ {1 + {{\left( {2\cos \left( t \right)} \right)}^3}} \right]\left( { - 2\sin \left( t \right)} \right)\,dt}}\\ & = \int_{{ - \,\,\frac{1}{2}\pi }}^{{\frac{1}{2}\pi }}{{ - 2\sin \left( t \right) - 16{{\cos }^3}\left( t \right)\sin \left( t \right)\,dt}}\\ & = \left. {\left( {2\cos \left( t \right) + 4{{\cos }^4}\left( t \right)} \right)} \right|_{ - \,\frac{1}{2}\pi }^{\frac{1}{2}\pi } = \underline 0 \end{align*}\] \[\begin{align*}\int\limits_{{{C_2}}}{{1 + {x^3}\,dx}} & = \int_{0}^{1}{{\left[ {1 + {{\left( { - 3t} \right)}^3}} \right]\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{ - 3 + 81{t^3}\,dt}}\\ & = \left. {\left( { - 3t + \frac{{81}}{4}{t^4}} \right)} \right|_0^1 = \underline {\frac{{69}}{4}} \end{align*}\]

Don’t forget to correctly deal with the differentials when converting the line integral into a “standard” integral.

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Okay to finish this problem out all we need to do is add up the line integrals over these curves to get the full line integral.

\[ \displaystyle \int\limits_{C}{{1 + {x^3}\,dx}} = \left( 0 \right) + \left( {\frac{{69}}{4}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{69}}{4}}}\]

Note that we put parenthesis around the result of each individual line integral simply to illustrate where it came from and they aren’t needed in general of course.