I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 16.5 : Fundamental Theorem for Line Integrals
1. Evaluate \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\) where \(f\left( {x,y} \right) = {x^3}\left( {3 - {y^2}} \right) + 4y\) and \(C\) is given by \(\vec r\left( t \right) = \left\langle {3 - {t^2},5 - t} \right\rangle \) with \( - 2 \le t \le 3\).
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Start SolutionThere really isn’t all that much to do with this problem. We are integrating over a gradient vector field and so the integral is set up to use the Fundamental Theorem for Line Integrals.
To do that we’ll need the following two “points”.
\[\vec r\left( { - 2} \right) = \left\langle { - 1,7} \right\rangle \hspace{0.25in}\hspace{0.25in}\vec r\left( 3 \right) = \left\langle { - 6,2} \right\rangle \]Remember that we are thinking of these as the position vector representations of the points \(\left( { - 1,7} \right)\) and \(\left( { - 6,2} \right)\) respectively.
Show Step 2Now simply apply the Fundamental Theorem to evaluate the integral.
\[ \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}} = f\left( {\vec r\left( 3 \right)} \right) - f\left( {\vec r\left( { - 2} \right)} \right) = f\left( { - 6,2} \right) - f\left( { - 1,7} \right) = 224 - 74 = \require{bbox} \bbox[2pt,border:1px solid black]{{150}}\]