Calculus III - Fundamental Theorem for Line Integrals

Show Mobile Notice Show All Notes Hide All Notes

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 16.5 : Fundamental Theorem for Line Integrals

Back to Problem List

2. Evaluate \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\) where \(f\left( {x,y} \right) = y{{\bf{e}}^{{x^{\,2}} - 1}} + 4x\sqrt y \) and \(C\) is given by \(\vec r\left( t \right) = \left\langle {1 - t,2{t^2} - 2t} \right\rangle \) with \(0 \le t \le 2\).

Show All Steps Hide All Steps

Start Solution

There 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”.

\[\vec r\left( 0 \right) = \left\langle {1,0} \right\rangle \hspace{0.25in}\hspace{0.25in}\vec r\left( 2 \right) = \left\langle { - 1,4} \right\rangle \]

Remember that we are thinking of these as the position vector representations of the points \(\left( {1,0} \right)\) and \(\left( { - 1,4} \right)\) respectively.

Show Step 2

Now simply apply the Fundamental Theorem to evaluate the integral.

\[ \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}} = f\left( {\vec r\left( 2 \right)} \right) - f\left( {\vec r\left( 0 \right)} \right) = f\left( { - 1,4} \right) - f\left( {1,0} \right) = - 4 - 0 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 4}}\]