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### Section 16.5 : Fundamental Theorem for Line Integrals

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 $$0 \le t \le 2$$.

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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.

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Now simply apply the Fundamental Theorem to evaluate the integral.

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