Fundamental Theorem for Line Integrals — Question 5

Question 5 – Solution

Strategy. First use the endpoints, then verify that the parameter integrand is the derivative of $f(\mathbf r(t))$.

Step 1: Endpoint method The curve begins at $(1,0)$ and ends at $(0,1)$. Both endpoint values are zero: $$f(1,0)=0, \qquad f(0,1)=0.$$ Therefore the theorem gives $\boxed{0}$.

Step 2: Direct method Since $\nabla f=\langle 2xy,x^2\rangle$, $$\begin{align*} \nabla f(\mathbf r(t))\cdot\mathbf r'(t) &=\langle 2\cos t\sin t,\cos^2t\rangle \cdot\langle-\sin t,\cos t\rangle\\ &=-2\cos t\sin^2t+\cos^3t. \end{align*}$$ This is $$\frac{d}{dt}\big(\cos^2t\sin t\big) =\frac{d}{dt}f(\mathbf r(t)).$$ Hence $$\int_0^{\pi/2}\frac{d}{dt}(\cos^2t\sin t)\,dt =\big[\cos^2t\sin t\big]_0^{\pi/2}=\boxed{0}.$$

Verification The two methods agree, and the displayed antiderivative explicitly exhibits the chain rule behind the theorem.