Line Integrals - Part I — Question 2

Question 2 – Solution

Strategy. Parametrize the graph by $\mathbf r(x)=\langle x,x^2\rangle$ so the given scalar function is already the parameter.

Step 1: Parameter and speed $$\mathbf r(x)=\langle x,x^2\rangle,\qquad 0\le x\le 1,$$ $$\mathbf r'(x)=\langle 1,2x\rangle,\qquad ds=\sqrt{1+4x^2}\,dx.$$

Step 2: Evaluate Let $u=1+4x^2$, so $du=8x\,dx$. Then $$\begin{align*} \int_Cx\,ds &=\int_0^1x\sqrt{1+4x^2}\,dx\\ &=\frac 18\int_1^5u^{1/2}\,du =\frac 1{12}\left[u^{3/2}\right]_1^5 =\boxed{\frac{5\sqrt 5-1}{12}}. \end{align*}$$

Verification Both $x$ and $ds$ are nonnegative, so the answer must be positive. Also $0\le x\le 1$ implies the integral is no larger than the curve length; numerically the result is about $0.848$, consistent with that bound.