Calculation of a line-integral

Question

How do I calculate $\int_L (y+x)dl,$ where $$ L: \begin{cases}x = 1+t+\frac{2}{3}t^\frac{3}{2}\\y= t-\frac{2}{3}t^\frac{3}{2}-1\\ 0 ≤ t ≤ 3 \end{cases}$$

Does anyone know how to do this?

Answer 1

As mentioned in the comments, the value $dl$ is nothing but $\sqrt{(dx(t))^2+(dy(t))^2}dt$. Comes intuitively from the Pythagorean theorem as we are approximating the curve $L$ with infinitely small tangent vectors(and they have infinitely small components $dx$ and $dy$), just as we would split the $x$-axis into very small pieces with a normal Riemann integral. In fact a normal integral is nothing but a line-integral along the path $y=0$. And for $x$ and $y$ we just plug in their parametrizations and integrate along the curve. So calculating: $$\int_{L}(y+x)dl$$ $$=\int_{0}^{3}(t-\frac{2}{3}t^{\frac{3}{2}}-1+1+t+\frac{2}{3}t^{\frac{3}{2}})\sqrt{(\frac{d}{dt}(1+t+\frac{2}{3}t^{\frac{3}{2}}))^2+(\frac{d}{dt}(t-\frac{2}{3}t^{\frac{3}{2}}-1))^2}dt$$ $$=\int_{0}^{3}2t\sqrt{2t+2}dt$$ Substituting $u=2t+2$ gives the following: $$\int_{0}^{3}2t\sqrt{2t+2}dt=\frac{1}{2}\int_{2}^{8}(u-2)\sqrt{u}du=\frac{1}{2}\int_{2}^{8}u^{\frac{3}{2}}du-\int_{2}^{8}\sqrt{u}du$$ These should be easy enough integrals to do. Final answer is: $$\int_{L}(y+x)=\frac{232\sqrt{2}}{15}$$