This is from a past exam paper that I'm using for revision I've calculated the answer as $21.75$ but I'm not sure I've done it right.
The question is a follows:
Calculate the line integral
$\int F.dr$ where $ F = (x^3,yx)$ and $y=3x$ also $-1<x<2$.
My working so far is as follows :
$F= (x^3,3x^2) \,\, \text{and} \,\, dr = (dx,3dx)$
$\int_ {-1}^{2}x^3 dx+6\int_{-1}^{2}x^2dx$
and from that I get the value of $21.75$.
Another way of doing the integral is to explicitly parameterise the path, and turn everything into functions of the parameter. It's a handy method when dealing with more complex paths.
The line segment $y=3x, -1 \le x \le 2$, i.e. between $(-1,-3)$ and $(2,6)$ can be parameterised as $x=t,y=3t$ where $t:-1\rightarrow2$. Then
$$\begin{array}{rcl}\displaystyle\int_C \vec F\cdot d\vec r &=& \displaystyle\int_C x^3~\dfrac{dx}{dt}+yx~\dfrac{dy}{dt}~dt\\&=&\displaystyle\int_{-1}^2t^3(1)+3t(t)(3)~dt\\&=&\displaystyle\int_{-1}^2t^3+9t^2~dt\\&=&\left[\dfrac{t^4}{4}+3t^3\right]_{-1}^2\\&=&(4+24)-(\dfrac14-3)=\dfrac{123}4=30.75\end{array}$$
This differs from your answer because you had $yx~dy = 6x^2$ instead of $(3x)(x)(3) = 9x^2$.
