Parameterize [tex]C[/tex] by
[tex]\mathbf r(t)=(x(t),y(t))=(1-t)(7,3)+t(0,0)=(7-7t,3-3t)[/tex]
where [tex]0\le t\le1[/tex]. Then
[tex]\mathrm d\mathbf r=(-7\,\mathbf i-3\,\mathbf j)\,\mathrm dt[/tex]
and the line integral is equivalent to
[tex]\displaystyle\int_C(4y^2\,\mathbf i+x\,\mathbf j)\cdot\mathrm d\mathbf r=\int_0^1(4(3-3t)^2\,\mathbf i+(7-7t)\,\mathbf j)\cdot(-7\,\mathbf i-3\,\mathbf j)\,\mathrm dt[/tex]
[tex]\displaystyle=\int_0^1(-252t^2+525t-273)\,\mathrm dt=-\dfrac{189}2[/tex]
