[tex]\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}[/tex]
The first line segment can be parameterized by [tex]\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle[/tex] with [tex]0\le t\le1[/tex]. Denote this first segment by [tex]C_1[/tex]. Then
[tex]\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt[/tex]
[tex]=108[/tex]
The second line segment ([tex]C_2[/tex]) can be described by [tex]\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle[/tex], again with [tex]0\le t\le1[/tex]. Then
[tex]\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt[/tex]
[tex]=-\dfrac{229}3[/tex]
Finally,
[tex]\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3[/tex]
