You can simplify this considerably. The field is $$(x^2,x+y)=(x^2,y)+(0,x)$$ Note that the first component is conservative, so its line integral over a closed path is $0$. Therefore we are left with evaluating $$\int_C{xdy}$$ where $C$ is the right triangle. Now if you know Green's Theorem, then: By Green's Theorem, this will just be the area of the triangle. So the answer is $${1\over 2}\cdot4\cdot10=20$$
Otherwise, you should not have too much trouble computing that line integral along the triangle. Note that you can do one more simplification: it will be zero along the first edge, because the field $(0,x)$ is orthogonal to the line $y=0$.
I've solved the two integral by two distinct methods,
$$\int_{C_1}=\int_{0}^2(9x)dx+\int_0^2x^2d(9x)=\int_0^29x+x^2\cdot9dx=42.$$
$$\int_{C_2}=\int_{2}^5(-x^2+8x+6)dx+\int_2^5x^2(-2x+8)dx$$
$$=\int_2^{5}7x^2+8x+6-2x^3dx=383-\frac{625}{2}.$$
Putting $\int_C Fdr=\int_{C_{1}}Fdr+\int_{C_{2}}Fdr$ for $F(x,y)=(y,x^2)$, then $F(x(t),y(t))=(y(t),(x(t))^2)$.
Thus on $C_1$ we have: $F(t)=F(x(t),y(t))=(9t,(t)^2)$ since $r(t)=(x(t),y(t))=(t,9t)$.
And on $C_2$ we have: $F(t)=F(x(t),y(t))=(-t^2+8t+6,(t)^2)$ since $r(t)=(x(t),y(t))=(t,-t^2+8t+6)$.
And limits in integrals are the same because $x(t)$ is equals to $t$ in both cases. In your book you can find the following method and compute as follows since the functions and paths satisfy the requiered hypothesis:
$$\int_{C_{1}}Fdr=\int_{0}^2 F(x(t),y(t))\cdot r'(t)dt=\int_0^2(9t,t^2)\cdot (1,9)dt=\int_0^2 9t+9t^2dt=42.$$
$$\int_{C_{2}}Fdr=\int_{2}^5 F(x(t),y(t))\cdot r'(t)dt=\int_2^5(-t^2+8t+6,t^2)\cdot (1,-2t+8)dt=$$
$$=\int_2^5 -2t^3+7t^2+8t+6dt=\left[-\frac{2t^4}{4}+\frac{7t^3}{3}+\frac{8t^2}{2}+6t\right]_{t=2}^{t=5}=$$
$$=-\frac{625}{2}+\frac{875}{3}+100+30+8-\frac{56}{3}-16-12=-\frac{625}{2}+383.$$
Best Answer
You forgot to include the radius. The substitution $x= \cos t$, $y= \sin t$ parametrizes a circle of radius $1$.
But $C$ is a subset of a circle of radius $4$.