Calculate integral: $$\int_D = (x^2 + y)dx + (x-y)dy$$ where $D$ is an arc of a parabola $x = y^2$ oriented in such way that it starts in $(1,1)$ and ends in $(1,-1)$.
My solution:
We have: $x=y^2$ for $y \in (-1,1)$. We can calculate derivatives: $dx=2y \ dy$
The we organize our integral in orientation that contrary to the orientation of the arc of a parabola. Therefore we have that: $$\int_D (x^2 + y)dx + (x-y)dy = - \int_{-1}^{1} ((y^2)^2 + y)2ydy + (y^2-y)dy = $$ $$= - \int_{-1}^{1} (2y^5 + 2y^2)dy + (y^2-y)dy = \int_{-1}^{1} (- 2y^5 - 3y^2 + y)dy =$$ $$= -2 \int_{-1}^{1} y^5 \ dy -3 \int_{-1}^{1} y^2 \ dy + \int_{-1}^{1} y dy =$$ $$= -2 \left[ \frac{y^6}{6} \right]_{-1}^{1} -3 \left[ \frac{y^3}{3} \right]_{-1}^{1} + \left[ \frac{y^2}{2} \right]_{-1}^{1} =$$ $$= -3 \left[ \frac{y^3}{3} \right]_{-1}^{1} = -3 \left( \frac{1}{3} + \frac{1}{3} \right) = -3 \cdot \frac{2}{3} = -2$$
Is my solution correct?
As is mentioned in the comment, the user of the original post made a few mistakes in the title and in the description of the question. The calculation of the user is correct; I gave a solution to the mistyped integral.
Your calculation shows that $$ \int_{D} {(x^2 + \color{red}{y})\,\mathrm{d}x + (x - y)\,\mathrm{d}y} = -2, $$ which is correct; however, your calculation says nothing about $$ \int_{D} {(x^2 + \color{red}{y^2})\,\mathrm{d}x + (x - y)\,\mathrm{d}y}. $$
For your reference, I give a brief solution to the integral shown in the title. $$ \begin{aligned} & \int_{D} {(x^2 + \color{red}{y^2})\,\mathrm{d}x + (x - y)\,\mathrm{d}y} \\ = {} & \int_{1}^{-1} {(((\color{blue}{y^2})^2 + \color{red}{y^2})\,\mathrm{d}(\color{blue}{y^2}) + (\color{blue}{y^2} - y)\,\mathrm{d}y)} \\ = {} & \int_{1}^{-1} {((y^4 + y^2)\cdot 2y\,\mathrm{d}y + (\color{blue}{y^2} - y)\,\mathrm{d}y)} \\ = {} & \int_{1}^{-1} {(2y^5 + 2y^3 + y^2 - y)\,\mathrm{d}y} \\ = {} & {-\int_{-1}^{1} {(2y^5 + 2y^3 + y^2 - y)\,\mathrm{d}y}} \\ = {} & {-\int_{-1}^{1} {(2y^5 + 2y^3 - y)\,\mathrm{d}y}} - \int_{-1}^{1} {y^2\,\mathrm{d}y} \\ = {} & 0 - 2\int_{0}^{1} {y^2\,\mathrm{d}y} \\ = {} & \frac{-2}{3}. \\ \end{aligned} $$
I have used the following two properties to simplify the calculation (in which $a$ is positive and $f$ is a continuous function on $[-a, a]$):
Suppose that $f(t) = -f(-t)$ for $t \in [-a, a]$. Then $$ \int_{-a}^{a} {f(t)\,\mathrm{d}t} = 0. $$
Suppose that $f(t) = f(-t)$ for $t \in [-a, a]$. Then $$ \int_{-a}^{a} {f(t)\,\mathrm{d}t} = 2\int_{0}^{a} {f(t)\,\mathrm{d}t}. $$