Answer
The area of the shaded region is $5/6$.
Work Step by Step
With respect to $y$, the formulas of the curves are
- Curve $x+y=2$: $x=2-y$
- Curve $y=x^2$: $x=\pm\sqrt y=\pm y^{1/2}$
However, from the graph, we can see that we work only with the positive values of $x$, so there is no need for the negative sign, meaning that $x=y^{1/2}$.
In this exercise, it might be easier if we carry out integration with respect to $y$ by the formula $$A=\int^d_c[f(y)-g(y)]dy$$ to calculate the area, in which $f$ denotes the right-hand curve and $g$ the left-hand curve.
Here, the shaded region is bounded on the right by the curve $x=2-y$ and on the left by the curve $x=y^{1/2}$ and runs from $y=0$ to $y=1$.
Therefore, the area of the shaded region is $$A=\int^{1}_{0}(2-y-y^{1/2})dy$$ $$A=\Big(2y-\frac{y^2}{2}-\frac{2y^{3/2}}{3}\Big)\Big]^{1}_{0}$$ $$A=2-\frac{1}{2}-\frac{2}{3}$$ $$A=\frac{5}{6}$$
