Understand the curves

We have:

• Parabola: $y^2 = x$ i.e. $x = y^2$
• V-shaped graph: $y = |x|$

Since area is easier to handle using intersections, let us rewrite $y = |x|$ in parts:

• For $x \ge 0$, $y = x$
• For $x < 0$, $y = -x$

Also note that $y = |x| \ge 0$, so only nonnegative $y$ values matter.

Find intersection points

Since on the parabola $x = y^2 \ge 0$, we only need the branch $y = x$ of $y = |x|$.

So solve: $y = x, \quad x = y^2$

Substitute $x=y$ into $x=y^2$: $y = y^2$ $y(y-1)=0$

Hence: $y=0 \quad \text{or} \quad y=1$

Therefore intersection points are:

• $(0,0)$
• $(1,1)$

Determine the enclosed region

Between $y=0$ and $y=1$:

• Line $y=x$ gives $x=y$
• Parabola gives $x=y^2$

Since for $0 \le y \le 1$, we have: $y^2 \le y$ the parabola lies to the left and the line lies to the right.

So the enclosed area is: $A = \int_0^1 \big(x_{\text{right}} - x_{\text{left}}\big)\,dy = \int_0^1 (y - y^2)\,dy$

Compute the integral

$A = \int_0^1 (y-y^2)\,dy$ $= \left[\frac{y^2}{2} - \frac{y^3}{3}\right]_0^1$ $= \frac{1}{2} - \frac{1}{3}$ $= \frac{1}{6}$

Check options

• A: $\frac{1}{6}$ ✅
• B: $\frac{1}{3}$
• C: $\frac{2}{3}$
• D: $1$

Therefore the correct option is A.