To find the area enclosed between the curves y2 = x and y = |x|, we first need to understand the shapes of these curves and find their points of intersection.

Curve Analysis

• Curve 1: y2 = x
  This equation represents a parabola opening to the right along the positive x-axis. For any value of x > 0, there are two corresponding y-values: y = \(\sqrt{x}\) (the upper half) and y = -\(\sqrt{x}\) (the lower half). This curve exists only for x ≥ 0.
• Curve 2: y = |x|
  This equation represents the absolute value function. It forms a 'V' shape.
  • For x ≥ 0, y = |x| simplifies to y = x.
  • For x < 0, y = |x| simplifies to y = -x.
  Importantly, y = |x| implies that y ≥ 0 for all x.

Intersection Points Analysis

We need to find where the two curves intersect. Since y = |x| requires y ≥ 0, we only need to consider the part of the parabola y2 = x where y ≥ 0. This is the upper branch, y = \(\sqrt{x}\).

Furthermore, the equation y2 = x requires x ≥ 0. Therefore, any intersection must occur in the first quadrant (where both x ≥ 0 and y ≥ 0).

In the first quadrant (where x ≥ 0), the equation y = |x| becomes y = x.

So, we need to find the intersection points of y = \(\sqrt{x}\) and y = x.

Set the expressions for y equal:

\(\sqrt{x} = x\)

Square both sides:

x = x2

Rearrange the equation:

x2 - x = 0

Factor out x:

x(x - 1) = 0

This gives two possible values for x:

• x = 0. Substituting into y = x gives y = 0. Point: (0, 0).
• x = 1. Substituting into y = x gives y = 1. Point: (1, 1).

The curves intersect at (0, 0) and (1, 1).

Area Calculation Using Integration

The enclosed area lies between x = 0 and x = 1. In this interval, we need to determine which function is the upper boundary and which is the lower boundary.

Let's test a value between 0 and 1, for example, x = 1/4:

• For y = \(\sqrt{x}\), y = \(\sqrt{1/4}\) = 1/2.
• For y = x, y = 1/4.

Since 1/2 > 1/4, the curve y = \(\sqrt{x}\) is above the curve y = x in the interval (0, 1).

Integration with respect to x

The area (A) is the integral of the upper curve minus the lower curve, from x = 0 to x = 1:

A = \(\int_{0}^{1} (\text{upper curve} - \text{lower curve}) \, dx\)

A = \(\int_{0}^{1} (\sqrt{x} - x) \, dx\)

A = \(\int_{0}^{1} (x^{1/2} - x) \, dx\)

Now, we evaluate the integral:

A = \(\left[ \frac{x^{1/2 + 1}}{1/2 + 1} - \frac{x^{2}}{2} \right]_{0}^{1}\)

A = \(\left[ \frac{x^{3/2}}{3/2} - \frac{x^{2}}{2} \right]_{0}^{1}\)

A = \(\left[ \frac{2}{3} x^{3/2} - \frac{1}{2} x^{2} \right]_{0}^{1}\)

Evaluate at the limits:

A = \(\left( \frac{2}{3} (1)^{3/2} - \frac{1}{2} (1)^{2} \right) - \left( \frac{2}{3} (0)^{3/2} - \frac{1}{2} (0)^{2} \right)\)

A = \(\left( \frac{2}{3} - \frac{1}{2} \right) - (0)\)

A = \(\frac{4}{6} - \frac{3}{6} = \frac{1}{6}\)

Integration with respect to y

Alternatively, we can integrate with respect to y. We need to express x in terms of y.

• From y2 = x, we have x = y2.
• From y = |x|, since we are in the first quadrant where y ≥ 0, we have y = x, which means x = y.

The intersection points (0, 0) and (1, 1) mean y ranges from 0 to 1.

In the interval y ∈ (0, 1), we compare x = y and x = y2. For example, if y = 1/2, then x = 1/2 for the line and x = (1/2)2 = 1/4 for the parabola. Since 1/2 > 1/4, the line x = y is to the right of the parabola x = y2.

The area (A) is the integral of the right curve minus the left curve, from y = 0 to y = 1:

A = \(\int_{0}^{1} (\text{right curve} - \text{left curve}) \, dy\)

A = \(\int_{0}^{1} (y - y^{2}) \, dy\)

Evaluate the integral:

A = \(\left[ \frac{y^{2}}{2} - \frac{y^{3}}{3} \right]_{0}^{1}\)

A = \(\left( \frac{1^{2}}{2} - \frac{1^{3}}{3} \right) - \left( \frac{0^{2}}{2} - \frac{0^{3}}{3} \right)\)

A = \(\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}\)

Conclusion

Both methods of integration yield the same result. The area enclosed between the curves y2 = x and y = |x| is \(\frac{1}{6}\) square units.