We are asked to find the area enclosed between the curves:
1. \( y^2 = x \)
2. \( y = |x| \)
Step 1: Understand the curves
Curve 1: \( y^2 = x \)
This is a sideways parabola that opens to the right.
Domain: \( x \geq 0 \)
Range: \( y \in (-\infty, \infty) \)
Curve 2: \( y = |x| \)
This is a V-shaped graph with vertex at (0, 0).
It splits into two parts:
\( y = x \) when \( x \geq 0 \)
\( y = -x \) when \( x < 0 \)
But since \( y^2 = x \) only makes sense for \( x \geq 0 \),
we will only consider the portion of \( y = |x| \) for \( x \geq 0 \). So within the region of interest:
\( y = x \), because we’re only dealing with \( x \geq 0 \)
Step 2: Find the points of intersection
We solve: \[ y^2 = x \quad \text{and} \quad y = x \] Substitute \( y = x \) into \( y^2 = x \): \[ x^2 = x \Rightarrow x(x - 1) = 0 \Rightarrow x = 0, 1 \] So intersection points are:
\( (0, 0) \)
\( (1, 1) \)
Step 3: Set up the integral
We find the area between the curves from \( x = 0 \) to \( x = 1 \):
Top curve: \( y = x \)
Bottom curve (from \( y^2 = x \)): \( y = \sqrt{x} \)
So the vertical distance between the curves is: \[ x - \sqrt{x} \] \[ \text{Area} = \int_{0}^{1} (x - \sqrt{x}) \, dx \] Step 4: Solve the integral \[ \int_{0}^{1} (x - \sqrt{x}) \, dx = \int_{0}^{1} x \, dx - \int_{0}^{1} x^{1/2} \, dx \] \[ = \left[ \frac{x^2}{2} \right]_0^1 - \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \left( \frac{1}{2} - \frac{2}{3} \right) \] \[ = \frac{3 - 4}{6} = -\frac{1}{6} \] Since area is positive: \( {\frac{1}{6}} \)