1. The problem is to define the region shaded on the graph which is bounded by the curve $y^2 = 4x$, the line $x=1$, and the x-axis ($y=0$). 2. First, understand the curve. The equation $y^2 = 4x$ describes a rightward-opening parabola with vertex at $(0,0)$. 3. The line $x=1$ is a vertical line cutting through the parabola. At $x=1$, the parabola has points where $y^2 = 4(1) = 4$, so $y = \pm 2$. 4. The x-axis is $y=0$, which is the boundary below the shaded region. 5. The shaded region is inside the parabola, above the x-axis, and to the left of the vertical line $x=1$. 6. To express this region analytically: - Since $y^2 = 4x$, then $x = \frac{y^2}{4}$. - The region lies between $x = \frac{y^2}{4}$ and $x=1$. - The vertical bounds for $y$ are from $0$ (x-axis) up to $2$ (top intersection with $x=1$). 7. Therefore, the region $R$ can be defined as: $$R = \{ (x,y) \mid \frac{y^2}{4} \leq x \leq 1,\ 0 \leq y \leq 2 \}$$ This describes all points between the parabola and the line $x=1$, above the $x$-axis and below $y=2$. Final answer: The region bounded by the curve $y^2 = 4x$, the line $x=1$, and the x-axis is $\boxed{\left\{(x,y) \mid \frac{y^2}{4} \leq x \leq 1,\ 0 \leq y \leq 2 \right\}}$.