To graph this, we are dealing with a curve, axes, and a vertical line. Let's break it down:

1. Equation of the curve:

The given equation is: $y = x^2 + 1$ This is a parabola that opens upwards with its vertex shifted one unit upwards (because of the "+1" term). The vertex is at (0,1).

2. Boundaries:

• The region is bounded by the x-axis ($y = 0$) and the y-axis ($x = 0$).
• There's also a boundary at the vertical line x = 2.

3. Graphing steps:

1. Plot the curve $y = x^2 + 1$:

   • At $x = 0$, $y = 1$.
   • At $x = 1$, $y = 2$ (since $1^2 + 1 = 2$).
   • At $x = 2$, $y = 5$ (since $2^2 + 1 = 5$).

2. Plot the axes:

   • The x-axis is the horizontal line where $y = 0$.
   • The y-axis is the vertical line where $x = 0$.

3. Draw the line $x = 2$:

   • This is a vertical line passing through $x = 2$ that intersects the curve at the point (2, 5).

4. Sketch the bounded region:

The region is bounded by:

• The curve $y = x^2 + 1$ from $x = 0$ to $x = 2$,
• The x-axis from $x = 0$ to $x = 2$,
• The y-axis at $x = 0$, and
• The vertical line $x = 2$.

The region in the first quadrant (QI) is enclosed by these elements.

Would you like me to generate a graph for this description or explain more details? Here are some questions to expand on this:

1. How can you find the area under a curve using integration?
2. What is the significance of the limits $x = 0$ and $x = 2$?
3. How do you set up an integral to calculate the area under a parabola?
4. What happens if the line $x = 2$ is moved further right, say to $x = 3$?
5. How would the area change if the equation was $y = x^2$ (without the +1)?

Tip: To find the area under a curve, we often use definite integrals with the limits of the given region.