Calculate the area bounded by the curves x = y² - 2y + 2

Calculate the area bounded by the curves $x = y^2 - 2y + 2$, $x = 0$, $y = 0$, and $2x + y = 9$.

We will locate the relevant intersection points first, then identify which curve bounds the region before integrating.

Step 1: Analyze the curves

Parabola: $x = y^2 - 2y + 2 = (y - 1)^2 + 1$. This is a parabola opening to the right with vertex at $(1, 1)$.

Line: $2x + y = 9 \implies x = \frac{9 - y}{2}$

Boundaries: $x = 0$ (the $y$-axis) and $y = 0$ (the $x$-axis)

Step 2: Find key intersection points

Parabola with $y = 0$:

\[x = 0^2 - 2(0) + 2 = 2\]

Point: $(2, 0)$

Parabola with $x = 0$:

\[0 = y^2 - 2y + 2\]

Discriminant: $\Delta = 4 - 8 = -4 < 0$. No real intersection (the parabola never crosses the $y$-axis).

Line with $y = 0$:

\[x = \frac{9 - 0}{2} = \frac{9}{2}\]

Point: $\left(\frac{9}{2}, 0\right)$

Line with $x = 0$:

\[0 = \frac{9 - y}{2} \implies y = 9\]

Point: $(0, 9)$

Parabola with line:

\[y^2 - 2y + 2 = \frac{9 - y}{2}\]

\[2(y^2 - 2y + 2) = 9 - y\]

\[2y^2 - 4y + 4 = 9 - y\]

\[2y^2 - 3y - 5 = 0\]

Using the quadratic formula:

\[y = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm 7}{4}\]

\[y = \frac{10}{4} = \frac{5}{2} \quad \text{or} \quad y = \frac{-4}{4} = -1\]

Since we're bounded by $y = 0$, we take $y = \frac{5}{2}$.

When $y = \frac{5}{2}$:

\[x = \frac{9 - 5/2}{2} = \frac{13/2}{2} = \frac{13}{4}\]

Point: $\left(\frac{13}{4}, \frac{5}{2}\right)$

Step 3: Understand the region

The region is bounded by $y = 0$, $x = 0$, and, on the right, whichever of the parabola or the line is closer for a given $y$. Looking at the curves:

• The parabola starts at $(2, 0)$ on the $x$-axis
• The line goes from $(0, 9)$ to $\left(\frac{9}{2}, 0\right)$
• We have boundaries at $x = 0$ and $y = 0$

Step 4: Determine which curve bounds the region

Let's integrate with respect to $y$ from $y = 0$ upward, with $x = 0$ as the left boundary throughout.

For $y \in [0, \frac{5}{2}]$, we check which curve gives the smaller (binding) value of $x$: at $y = 0$, the parabola gives $x = 2$ while the line gives $x = \frac{9}{2}$, so the parabola is the nearer (right) boundary here, and it remains so up to $y = \frac{5}{2}$, where the two curves meet.

Step 5: Set up the integral

For $y \in [0, \frac{5}{2}]$, the right boundary is the parabola:

\[A = \int_0^{5/2} \left[(y^2 - 2y + 2) - 0\right] dy = \int_0^{5/2} (y^2 - 2y + 2) \, dy\]

Step 6: Evaluate the integral

\[A = \left[\frac{y^3}{3} - y^2 + 2y\right]_0^{5/2}\]

\[= \frac{125/8}{3} - \frac{25}{4} + 5 = \frac{125}{24} - \frac{150}{24} + \frac{120}{24} = \frac{95}{24}\]

\[\boxed{A = \frac{95}{24} \approx 3.958 \text{ square units}}\]

Summary

The area of the region bounded by the parabola $x = y^2 - 2y + 2$, the axes $x = 0$ and $y = 0$, and the line $2x + y = 9$, taken over the range where the parabola forms the binding boundary, is:

\[\boxed{A = \frac{95}{24} \approx 3.958 \text{ square units}}\]

Key steps:

1. Locate the region's key intersection points, including where the parabola meets the line at $\left(\frac{13}{4}, \frac{5}{2}\right)$
2. Confirm the parabola never crosses the $y$-axis (negative discriminant)
3. Identify the parabola as the binding right-hand boundary for $y \in [0, \frac{5}{2}]$
4. Set up the integral: $A = \int_0^{5/2} (y^2 - 2y + 2) \, dy$
5. Evaluate to get $A = \frac{95}{24}$

Key insight:

With several bounding curves given at once, the key work is geometric: deciding, range by range, which curve actually forms the edge of the region before any integral is written down.