Calculate the area bounded by the curves y² = 2x + 1

Calculate the area bounded by the curves $y^2 = 2x + 1$ and $y = x - 1$.

We will integrate with respect to $y$, since the parabola is oriented horizontally.

Step 1: Find intersection points

The curves intersect where both equations are satisfied:

\[y^2 = 2x + 1 \quad \text{and} \quad y = x - 1\]

From the second equation: $x = y + 1$. Substituting into the first:

\[y^2 = 2(y + 1) + 1 = 2y + 2 + 1 = 2y + 3\]

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

Factoring:

\[(y - 3)(y + 1) = 0\]

Therefore: $y = 3$ or $y = -1$

When $y = 3$: $x = 3 + 1 = 4$, so point is $(4, 3)$

When $y = -1$: $x = -1 + 1 = 0$, so point is $(0, -1)$

The curves intersect at $(0, -1)$ and $(4, 3)$.

Step 2: Express both curves as functions of $y$

Since we have two values of $y$, it's easier to integrate with respect to $y$.

From parabola: $y^2 = 2x + 1 \implies x = \frac{y^2 - 1}{2}$

From line: $y = x - 1 \implies x = y + 1$

Step 3: Determine which curve is to the right

For $y$ between $-1$ and $3$, let's check at $y = 0$:

Parabola: $x = \frac{0 - 1}{2} = -\frac{1}{2}$. Line: $x = 0 + 1 = 1$

So the line is to the right of the parabola. Let's verify at $y = 2$:

Parabola: $x = \frac{4 - 1}{2} = \frac{3}{2}$. Line: $x = 2 + 1 = 3$

Yes, the line $x = y + 1$ is to the right of the parabola $x = \frac{y^2 - 1}{2}$ for $y \in [-1, 3]$.

Step 4: Set up the integral

\[A = \int_{-1}^{3} \left[(y + 1) - \frac{y^2 - 1}{2}\right] dy\]

\[= \int_{-1}^{3} \left[y + 1 - \frac{y^2}{2} + \frac{1}{2}\right] dy\]

\[= \int_{-1}^{3} \left[y + \frac{3}{2} - \frac{y^2}{2}\right] dy\]

Step 5: Evaluate the integral

\[A = \left[\frac{y^2}{2} + \frac{3y}{2} - \frac{y^3}{6}\right]_{-1}^{3}\]

At $y = 3$:

\[\frac{9}{2} + \frac{9}{2} - \frac{27}{6} = \frac{9}{2} + \frac{9}{2} - \frac{9}{2} = \frac{9}{2}\]

At $y = -1$:

\[\frac{1}{2} + \frac{-3}{2} - \frac{-1}{6} = \frac{1}{2} - \frac{3}{2} + \frac{1}{6} = -1 + \frac{1}{6} = -\frac{5}{6}\]

Therefore:

\[A = \frac{9}{2} - \left(-\frac{5}{6}\right) = \frac{9}{2} + \frac{5}{6} = \frac{27}{6} + \frac{5}{6} = \frac{32}{6} = \frac{16}{3}\]

\[\boxed{A = \frac{16}{3} \text{ square units}}\]

Step 6: Numerical value

\[A = \frac{16}{3} \approx 5.333 \text{ square units}\]

Verification

Let's verify our calculation at $y = 3$:

\[\frac{9}{2} + \frac{9}{2} - \frac{27}{6} = 9 - \frac{27}{6} = \frac{54 - 27}{6} = \frac{27}{6} = \frac{9}{2} \quad \checkmark\]

And the final result:

\[\frac{9}{2} + \frac{5}{6} = \frac{27 + 5}{6} = \frac{32}{6} = \frac{16}{3} \quad \checkmark\]

Summary

The area bounded by the parabola $y^2 = 2x + 1$ and the line $y = x - 1$ is:

\[\boxed{A = \frac{16}{3} \approx 5.333 \text{ square units}}\]

Key steps:

1. Find intersections: $y^2 - 2y - 3 = 0 \implies y = -1, 3$
2. Points: $(0, -1)$ and $(4, 3)$
3. Express as functions of $y$: $x = y + 1$ (line) and $x = \frac{y^2 - 1}{2}$ (parabola)
4. Line is to the right: $A = \int_{-1}^{3} \left[(y+1) - \frac{y^2-1}{2}\right] dy$
5. Evaluate: $\frac{9}{2} - (-\frac{5}{6}) = \frac{16}{3}$

Key insight:

Integrating with respect to $y$ is more convenient here since the parabola is oriented horizontally (opens to the right).