the area of the region bounded by y^2 = 2x+1 and x-y = 1 is

We are tasked with finding the area of the region bounded by the curves \( y^2 = 2x + 1 \) and \( x - y = 1 \).

### Step 1: Express the second equation in terms of \(x\) and \(y\)
The second equation is:
\[
x - y = 1 \implies x = y + 1
\]
Thus, the second curve can be expressed as \(x = y + 1\).

### Step 2: Solve for the points of intersection
To find the limits of integration, we need to determine where the two curves intersect. Substituting \(x = y + 1\) into the first equation \( y^2 = 2x + 1 \), we get:
\[
y^2 = 2(y + 1) + 1
\]
Simplifying:
\[
y^2 = 2y + 2 + 1
\]
\[
y^2 = 2y + 3
\]
Rearrange this to form a quadratic equation:
\[
y^2 - 2y - 3 = 0
\]
Factoring:
\[
(y - 3)(y + 1) = 0
\]
So, the solutions are \( y = 3 \) and \( y = -1 \).

### Step 3: Set up the area integral
The region's area is bounded by the two curves from \( y = -1 \) to \( y = 3 \). The area can be found by integrating the difference between the curves \(x = \frac{y^2 - 1}{2}\) (from the first equation) and \(x = y + 1\) (from the second equation).

The area \(A\) is given by:
\[
A = \int_{y=-1}^{y=3} \left[ (y + 1) - \frac{y^2 - 1}{2} \right] dy
\]
Simplifying the integrand:
\[
A = \int_{y=-1}^{y=3} \left( y + 1 - \frac{y^2}{2} + \frac{1}{2} \right) dy
\]
\[
A = \int_{y=-1}^{y=3} \left( y + \frac{3}{2} - \frac{y^2}{2} \right) dy
\]

### Step 4: Compute the integral
Now, compute the integral term by term:
\[
A = \int_{y=-1}^{y=3} y \, dy + \int_{y=-1}^{y=3} \frac{3}{2} \, dy - \int_{y=-1}^{y=3} \frac{y^2}{2} \, dy
\]

1. \(\int_{y=-1}^{y=3} y \, dy = \left[ \frac{y^2}{2} \right]_{y=-1}^{y=3} = \frac{9}{2} - \frac{1}{2} = 4\)
2. \(\int_{y=-1}^{y=3} \frac{3}{2} \, dy = \frac{3}{2} \cdot (3 - (-1)) = \frac{3}{2} \cdot 4 = 6\)
3. \(\int_{y=-1}^{y=3} \frac{y^2}{2} \, dy = \left[ \frac{y^3}{6} \right]_{y=-1}^{y=3} = \frac{27}{6} - \frac{(-1)^3}{6} = \frac{27}{6} + \frac{1}{6} = \frac{28}{6} = \frac{14}{3}\)

### Step 5: Final calculation
Now subtract and combine all terms:
\[
A = 4 + 6 - \frac{14}{3}
\]
\[
A = 10 - \frac{14}{3} = \frac{30}{3} - \frac{14}{3} = \frac{16}{3}
\]

Thus, the area of the region is \( \frac{16}{3} \) square units.

---

Would you like more details on any of these steps or have any questions?

Here are 5 related questions:
1. How would the area change if the curves were translated?
2. Can we generalize this method for any parabolic region?
3. How does the choice of integration method (horizontal vs. vertical) affect the outcome?
4. What if the curves are rotated before calculating the area?
5. How can you visualize this region in a graph?

**Tip:** Double-check your limits of integration when solving area problems, as mistakes here can affect the entire solution!

Explore the step-by-step solution to finding the area of the region bounded by the curves y^2 = 2x + 1 and x - y = 1. This calculus problem involves integration techniques and quadratic equations, suitable for students in Grades 11-12.

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