Sketch the region enclosed by the given curves. Decide whether to integrate with respect to a or y. Then find the area of the region. y=7+ √x,y=7+x

To find the area of the region enclosed by the curves \( y = 7 + \sqrt{x} \) and \( y = 7 + x \), we will follow these steps:

### Step 1: Sketch the Region
- The first curve is \( y = 7 + \sqrt{x} \), a transformed square root function that shifts the graph of \( \sqrt{x} \) upwards by 7 units.
- The second curve is \( y = 7 + x \), a straight line with slope 1 that intersects the y-axis at \( y = 7 \).

The region enclosed by these curves will be bounded by these two functions for specific values of \( x \), starting where the curves intersect.

### Step 2: Find Points of Intersection
To find where the two curves intersect, set the equations equal to each other:
\[
7 + \sqrt{x} = 7 + x
\]
Simplifying:
\[
\sqrt{x} = x
\]
Squaring both sides:
\[
x = x^2
\]
Rearranging the equation:
\[
x^2 - x = 0
\]
Factoring:
\[
x(x - 1) = 0
\]
So, \( x = 0 \) or \( x = 1 \).

Thus, the curves intersect at \( x = 0 \) and \( x = 1 \).

### Step 3: Decide Whether to Integrate with Respect to \( x \) or \( y \)
Since both curves are expressed as functions of \( x \) (i.e., \( y \) is in terms of \( x \)), it's easiest to integrate with respect to \( x \).

### Step 4: Set Up the Integral
To find the area between the curves, we subtract the lower function \( y = 7 + \sqrt{x} \) from the upper function \( y = 7 + x \) and integrate over the interval from \( x = 0 \) to \( x = 1 \):

\[
\text{Area} = \int_0^1 \left( (7 + x) - (7 + \sqrt{x}) \right) dx
\]
Simplifying the integrand:
\[
\text{Area} = \int_0^1 (x - \sqrt{x}) dx
\]

### Step 5: Compute the Integral
Now, integrate each term:
\[
\int_0^1 x \, dx - \int_0^1 \sqrt{x} \, dx
\]

The first integral is straightforward:
\[
\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}
\]

The second integral can be rewritten as:
\[
\int_0^1 \sqrt{x} \, dx = \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3}
\]

Thus, the area is:
\[
\text{Area} = \frac{1}{2} - \frac{2}{3}
\]

Now, finding the common denominator:
\[
\text{Area} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6}
\]

Since the area cannot be negative, we take the absolute value:
\[
\text{Area} = \frac{1}{6}
\]

### Final Answer:
The area of the region enclosed by the curves is \( \frac{1}{6} \) square units.

Would you like a detailed sketch of the region, or do you have any questions? 

Here are 5 related questions:
1. What is the general method to find the area between two curves?
2. How would the solution change if the curves were inverted?
3. What happens if the curves intersect at more than two points?
4. How do we decide whether to integrate with respect to \( x \) or \( y \)?
5. How can we apply this method to 3D surfaces?

**Tip:** Always check the limits of integration and which function is on top to avoid negative areas!

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=7+√x, y=7+x

This problem covers how to find the area between the curves y = 7 + √x and y = 7 + x. It involves solving for the intersection points, setting up the integral, and computing the area. The detailed solution applies the Fundamental Theorem of Calculus, and it's ideal for students in Grades 11-12.

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