To find the area of the region in the first quadrant that is bounded by the curve $y = \sqrt{x}$, the line $y = x - 2$, and the x-axis, we can follow these steps:

1. Determine the points of intersection:

The region of interest lies between the curve $y = \sqrt{x}$ and the line $y = x - 2$. To find the points of intersection, set the two equations equal to each other:

$\sqrt{x} = x - 2$

Square both sides to eliminate the square root:

$x = (x - 2)^2$

Expand the right side:

$x = x^2 - 4x + 4$

Rearrange the equation:

$x^2 - 5x + 4 = 0$

2. Solve the quadratic equation:

To solve the quadratic equation $x^2 - 5x + 4 = 0$, use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -5$, and $c = 4$:

$x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}$

$x = \frac{5 \pm \sqrt{25 - 16}}{2}$

$x = \frac{5 \pm \sqrt{9}}{2}$

$x = \frac{5 \pm 3}{2}$

This gives two solutions:

$x = 4 \quad \text{and} \quad x = 1$

So, the points of intersection are at $x = 1$ and $x = 4$.

3. Set up the integral to find the area:

The area $A$ is the integral of the difference between the two functions from $x = 1$ to $x = 4$:

$A = \int_{1}^{4} \left[(x - 2) - \sqrt{x}\right] \, dx$

4. Evaluate the integral:

First, integrate the function $x - 2 - \sqrt{x}$:

$\int (x - 2) \, dx = \frac{x^2}{2} - 2x$

$\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3}x^{3/2}$

So, the area is:

$A = \left[\frac{x^2}{2} - 2x - \frac{2}{3}x^{3/2}\right]_{1}^{4}$

5. Calculate the definite integral:

Substitute $x = 4$ into the antiderivative:

$\left(\frac{4^2}{2} - 2(4) - \frac{2}{3}(4)^{3/2}\right) = \left(\frac{16}{2} - 8 - \frac{2}{3}(8)\right) = \left(8 - 8 - \frac{16}{3}\right) = -\frac{16}{3}$

Substitute $x = 1$ into the antiderivative:

$\left(\frac{1^2}{2} - 2(1) - \frac{2}{3}(1)^{3/2}\right) = \left(\frac{1}{2} - 2 - \frac{2}{3}\right) = \left(\frac{1}{2} - 2 - \frac{2}{3}\right) = -\frac{13}{6}$

Now, subtract the result at $x = 1$ from the result at $x = 4$:

$A = \left(-\frac{16}{3}\right) - \left(-\frac{13}{6}\right) = -\frac{16}{3} + \frac{13}{6} = -\frac{32}{6} + \frac{13}{6} = -\frac{19}{6}$

So, the area is $\frac{19}{6}$.

Thus, the area of the region is $\frac{19}{6}$ square units.

Would you like more details on any part of this solution?

8 Relative Questions:

1. How do you find the points of intersection for different functions?
2. What is the significance of integrating the difference between two curves?
3. Can you explain how to handle square roots when solving equations?
4. How do you interpret the geometric meaning of definite integrals?
5. What are some common mistakes when solving for areas between curves?
6. How can we use symmetry to simplify area calculations?
7. How do changes in the limits of integration affect the area?
8. What other methods can be used to find areas between curves?

Tip:

When solving for areas between curves, always check that your limits of integration correspond to the points where the curves intersect. This ensures the calculated area is accurate and within the correct bounds.