Using integration find the area of the region bounded by the curves y=root of 5-x² and y= lx-1l

To find the area of the region bounded by the curves \( y = \sqrt{5 - x^2} \) and \( y = |x - 1| \), we first need to understand the shapes of these curves and where they intersect. The curve \( y = \sqrt{5 - x^2} \) represents a semicircle with a radius of \( \sqrt{5} \) centered at the origin, while \( y = |x - 1| \) is a V-shaped graph that opens upwards with its vertex at the point (1, 0).

Step 1: Finding Points of Intersection

To determine the area between the two curves, we start by finding their points of intersection. We set the equations equal to each other:

\( \sqrt{5 - x^2} = |x - 1| \)

Since \( |x - 1| \) can be split into two cases, we will analyze both:

Case 1: \( x - 1 \geq 0 \) (i.e., \( x \geq 1 \))

In this case, \( |x - 1| = x - 1 \). Thus, we have:

\( \sqrt{5 - x^2} = x - 1 \)

Squaring both sides gives:

\( 5 - x^2 = (x - 1)^2 \)

Expanding the right side:

\( 5 - x^2 = x^2 - 2x + 1 \)

Rearranging terms leads to:

\( 2x^2 - 2x - 4 = 0 \)

Dividing by 2 simplifies this to:

\( x^2 - x - 2 = 0 \)

Factoring gives:

\( (x - 2)(x + 1) = 0 \)

Thus, \( x = 2 \) (valid since \( x \geq 1 \)) and \( x = -1 \) (not valid in this case).

Case 2: \( x - 1 < 0 \) (i.e., \( x < 1 \))

Here, \( |x - 1| = -(x - 1) = 1 - x \). Setting the equations equal gives:

\( \sqrt{5 - x^2} = 1 - x \)

Squaring both sides results in:

\( 5 - x^2 = (1 - x)^2 \)

Expanding the right side yields:

\( 5 - x^2 = 1 - 2x + x^2 \)

Rearranging gives:

\( 2x^2 - 2x - 4 = 0 \)

Again, simplifying leads to:

\( (x - 2)(x + 1) = 0 \)

So, \( x = 2 \) (not valid here) and \( x = -1 \) (valid since \( x < 1 \)).

Step 2: Area Calculation

Now we have the points of intersection: \( x = -1 \) and \( x = 2 \). The area \( A \) between the curves can be calculated using the integral:

\( A = \int_{-1}^{2} \left( \sqrt{5 - x^2} - |x - 1| \right) dx \)

Since \( |x - 1| \) changes at \( x = 1 \), we split the integral into two parts:

\( A = \int_{-1}^{1} \left( \sqrt{5 - x^2} - (1 - x) \right) dx + \int_{1}^{2} \left( \sqrt{5 - x^2} - (x - 1) \right) dx \)

Calculating the First Integral

For \( x \) from -1 to 1:

\( A_1 = \int_{-1}^{1} \left( \sqrt{5 - x^2} - (1 - x) \right) dx \)

Calculating \( A_1 \):

\( A_1 = \int_{-1}^{1} \sqrt{5 - x^2} dx - \int_{-1}^{1} (1 - x) dx \)

The first integral, \( \int_{-1}^{1} \sqrt{5 - x^2} dx \), represents the area of a semicircle and can be computed as:

\( \frac{1}{2} \pi r^2 = \frac{1}{2} \pi ( \sqrt{5} )^2 = \frac{5\pi}{2} \)

The second integral, \( \int_{-1}^{1} (1 - x) dx \), evaluates to:

\( \left[ x - \frac{x^2}{2} \right]_{-1}^{1} = \left( 1 - \frac{1}{2} \right) - \left( -1 - \frac{1}{2} \right) = \frac{1}{2} + \frac{3}{2} = 2 \)

Thus, \( A_1 = \frac{5\pi}{2} - 2 \).

Calculating the Second Integral

For \( x \) from 1 to 2:

\( A_2 = \int_{1}^{2} \left( \sqrt{5 - x^2} - (x - 1) \right) dx \)

Calculating \( A_2 \):

\( A_2 = \int_{1}^{2} \sqrt{5 - x^2} dx - \int_{1}^{2} (x - 1) dx \)

The second integral evaluates to:

\( \left[ \frac{x^2}{2} - x \right]_{1}^{2} = \left( 2 - 2 \right) - \left( \frac{1}{2} - 1 \right) = 0 + \frac{1}{2} = \frac