Question

The area of the region in the first quadrant enclosed by the curves\( y=√x,y=-x+6 \)and the x-axis is 

• A. \(\dfrac{22}{7}\)
• B. \(\dfrac{22}{3}\)
• C. \(12\)
• D. \(24\)
• E. \(8\)

Answer 1

The Correct Option is B

Step 1: Find the points of intersection:

1. Between \( y = \sqrt{x} \) and \( y = -x + 6 \): \[ \sqrt{x} = -x + 6 \] \[ x = (-x + 6)^2 \] \[ x = x^2 - 12x + 36 \] \[ x^2 - 13x + 36 = 0 \] \[ (x-4)(x-9) = 0 \] Only \( x = 4 \) is in the first quadrant.

2. Between \( y = \sqrt{x} \) and x-axis (\( y=0 \)): \[ \sqrt{x} = 0 \Rightarrow x = 0 \]

3. Between \( y = -x + 6 \) and x-axis: \[ -x + 6 = 0 \Rightarrow x = 6 \]

Step 2: Set up the integral for the area:

The region is bounded by: - From \( x = 0 \) to \( x = 4 \): between \( y = \sqrt{x} \) and x-axis - From \( x = 4 \) to \( x = 6 \): between \( y = -x + 6 \) and x-axis

Total area \( A = A_1 + A_2 \): \[ A_1 = \int_0^4 \sqrt{x} \, dx = \left[ \frac{2}{3}x^{3/2} \right]_0^4 = \frac{2}{3}(8) = \frac{16}{3} \] \[ A_2 = \int_4^6 (-x + 6) \, dx = \left[ -\frac{x^2}{2} + 6x \right]_4^6 = (-18 + 36) - (-8 + 24) = 18 - 16 = 2 \]

Step 3: Calculate total area: \[ A = \frac{16}{3} + 2 = \frac{16}{3} + \frac{6}{3} = \frac{22}{3} \]

Conclusion: The area of the region is \(\boxed{B}\) \(\left( \frac{22}{3} \right)\).

Answer 2

First, let's find the points of intersection of the curves.

The intersection of \( y = \sqrt{x} \) and \( y = -x + 6 \) occurs when \( \sqrt{x} = -x + 6 \). 

Squaring both sides, we get \( x = x^2 - 12x + 36 \), which simplifies to \( x^2 - 13x + 36 = 0 \). 

This factors as \( (x-4)(x-9) = 0 \), so \( x = 4 \) or \( x = 9 \). Since \( y = \sqrt{x} \), we must have \( x \geq 0 \), and when \( x=4 \), \( y = \sqrt{4} = 2 \), and when \( x=9 \), \( y=\sqrt{9}=3 \). 

However, when \( x=9 \), \( y=-9+6=-3 \), which is inconsistent with \( y = \sqrt{x} \). 

Therefore, the intersection point is \( (4, 2) \).

The intersection of \( y = \sqrt{x} \) and the x-axis is \( (0, 0) \).

The intersection of \( y = -x + 6 \) and the x-axis is found by setting \( y=0 \): \( 0 = -x + 6 \), so \( x = 6 \). The intersection point is \( (6, 0) \).

The area is given by the integral of the difference between the two curves from \( x=0 \) to \( x=4 \).

\[ A = \int_0^4 \sqrt{x} \, dx + \int_4^6 (-x+6) \, dx \] \[ A = \left[ \frac{2}{3}x^{3/2} \right]_0^4 + \left[ -\frac{1}{2}x^2 + 6x \right]_4^6 \] \[ A = \frac{2}{3}(4^{3/2}) - 0 + \left( -\frac{1}{2}(6)^2 + 6(6) \right) - \left( -\frac{1}{2}(4)^2 + 6(4) \right) \] \[ A = \frac{2}{3}(8) + (-18 + 36) - (-8 + 24) \] \[ A = \frac{16}{3} + 18 - 16 \] \[ A = \frac{16}{3} + 2 = \frac{16 + 6}{3} = \frac{22}{3} \]

Therefore, the area of the region is \( \frac{22}{3} \).

Final Answer: The final answer is \( {\frac{22}{3}} \).