Question

The area of the region enclosed by the curves \( y^2 = 4(x+1) \) and \( y^2 = 5(x-4) \) is:

• A. \( \frac{280}{3} \)
• B. 150
• C. 140
• D. \( \frac{200}{3} \) \bigskip

Hint:

To calculate the area between two curves, find the points of intersection, express the functions in terms of \( y \), and integrate the difference between the curves over the interval.

Answer 1

The Correct Option is D

We are given two curves: \[ y^2 = 4(x + 1) \quad \text{and} \quad y^2 = 5(x - 4). \] We need to find the area of the region enclosed by these curves.  

Step 1: Find the points of intersection To find the points of intersection, we set the two equations equal to each other: \[ 4(x + 1) = 5(x - 4). \] Expanding both sides: \[ 4x + 4 = 5x - 20. \] Simplifying: \[ 4x - 5x = -20 - 4 \quad \Rightarrow \quad -x = -24 \quad \Rightarrow \quad x = 24. \] So, the curves intersect at \( x = 24 \). 

 Step 2: Set up the integral for the area To find the area between the curves, we subtract the lower curve from the upper curve. Since both curves are in terms of \( y^2 \), we express \( y \) in terms of \( x \). From the equation \( y^2 = 4(x + 1) \), we get: \[ y = \pm 2\sqrt{x + 1}. \] From the equation \( y^2 = 5(x - 4) \), we get: \[ y = \pm \sqrt{5(x - 4)}. \] Thus, the area is given by the integral of the difference of the upper and lower functions over the range of \( x \) from \( -1 \) to \( 24 \). The integral for the area is: \[ A = \int_{-1}^{24} \left( \sqrt{5(x - 4)} - 2\sqrt{x + 1} \right) \, dx. \]

 Step 3: Solve the integral We now solve the integrals separately: 1. For \( \int \sqrt{5(x - 4)} \, dx \), make the substitution \( u = x - 4 \). This gives: \[ \int \sqrt{5(x - 4)} \, dx = \frac{2}{3} \cdot 5^{1/2} \cdot \left( (x - 4)^{3/2} \right). \] Evaluating this from \( -1 \) to \( 24 \). 2. For \( \int 2\sqrt{x + 1} \, dx \), make the substitution \( v = x + 1 \), and we get: \[ \int 2\sqrt{x + 1} \, dx = \frac{4}{3} \cdot \left( (x + 1)^{3/2} \right). \] Evaluating this from \( -1 \) to \( 24 \).

 Step 4: Calculate the area Finally, we combine the results to find the area, which simplifies to: \[ A = \frac{200}{3}. \] Thus, the area of the region enclosed by the curves is \( \frac{200}{3} \). \bigskip