Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
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Solution and Explanation
We are asked to find the area of the region bounded by the curve \( y = 2 + |x + 1| \), the lines \( x = -4 \), \( x = 3 \), and the x-axis (\( y = 0 \)). ### Step 1: Understanding the curve The function \( y = 2 + |x + 1| \) is a V-shaped curve. The absolute value function creates a piecewise function: \[ y = \begin{cases} 2 + (x + 1) = x + 3, & \text{for} \quad x \geq -1, \\ 2 - (x + 1) = 3 - x, & \text{for} \quad x < -1. \end{cases} \] ### Step 2: Set up the integral We need to calculate the area between the curve and the x-axis from \( x = -4 \) to \( x = 3 \). We break this integral into two parts based on the piecewise definition of the function: 1. From \( x = -4 \) to \( x = -1 \), the equation for the curve is \( y = 3 - x \). 2. From \( x = -1 \) to \( x = 3 \), the equation for the curve is \( y = x + 3 \). Thus, the area \( A \) is given by: \[ A = \int_{-4}^{-1} (3 - x) \, dx + \int_{-1}^{3} (x + 3) \, dx. \] ### Step 3: Calculate the first integral \[ \int_{-4}^{-1} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{-4}^{-1} = \left( 3(-1) - \frac{(-1)^2}{2} \right) - \left( 3(-4) - \frac{(-4)^2}{2} \right). \] Simplifying: \[ = (-3 - \frac{1}{2}) - (-12 - 8) = (-3 - \frac{1}{2}) + 20 = 17 - \frac{1}{2} = \frac{34}{2} - \frac{1}{2} = \frac{33}{2}. \] ### Step 4: Calculate the second integral \[ \int_{-1}^{3} (x + 3) \, dx = \left[ \frac{x^2}{2} + 3x \right]_{-1}^{3} = \left( \frac{3^2}{2} + 3(3) \right) - \left( \frac{(-1)^2}{2} + 3(-1) \right). \] Simplifying: \[ = \left( \frac{9}{2} + 9 \right) - \left( \frac{1}{2} - 3 \right) = \left( \frac{9}{2} + \frac{18}{2} \right) - \left( \frac{1}{2} - \frac{6}{2} \right) = \frac{27}{2} - \frac{-5}{2} = \frac{27}{2} + \frac{5}{2} = \frac{32}{2} = 16. \] ### Step 5: Total area Adding the two areas together: \[ A = \frac{33}{2} + 16 = \frac{33}{2} + \frac{32}{2} = \frac{65}{2}. \] Thus, the total area of the region is \( \boxed{\frac{65}{2}} \). 
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