Question

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 $y = 0$.

Hint:

For piecewise functions, break the integral into separate intervals based on the different expressions for the function. Then calculate the integrals separately and add the results.

Answer 1

The objective is to determine the area of the region enclosed by the curve \( y = 2 + |x + 1| \), the vertical lines \( x = -4 \) and \( x = 3 \), and the x-axis (\( y = 0 \)).

Step 1: Characterize the curve The function \( y = 2 + |x + 1| \) represents a V-shaped graph. The absolute value function defines it as 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: Formulate the integral To find the area between the curve and the x-axis from \( x = -4 \) to \( x = 3 \), the integral is divided into two segments based on the piecewise function definition: 1. For the interval \( x \in [-4, -1] \), the curve is defined by \( y = 3 - x \). 2. For the interval \( x \in [-1, 3] \), the curve is defined by \( y = x + 3 \). Consequently, the total area \( A \) is calculated as: \[ A = \int_{-4}^{-1} (3 - x) \, dx + \int_{-1}^{3} (x + 3) \, dx. \] 
 Step 3: Evaluate 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 the expression: \[ = (-3 - \frac{1}{2}) - (-12 - 8) = (-3 - \frac{1}{2}) + 20 = 17 - \frac{1}{2} = \frac{34}{2} - \frac{1}{2} = \frac{33}{2}. \] 
Step 4: Evaluate 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 the expression: \[ = \left( \frac{9}{2} + 9 \right) - \left( \frac{1}{2} - 3 \right) = \left( \frac{9}{2} + \frac{18}{2} \right) - \left( \frac{1}{2} - \frac{6}{2} \right) = \frac{27}{2} - \frac{-5}{2} = \frac{27}{2} + \frac{5}{2} = \frac{32}{2} = 16. \]
Step 5: Compute the total area Summing the results of the two integrals: \[ A = \frac{33}{2} + 16 = \frac{33}{2} + \frac{32}{2} = \frac{65}{2}. \] 
The total area of the specified region is \( \boxed{\frac{65}{2}} \).