Question

In a rough sketch, mark the region bounded by \( y = 1 + |x + 1| \), \( x = -2 \), \( x = 2 \), and \( y = 0 \). Using integration, find the area of the marked region.

Hint:

To find the area under a curve with an absolute value function, split the integral at the point where the expression inside the absolute value changes sign.

Answer 1

We need to find the area under the curve \( y = 1 + |x+1| \) between \( x = -2 \) and \( x = 2 \). This function has two parts due to the absolute value function: For \( x \in [-2, -1] \), \( |x+1| = -(x+1) \), so the equation becomes: \[ y = 1 - (x+1) = 1 - x - 1 = -x \] For \( x \in [-1, 2] \), \( |x+1| = x + 1 \), so the equation becomes: \[ y = 1 + (x + 1) = x + 2 \] Now we can calculate the area using integrals for both intervals. For the interval \( [-2, -1] \): \[ A_1 = \int_{-2}^{-1} (-x) dx = \left[ \frac{-x^2}{2} \right]_{-2}^{-1} = \frac{-(-1)^2}{2} - \frac{-(-2)^2}{2} = \frac{-1}{2} - \frac{-4}{2} = \frac{3}{2} \] For the interval \( [-1, 2] \): \[ A_2 = \int_{-1}^{2} (x+2) dx = \left[ \frac{x^2}{2} + 2x \right]_{-1}^{2} = \left( \frac{(2)^2}{2} + 2(2) \right) - \left( \frac{(-1)^2}{2} + 2(-1) \right) \] \[ = \left( \frac{4}{2} + 4 \right) - \left( \frac{1}{2} - 2 \right) = (2 + 4) - \left( \frac{1}{2} - 2 \right) \] \[ = 6 - \left( -\frac{3}{2} \right) = 6 + \frac{3}{2} = \frac{15}{2} \] Thus, the total area is: \[ A = A_1 + A_2 = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9 \] \begin{center} \includegraphics[width=0.7\textwidth]{ig4.png} \end{center}