The function \( y = |x + 3| \) can be defined piecewise:
\[y = \begin{cases} x + 3 & \text{if} \, x \geq -3,\\-(x + 3) & \text{if} \, x<-3.\end{cases}\]
To find the area between \( x = -6 \) and \( x = 0 \), the integral must be split at \( x = -3 \) due to the piecewise definition of the function.
1. For the interval \( [-6, -3] \), the function is \( y = -(x + 3) \). The integral for this region is:
\[A_1 = \int_{-6}^{-3} -(x + 3) \, dx.\]
2. For the interval \( [-3, 0] \), the function is \( y = x + 3 \). The integral for this region is:
\[A_2 = \int_{-3}^{0} (x + 3) \, dx.\]
Calculating the integrals:
For \( A_1 \):
\[A_1 = \int_{-6}^{-3} -(x + 3) \, dx = - \left[ \frac{x^2}{2} + 3x \right]_{-6}^{-3}\]
\[= - \left[ \left( \frac{(-3)^2}{2} + 3(-3) \right) - \left( \frac{(-6)^2}{2} + 3(-6) \right) \right]\]
\[= - \left[ \left( \frac{9}{2} - 9 \right) - \left( \frac{36}{2} - 18 \right) \right]\]
\[= - \left[ -\frac{9}{2} - (18 - 18) \right] = - \left[ -\frac{9}{2} - 0 \right] = \frac{9}{2}.\]
For \( A_2 \):
\[A_2 = \int_{-3}^{0} (x + 3) \, dx = \left[ \frac{x^2}{2} + 3x \right]_{-3}^{0}\]
\[= \left[ \left( \frac{(0)^2}{2} + 3(0) \right) - \left( \frac{(-3)^2}{2} + 3(-3) \right) \right]\]
\[= \left[ 0 - \left( \frac{9}{2} - 9 \right) \right] = \left[ 0 - \left( -\frac{9}{2} \right) \right] = \frac{9}{2}.\]
The total area is the sum of \( A_1 \) and \( A_2 \):
\[A = A_1 + A_2 = \frac{9}{2} + \frac{9}{2} = 9.\]
The area between the curve and the x-axis from \( x = -6 \) to \( x = 0 \) is 9.