Using integration, find the area of the region bounded by the following curves, after making a rough sketch:
(i) `y=1 + | x+1| , x=-3, x=3, y=0`
(ii) `y=1 + |x+1| , x=-2, x=3, y=0.`

Answer

To find the area of the region bounded by the curves given in the question, we will follow these steps:

### Part (i): Area bounded by `y = 1 + |x + 1|`, `x = -3`, `x = 3`, and `y = 0`

1. **Understanding the function**:
- The function `y = 1 + |x + 1|` can be broken down into two cases:
- For `x < -1`, `|x + 1| = -(x + 1) = -x - 1`, so `y = 1 - x - 1 = -x`.
- For `x >= -1`, `|x + 1| = x + 1`, so `y = 1 + x + 1 = x + 2`.

2. **Finding intersection points**:
- The curves intersect the x-axis (where `y = 0`):
- For `y = -x`, setting `-x = 0` gives `x = 0`.
- For `y = x + 2`, setting `x + 2 = 0` gives `x = -2`.

3. **Sketching the graph**:
- The graph consists of two linear pieces:
- From `x = -3` to `x = -1`, the line is `y = -x` (which goes from `(−3, 3)` to `(−1, 1)`).
- From `x = -1` to `x = 3`, the line is `y = x + 2` (which goes from `(−1, 1)` to `(3, 5)`).

4. **Setting up the integral**:
- The area can be split into two parts:
- Area A1 from `x = -3` to `x = -1`:
\[ A_1 = \int_{-3}^{-1} (-x) \, dx \]
- Area A2 from `x = -1` to `x = 3`:
\[ A_2 = \int_{-1}^{3} (x + 2) \, dx \]

5. **Calculating A1**:
\[ A_1 = \int_{-3}^{-1} (-x) \, dx = \left[-\frac{x^2}{2}\right]_{-3}^{-1} = \left[-\frac{(-1)^2}{2} + \frac{(-3)^2}{2}\right] = \left[-\frac{1}{2} + \frac{9}{2}\right] = \frac{8}{2} = 4 \]

6. **Calculating A2**:
\[ A_2 = \int_{-1}^{3} (x + 2) \, dx = \left[\frac{x^2}{2} + 2x\right]_{-1}^{3} = \left[\frac{3^2}{2} + 2(3)\right] - \left[\frac{(-1)^2}{2} + 2(-1)\right] \]
\[ = \left[\frac{9}{2} + 6\right] - \left[\frac{1}{2} - 2\right] = \left[\frac{9}{2} + \frac{12}{2}\right] - \left[\frac{1}{2} - \frac{4}{2}\right] = \left[\frac{21}{2}\right] - \left[-\frac{3}{2}\right] = \frac{21}{2} + \frac{3}{2} = \frac{24}{2} = 12 \]

7. **Total Area**:
\[ \text{Total Area} = A_1 + A_2 = 4 + 12 = 16 \text{ square units} \]

### Part (ii): Area bounded by `y = 1 + |x + 1|`, `x = -2`, `x = 3`, and `y = 0`

1. **Sketching the graph**:
- The graph remains the same as in part (i), but now we only consider the area from `x = -2` to `x = 3`.

2. **Setting up the integral**:
- Area A1 from `x = -2` to `x = -1`:
\[ A_1 = \int_{-2}^{-1} (-x) \, dx \]
- Area A2 from `x = -1` to `x = 3`:
\[ A_2 = \int_{-1}^{3} (x + 2) \, dx \]

3. **Calculating A1**:
\[ A_1 = \int_{-2}^{-1} (-x) \, dx = \left[-\frac{x^2}{2}\right]_{-2}^{-1} = \left[-\frac{(-1)^2}{2} + \frac{(-2)^2}{2}\right] = \left[-\frac{1}{2} + \frac{4}{2}\right] = \frac{3}{2} \]

4. **Calculating A2**:
\[ A_2 = \int_{-1}^{3} (x + 2) \, dx = \left[\frac{x^2}{2} + 2x\right]_{-1}^{3} = \left[\frac{3^2}{2} + 2(3)\right] - \left[\frac{(-1)^2}{2} + 2(-1)\right] \]
\[ = \left[\frac{9}{2} + 6\right] - \left[\frac{1}{2} - 2\right] = \left[\frac{9}{2} + \frac{12}{2}\right] - \left[\frac{1}{2} - \frac{4}{2}\right] = \left[\frac{21}{2}\right] - \left[-\frac{3}{2}\right] = \frac{21}{2} + \frac{3}{2} = \frac{24}{2} = 12 \]

5. **Total Area**:
\[ \text{Total Area} = A_1 + A_2 = \frac{3}{2} + 12 = \frac{3}{2} + \frac{24}{2} = \frac{27}{2} = 13.5 \text{ square units} \]

### Final Answers:
- (i) The area is **16 square units**.
- (ii) The area is **13.5 square units**.