A metal rod extends on the `x-`axis from `x=0m` to `x=4m`. Its linear density `rho` (mass`//`length) id not uniform and varies with the distance `x` from the origin according to equation `rho=(1.5+2x)kg//m` What is the mass of rod`:-`

Options

A. `16kg`
B. `22kg`
C. `28kg`
D. `32kg`

Explanation

To find the mass of the metal rod, we will follow these steps:

### Step 1: Understand the linear density function
The linear density \(\rho\) of the rod is given as a function of \(x\):
\[ \rho(x) = 1.5 + 2x \quad \text{(in kg/m)} \]

### Step 2: Set up the mass element
The mass element \(dm\) of a small segment of the rod of length \(dx\) at position \(x\) can be expressed as:
\[ dm = \rho(x) \, dx = (1.5 + 2x) \, dx \]

### Step 3: Integrate to find the total mass
To find the total mass \(M\) of the rod, we need to integrate \(dm\) from \(x = 0\) to \(x = 4\):
\[ M = \int_{0}^{4} dm = \int_{0}^{4} (1.5 + 2x) \, dx \]

### Step 4: Perform the integration
Now, we will integrate the function:
\[ M = \int_{0}^{4} (1.5 + 2x) \, dx = \int_{0}^{4} 1.5 \, dx + \int_{0}^{4} 2x \, dx \]
Calculating each integral:
1. \(\int_{0}^{4} 1.5 \, dx = 1.5x \bigg|_{0}^{4} = 1.5 \times 4 - 1.5 \times 0 = 6\)
2. \(\int_{0}^{4} 2x \, dx = x^2 \bigg|_{0}^{4} = 4^2 - 0^2 = 16\)

Adding these results together:
\[ M = 6 + 16 = 22 \, \text{kg} \]

### Step 5: Final answer
The total mass of the rod is:
\[ \boxed{22 \, \text{kg}} \]

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