A rod of length L and cross section area A has variable density according to the relation `rho (x)=rho_(0)+kx` for `0 le x le L/2` and `rho(x)=2x^(2)` for `L/2 le x le L` where `rho_(0)` and k are constants. Find the mass of the rod.

Explanation

To find the mass of the rod with variable density, we will break the problem into two parts based on the given density functions for different segments of the rod.

### Step 1: Define the mass of the rod
The mass \( m \) of the rod can be calculated using the formula:
\[ m = \int \rho(x) \, dV \]
where \( \rho(x) \) is the density at position \( x \) and \( dV \) is the differential volume element. For a rod with a cross-sectional area \( A \), the differential volume element can be expressed as \( A \, dx \).

### Step 2: Break the integral into two parts
The rod is divided into two segments based on the given density functions:
1. From \( 0 \) to \( \frac{L}{2} \) where \( \rho(x) = \rho_0 + kx \)
2. From \( \frac{L}{2} \) to \( L \) where \( \rho(x) = 2x^2 \)

Thus, we can write the mass as:
\[ m = \int_0^{L/2} \rho_0 + kx \, A \, dx + \int_{L/2}^{L} 2x^2 \, A \, dx \]

### Step 3: Calculate the first integral
For the first part of the rod:
\[ m_1 = \int_0^{L/2} (\rho_0 + kx) A \, dx \]
This can be separated into two integrals:
\[ m_1 = A \left( \int_0^{L/2} \rho_0 \, dx + \int_0^{L/2} kx \, dx \right) \]
Calculating each integral:
1. \(\int_0^{L/2} \rho_0 \, dx = \rho_0 \cdot \frac{L}{2}\)
2. \(\int_0^{L/2} kx \, dx = k \cdot \frac{(L/2)^2}{2} = \frac{kL^2}{8}\)

Thus, we have:
\[ m_1 = A \left( \rho_0 \cdot \frac{L}{2} + \frac{kL^2}{8} \right) \]

### Step 4: Calculate the second integral
For the second part of the rod:
\[ m_2 = \int_{L/2}^{L} 2x^2 \, A \, dx \]
Calculating this integral:
\[ m_2 = A \cdot 2 \int_{L/2}^{L} x^2 \, dx \]
The integral \(\int x^2 \, dx\) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from \(L/2\) to \(L\):
\[ \int_{L/2}^{L} x^2 \, dx = \left[ \frac{L^3}{3} - \frac{(L/2)^3}{3} \right] = \frac{L^3}{3} - \frac{L^3}{24} = \frac{8L^3 - L^3}{24} = \frac{7L^3}{24} \]
Thus:
\[ m_2 = A \cdot 2 \cdot \frac{7L^3}{24} = \frac{7AL^3}{12} \]

### Step 5: Combine the masses
Now, we can combine both parts to find the total mass of the rod:
\[ m = m_1 + m_2 = A \left( \rho_0 \cdot \frac{L}{2} + \frac{kL^2}{8} \right) + \frac{7AL^3}{12} \]

### Final expression for the mass of the rod
Thus, the total mass of the rod is:
\[ m = A \left( \rho_0 \cdot \frac{L}{2} + \frac{kL^2}{8} + \frac{7L^3}{12} \right) \]