4∫ 
2
3
​
  
2+t 
6
 + 
t 
6
 
1
​
 
​
 dt

Let's evaluate the integral:

\[
\int_{2}^{4} \left(\frac{2 + t}{6} + \frac{t}{6}\right) \, dt
\]

### Step-by-step Solution:

1. **Simplify the integrand**:
   Combine the terms inside the integral:
   \[
   \frac{2 + t}{6} + \frac{t}{6} = \frac{2 + 2t}{6}
   \]
   This can be further simplified to:
   \[
   \frac{2 + 2t}{6} = \frac{2}{6} + \frac{2t}{6} = \frac{1}{3} + \frac{t}{3}
   \]

2. **Rewrite the integral**:
   \[
   \int_{2}^{4} \left(\frac{1}{3} + \frac{t}{3}\right) \, dt
   \]
   Factor out \(\frac{1}{3}\):
   \[
   \frac{1}{3} \int_{2}^{4} (1 + t) \, dt
   \]

3. **Evaluate the integral**:
   Break this into two integrals:
   \[
   \frac{1}{3} \left[\int_{2}^{4} 1 \, dt + \int_{2}^{4} t \, dt\right]
   \]

   - For \(\int_{2}^{4} 1 \, dt\):
     \[
     \int_{2}^{4} 1 \, dt = t \bigg|_{2}^{4} = 4 - 2 = 2
     \]

   - For \(\int_{2}^{4} t \, dt\):
     \[
     \int_{2}^{4} t \, dt = \frac{t^2}{2} \bigg|_{2}^{4} = \frac{4^2}{2} - \frac{2^2}{2} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6
     \]

4. **Combine the results**:
   \[
   \frac{1}{3} \left[2 + 6\right] = \frac{1}{3} \cdot 8 = \frac{8}{3}
   \]

### Final Answer:
\[
\int_{2}^{4} \left(\frac{2 + t}{6} + \frac{t}{6}\right) \, dt = \frac{8}{3}
\]

Would you like a more detailed explanation of any part of this process, or do you have any questions?

### Related Questions:
1. How do you evaluate definite integrals with polynomial functions?
2. What is the process for combining like terms inside an integral?
3. How do you factor constants out of an integral?
4. How does the Fundamental Theorem of Calculus apply to definite integrals?
5. What is the geometric interpretation of definite integrals?

### Tip:
Always simplify the integrand before integrating—it can make the integration process faster and less error-prone.

This page provides a detailed step-by-step solution for evaluating the definite integral of (2 + t) / 6 + t / 6 over the interval [2, 4]. Learn how to simplify the integrand, apply the Fundamental Theorem of Calculus, and calculate the final result of 8/3. Suitable for advanced high school students in Grades 11-12.

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