To solve the given problem, we need to compute the average value of the function $f(t) = \frac{t^2}{t^3 + 4}$ over the time interval $2 \leq t \leq 5$. The formula for the average value of a function over an interval $[a, b]$ is:

$\text{Average value} = \frac{1}{b-a} \int_a^b f(t) \, dt$

Steps:

1. The interval is $[2, 5]$, so $a = 2$ and $b = 5$.

2. The formula becomes: $\text{Average sales} = \frac{1}{5 - 2} \int_2^5 \frac{t^2}{t^3 + 4} \, dt$

3. Simplify the coefficient: $\text{Average sales} = \frac{1}{3} \int_2^5 \frac{t^2}{t^3 + 4} \, dt$

4. The integral $\int \frac{t^2}{t^3 + 4} \, dt$ can be computed using substitution. Let $u = t^3 + 4$, so $du = 3t^2 \, dt$. Adjusting, we get: $\int \frac{t^2}{t^3 + 4} \, dt = \frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln |u| + C = \frac{1}{3} \ln |t^3 + 4| + C$

5. Apply the limits $t = 2$ and $t = 5$: [ \int_2^5 \frac{t^2}{t^3 + 4} , dt = \frac{1}{3} \left[\ln(t^3 + 4)\right]_2^5 = \frac{1}{3} \left[ \ln(5^3 + 4) - \ln(2^3 + 4) \right] ]

6. Compute the numerical values:

   • When $t = 5$: $5^3 + 4 = 125 + 4 = 129$
   • When $t = 2$: $2^3 + 4 = 8 + 4 = 12$

   So: $\ln(5^3 + 4) - \ln(2^3 + 4) = \ln(129) - \ln(12) = \ln\left(\frac{129}{12}\right)$

7. Simplify the result: $\int_2^5 \frac{t^2}{t^3 + 4} \, dt = \frac{1}{3} \ln\left(\frac{129}{12}\right)$

8. Multiply by $\frac{1}{3}$ for the average: $\text{Average sales} = \frac{1}{3} \cdot \frac{1}{3} \ln\left(\frac{129}{12}\right) = \frac{1}{9} \ln\left(\frac{129}{12}\right)$

Let me calculate the numerical approximation. math

Calculate the values

numerator = 129 denominator = 12 ln_value = math.log(numerator / denominator)

Final average sales value

average_sales = (1 / 9) * ln_value average_salesThe average sales over the time interval $2 \leq t \leq 5$ is approximately 0.264 (rounded to three decimal places).

Would you like further details or assistance with this computation?

Here are 5 questions for further exploration:

1. How does substitution simplify integrals in calculus?
2. What is the significance of the natural logarithm in solving integrals like this one?
3. Can the result be interpreted graphically over the interval $2 \leq t \leq 5$?
4. How would the result change if the interval was altered, e.g., $1 \leq t \leq 4$?
5. How do average values of functions relate to real-world scenarios?

Tip: Always recheck substitution steps in integrals to ensure the variable change is properly accounted for in limits and coefficients.