MTTF=∫
0
1,000,000
​
t⋅α(t−5000)
2
dt. when alpha = 1.2x10^-11

To calculate the **Mean Time to Failure (MTTF)** given the integral formula:

\[
\text{MTTF} = \int_{0}^{1,000,000} t \cdot \alpha (t - 5000)^2 \, dt
\]

where \(\alpha = 1.2 \times 10^{-11}\), let's break this down into steps:

1. **Substitute \(\alpha\)**: We will first substitute \(\alpha = 1.2 \times 10^{-11}\) into the integral:
\[
\text{MTTF} = \int_{0}^{1,000,000} t \cdot (1.2 \times 10^{-11}) (t - 5000)^2 \, dt
\]

2. **Expand the integrand**: The expression inside the integral can be expanded and simplified. The expression \((t - 5000)^2\) is a quadratic, and multiplying by \(t\) and \(\alpha\) will lead to further terms. Let's compute the expanded form:
\[
(t - 5000)^2 = t^2 - 2 \cdot 5000 \cdot t + 5000^2 = t^2 - 10000t + 25,000,000
\]
Now multiply by \(t\):
\[
t \cdot (t^2 - 10000t + 25,000,000) = t^3 - 10000t^2 + 25,000,000t
\]
Now multiply the entire expression by \(\alpha = 1.2 \times 10^{-11}\):
\[
\text{MTTF} = \int_{0}^{1,000,000} (1.2 \times 10^{-11}) (t^3 - 10000t^2 + 25,000,000t) \, dt
\]

3. **Split the integral**: We can now split the integral into three parts:
\[
\text{MTTF} = 1.2 \times 10^{-11} \left[ \int_{0}^{1,000,000} t^3 \, dt - 10000 \int_{0}^{1,000,000} t^2 \, dt + 25,000,000 \int_{0}^{1,000,000} t \, dt \right]
\]

4. **Solve each integral**:
   - \(\int_{0}^{1,000,000} t^3 \, dt = \frac{1,000,000^4}{4} = 2.5 \times 10^{23}\)
   - \(\int_{0}^{1,000,000} t^2 \, dt = \frac{1,000,000^3}{3} = 3.33 \times 10^{17}\)
   - \(\int_{0}^{1,000,000} t \, dt = \frac{1,000,000^2}{2} = 5 \times 10^{11}\)

5. **Substitute back and calculate**:
\[
\text{MTTF} = 1.2 \times 10^{-11} \left[ 2.5 \times 10^{23} - 10000 \cdot 3.33 \times 10^{17} + 25,000,000 \cdot 5 \times 10^{11} \right]
\]
Now, let's compute this. I will do the calculations.The Mean Time to Failure (MTTF) is approximately **2.96 × 10^{12}** units of time.

Would you like further details on any part of the calculation? Here are 5 related questions to consider:

1. How does the choice of \(\alpha\) impact the MTTF value?
2. What is the physical significance of \(t - 5000\) in this equation?
3. How would changing the limits of the integral affect the result?
4. Can MTTF be used for other systems' reliability analysis?
5. How can you validate the units in such integrals?

**Tip**: Always check the units of each term in the integrand to ensure they align with the desired final result.

MTTF=∫ 0 to 1,000,000 t⋅α(t−5000)^2 dt, when alpha = 1.2x10^-11

Learn how to calculate the Mean Time to Failure (MTTF) using integral calculus and polynomial expansion. This detailed solution explores the impact of α = 1.2x10^-11 on MTTF, solving the integral over a range of 0 to 1,000,000. Suitable for college-level students in engineering and mathematics.

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