Improper Integral of \( 2e^{-2x} \) from 0 to Infinity
The concept of improper integrals allows us to understand and analyze scenarios that involve infinite domains or unbounded functions. Today, we’ll delve into the improper integral \( \int_{0}^{\infty} 2e^{-2x} \, dx \) and determine whether it converges to a finite value.
Real-world Applications
This integral is not just a mathematical abstraction; it appears in various fields like statistics, physics, and engineering, often in the context of exponential decay processes. Understanding its behavior can provide insights into real-world phenomena.
Step 1: Converting to a Limit
Since the integral involves infinity, we use a limit approach to make it more manageable. We replace the infinity with a variable \( a \) and then take the limit as \( a \) approaches infinity:
\( \lim_{{a \to \infty}} \int_{0}^{a} 2e^{-2x} \, dx \)
Step 2: Finding the Antiderivative
The integral \( \int 2e^{-2x} \, dx \) can be solved using elementary integration techniques. The antiderivative of \( 2e^{-2x} \) is \( -e^{-2x} \).
\( \int 2e^{-2x} \, dx = -e^{-2x} + C \), where \( C \) is the constant of integration.
Step 3: Evaluating at the Boundaries
Substitute \( a \) and \( 0 \) into the antiderivative \( -e^{-2x} \) and subtract:
\( -e^{-2a} – (-e^{-2 \times 0}) \)
This simplifies to \( 1 – e^{-2a} \).
Step 4: Taking the Limit
Now we evaluate the limit as \( a \) approaches infinity:
\( \lim_{{a \to \infty}} (1 – e^{-2a}) \)
Step 5: Convergence Check
As \( a \) tends to infinity, \( e^{-2a} \) goes to zero. Therefore, \( 1 – e^{-2a} \) approaches \( 1 \). This means the improper integral converges, and its value is \( 1 \).
Final Thoughts
Calculus gives us the tools to tackle seemingly intractable problems involving infinite ranges and come up with finite, concrete solutions. In this case, the improper integral \( \int_{0}^{\infty} 2e^{-2x} \, dx \) converges to \( 1 \), which can be a valuable insight in various applied fields.
What’s Next?
Integrals like these are just the tip of the iceberg. There’s a wide array of more complicated integrals, each with their own real-world applications. So, why stop here? Keep exploring!