Problem 25 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Either evaluate the given improper integral or show that it diverges. $$ \int_{0}^{+\infty} x e^{-2 x} d x $$

Step-by-Step Solution

Answer

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Step 1 - Set up the improper integral

Rewrite the given integral

Step 2 - Evaluate the integral from 0 to a

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Step 3 Title

Content for Step 3

Key Concepts

Improper Integrals

Improper integrals are integrals with limits that approach infinity, negative infinity, or involve a point where the integrand becomes unbounded. Unlike standard definite integrals, improper integrals often appear in complex mathematical and physical problems.
In our example, we face the integral \[ \int_{0}^{+\infty} x e^{-2x} \, dx. \] The integration involves infinity as the upper limit, which classifies it as an improper integral.
When working with improper integrals, we split the integration into a limit process that approaches infinity. This helps ensure we properly account for the behavior of the function as it extends toward infinity.

Integration Techniques

Managing improper integrals requires specific techniques, often building upon standard integration methods. A common method is integration by parts, which is useful for products of functions.
The integration by parts formula is written as: \[ \int u \, dv = uv - \int v \, du. \]Here:

• Let \ u = x \ and \ dv = e^{-2x}dx.
• A quick differentiation and integration give us \ du = dx \ and \ v = \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}. \

Applying these into our integral, it transforms as: \[ \int_{0}^{+\infty} x e^{-2x} \, dx = -\frac{1}{2} x e^{-2x} \biggr|_{0}^{+\infty} + \int_{0}^{\infty} \frac{1}{2} e^{-2x} \, dx. \]This step-by-step method is crucial for breaking down complex integrals into more manageable parts.

Convergence and Divergence of Integrals

Convergence and divergence determine whether an improper integral has a finite value. If the evaluation leads to a finite number, it converges. If not, it diverges.
Returning to \ \int_{0}^{+\infty} x e^{-2x} \, dx, we need to check the limit: \[ -\frac{1}{2} x e^{-2x} \biggr|_{0}^{+\infty} + \int_{0}^{\infty} \frac{1}{2} e^{-2x} \, dx \]Evaluating the first part, we get: \[ -\frac{1}{2} x e^{-2x} \biggr|_{0}^{+\infty} = \lim_{b \to \infty} \left( -\frac{1}{2} b e^{-2b} - 0 \right) = 0. \]Next, evaluating the second integral gives: \[ \int_{0}^{\infty} \frac{1}{2} e^{-2x} \, dx = \frac{1}{2} \left( \frac{-1}{2} e^{-2x} \right) \biggr|_{0}^{+\infty} = \frac{1}{4}. \] Thus, combining these results, the improper integral converges to a finite value of \ \frac{1}{4}.
Understanding the behavior at the infinite limit is key for convergence and determining the validity of the integral.