Question

Evaluate the integrals. If the integral diverges, answer "diverges."\(\int_{0}^{\infty} x e^{-x} d x\)

Short answer

The integral converges and evaluates to 1.

Step-by-step solution

Identify the Integral Type

This is an improper integral because it evaluates a function from 0 to infinity.

Set Up the Limit for the Integral

First, express the integral with a limit: \[\int_{0}^{ ext{∞}} x e^{-x} dx = \lim_{b \to ext{∞}} \int_{0}^{b} x e^{-x} dx\]

Perform Integration by Parts

Apply integration by parts where \( u = x \) and \( dv = e^{-x} dx \).- Differentiate \( u \): \( du = dx \).- Integrate \( dv \): \( v = -e^{-x} \).Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), the integral becomes:\[\int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx\]

Evaluate the Second Integral

The remaining integral \( \int e^{-x} dx \) evaluates to \(-e^{-x} \). Substituting back gives:\[-x e^{-x} - e^{-x}\]

Evaluate the Definite Integral

Now evaluate the integral from 0 to b:\[\left. \left(-x e^{-x} - e^{-x} \right) \right|_{0}^{b}\]Substituting the bounds, we get:\[\lim_{b \to \infty} \left( -(b e^{-b} + e^{-b}) + (0 + 1) \right)\]

Evaluate the Limit

As \( b \to \infty \), both terms \( b e^{-b} \) and \( e^{-b} \) approach 0. So:\[\lim_{b \to \infty} (-(b e^{-b} + e^{-b})) = 0\]The value of the integral is \( 1 \), since the remaining term is:\[1 - 0 = 1\]

Conclusion

The improper integral \( \int_{0}^{\infty} x e^{-x} dx \) converges and evaluates to 1.

Key concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals where standard methods do not work easily. It is especially useful when the integrand is a product of two functions. The formula for integration by parts is derived from the product rule of differentiation and is given as:

• \[ \int u \, dv = uv - \int v \, du \]

For the current problem, we choose:

• \( u = x \) and thus \( du = dx \)
• \( dv = e^{-x} \, dx \), resulting in \( v = -e^{-x} \)

By substituting into the formula, we successfully transform the integral \( \int x e^{-x} dx \) into a form that is easier to evaluate. The choice of \( u \) and \( dv \) is crucial for simplifying the integral.

Convergence of Integrals

Convergence in integrals refers to whether an integral approaches a finite number as the limit of integration extends to infinity or a discontinuity is approached. An improper integral like \( \int_{0}^{\infty} x e^{-x} dx \) evaluates a function over an unbounded region, making it necessary to test for convergence.
To determine the convergence:

• First, express the integral with a limit: \( \lim_{b \to \infty} \int_{0}^{b} x e^{-x} dx \).
• If the limit results in a finite number, the integral converges.

It is important to distinguish between convergent and divergent integrals, as one reflects a defined area under the curve whereas the other does not.

Limit Evaluation

Limit evaluation is crucial in dealing with improper integrals since it handles the behavior of a function as the variable approaches infinity or a point of discontinuity. The improper integral \( \int_{0}^{b} x e^{-x} dx \) is initially evaluated, and the limit \( \lim_{b \to \infty} \) is taken to determine its value.
When evaluating the limit

• \(-\lim_{b \to \infty} (b e^{-b} + e^{-b})\)

we note that \( b e^{-b} \) and \( e^{-b} \) both tend towards zero as \( b \) becomes very large.
Thus, the entire expression simplifies to zero, confirming the convergence of the integral and resulting in a finite value.

Definite Integrals

Definite integrals calculate the signed area under a curve from one point to another. They have bounds of integration, such as in the case of our \( \int_{0}^{b} x e^{-x} dx \). The evaluation of definite integrals often involves substituting the upper and lower bounds back into the evaluated antiderivative.
In our specific example, the evaluation starts by finding:

• \[ \left. (-x e^{-x} - e^{-x}) \right|_{0}^{b} \]

Substitute into this expression to determine the function values at each bound and subtract appropriately:

• The value at \( b \): \( -(b e^{-b} + e^{-b}) \)
• The value at 0: \( -(0) + 1 = 1 \)

The final value, once the limit is evaluated for the upper bound tending to infinity, results in balancing out these values to achieve the correct definite integral value.