Problem 1 - Calculus for Biology and Medicine

Question

All the integrals in problem are improper and converge. Explain in each case why the integral is improper, and evaluate each integral. $$ \int_{0}^{\infty} 3 e^{-x} d x $$

Step-by-Step Solution

Answer

The integral converges to 3.

Step 1: Identify Why the Integral is Improper

An improper integral is an integral where the interval of integration is infinite or the integrand becomes infinite at one or more points in the interval. In this case, the integral is improper because the upper limit of integration is infinity (in). This means that we are integrating over an infinite interval.

Step 2: Set Up the Limit for the Improper Integral

To evaluate an improper integral with an infinite limit, we replace the infinity with a limit. We write the integral as\[\lim_{b \to \infty} \int_{0}^{b} 3 e^{-x} \, dx.\]

Step 3: Evaluate the Definite Integral

The integral of \(3 e^{-x}\) with respect to \(x\) is \(-3e^{-x}\). So, we evaluate the definite integral:\[\int_{0}^{b} 3 e^{-x} \, dx = \left[ -3e^{-x} \right]_{0}^{b} = -3e^{-b} + 3e^{0}.\]

Step 4: Calculate the Limit

Substitute the evaluated integral into the limit expression:\[\lim_{b \to \infty} \left(-3e^{-b} + 3\right).\]As \(b\) approaches infinity, \(e^{-b}\) approaches 0. Thus,\[-3e^{-b} \to 0.\]Therefore, the expression simplifies to\[\lim_{b \to \infty} \left(0 + 3\right) = 3.\]

Step 5: Conclusion

The integral \(\int_{0}^{\infty} 3 e^{-x} \, dx\) converges and the evaluated value is 3.

Key Concepts

Convergence of Integrals

Convergence of integrals is all about determining if an improper integral sums to a finite value. It happens when the area under the curve from a starting point to infinity can be measured and is not infinite. For example, the integral \( \int_{0}^{\infty} 3 e^{-x} \, dx \) converges. This means the total area under the curve from 0 to infinity is finite, specifically equal to 3. To check if an integral converges:

• Identify the type of improper integral.
• Analyze the behavior of the function as it approaches the bounds of integration.
• Use limits to determine if the result is finite.

Here, the function \(3e^{-x}\) decreases towards zero as \(x\) increases. This decay helps ensure the area isn't infinite, allowing the integral to converge.

Infinite Interval of Integration

An infinite interval of integration occurs when the integral spans from a point to infinity, such as \( \int_{0}^{\infty} 3 e^{-x} \, dx \). This makes the integral improper since we cannot directly compute it over an unbounded area. To handle an infinite interval:

• Replace the infinite limit with a variable.
• Calculate the integral over a finite range, say from 0 to \(b\), where \(b\) will later approach infinity.

This approach transforms the problem into one of evaluating a simple definite integral and then analyzing its behavior as the limit approaches infinity.

Limit Evaluation

In the context of improper integrals, evaluating limits is the key step to solving them. Given the integral \( \int_{0}^{b} 3 e^{-x} \, dx \), we first integrate normally from 0 to \(b\), which gives us \(-3e^{-b} + 3\).The limit process:

• Substitute the expression from the evaluated definite integral into a limit operation.
• Let \( b \to \infty \).
• Examine the behavior of each term as \(b\) increases.

For our particular example, as \(b\) approaches infinity, \(e^{-b}\) tends towards zero, simplifying the expression to 3. Thus, the evaluated limit shows that the area converges to a finite value, confirming the integral's convergence.