Problem 1 - Calculus for Biology and Medicine

Question

All the integrals in Problems \(1-16\) are improper and converge. Explain in each case why the integral is improper, andevaluate each integral. $$ \int_{0}^{\infty} 3 e^{-6 x} d x $$

Step-by-Step Solution

Answer

The integral converges to \(\frac{1}{2}\).

Step 1: Identify Why the Integral is Improper

An improper integral occurs when the interval of integration is infinite or when the integrand has an infinite discontinuity. In this case, the integral \(\int_{0}^{\infty} 3 e^{-6x} \, dx\) is improper due to the upper limit being infinity.

Step 2: Set Up the Limit

To evaluate an improper integral with an infinite limit, we replace the infinity symbol \(\infty\) with a limit. Rewrite the integral as \(\lim_{a \to \infty} \int_{0}^{a} 3 e^{-6x} \, dx\).

Step 3: Evaluate the Integral

First, find the antiderivative of the integrand. The antiderivative of \(3e^{-6x}\) is \(-\frac{1}{2}e^{-6x}\). This is determined because taking the derivative of \(-\frac{1}{2}e^{-6x}\) with respect to \(x\) yields \(3e^{-6x}\). Substitute the antiderivative back into the integral: \(\int 3 e^{-6x} \, dx = -\frac{1}{2}e^{-6x}\).

Step 4: Apply the Fundamental Theorem of Calculus

Apply the limits of integration by evaluating the antiderivative from \(0\) to \(a\): \[-\frac{1}{2}e^{-6a} - \left(-\frac{1}{2}e^{-6 \cdot 0}\right) = -\frac{1}{2}e^{-6a} + \frac{1}{2}\].

Step 5: Take the Limit

Now, take the limit as \(a\) approaches infinity: \[\lim_{a \to \infty} \left(-\frac{1}{2}e^{-6a} + \frac{1}{2} \right)\]. As \(a\) becomes very large, \(e^{-6a}\) approaches zero, so the expression simplifies to \[0 + \frac{1}{2} = \frac{1}{2}\].

Key Concepts

Infinite Limits

Improper integrals often involve infinite limits as seen with the integral \( \int_{0}^{\infty} 3 e^{-6 x} d x \).This means that part of the integration range spans infinitely—
that is why we replace the upper limit of infinity with a variable and take the limit of that variable as it approaches infinity.
When dealing with these types of integrals, it's important to recognize:

• The interval of integration goes toward infinity, typically marked by \( \infty \) as a limit.
• Using a limit allows us to work within a finite structure, which simplifies calculations.

In other words, infinite limits transform integrals that might otherwise be undefined, into a sequence limit that can often converge to a finite number.

Antiderivative Calculation

Finding the antiderivative is a key step when solving integrals, particularly improper ones.
The original integrand, \(3e^{-6x}\), requires us to discover a function whose derivative gives the original integrand.
In this case, the antiderivative is derived as \(-\frac{1}{2}e^{-6x}\).The calculation follows the process of integration, essentially reversing derivatives.

• Start by recognizing coefficients, such as the constant \(3\), which influence only the magnitude.
• The exponential function change (from \(e^{-6x}\) to \(-\frac{1}{2}e^{-6x}\)) results from the chain rule in calculus, adjusting for the inner derivative of the exponent.

The accuracy of this calculation ensures that upon differentiation, we return to the original function. This checks the correctness of the antiderivative.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges differentiation and integration,
paving the way to evaluate definite integrals efficiently.
The theorem states that if \(F\) is an antiderivative of \(f\), then \( \int_a^b f(x) \, dx = F(b) - F(a) \).In our example, after finding the antiderivative \(-\frac{1}{2}e^{-6x}\),
we apply the definite limits from \(0\) to \(a\):

• Calculate \(-\frac{1}{2}e^{-6a} + \frac{1}{2}\). This uses the limits \(0\) and \(a\).
• Realize that as \(a\) reaches infinity, \(e^{-6a} \) works toward zero, simplifying our expression to \(\frac{1}{2}\).

Ultimately, the theorem transforms our problem into an easier one, by allowing evaluation through straightforward subtraction and limit calculations.