Question

Explain why the integral is improper and determine whether it diverges or converges. Evaluate the integral if it converses. $$ \int_{-\infty}^{0} e^{2 x} d x $$

Short answer

The given improper integral is convergent and its value is \(\frac{1}{2}\).

Step-by-step solution

Identify the improper integral

The given integral, \(\int_{-\infty}^{0} e^{2 x} d x\), is an improper integral because the lower limit of integration is \(-\infty\).

Rewrite the integral with finite limits using a variable substitution

Start by changing the lower limit of integration from \(-\infty\) to a variable \(t\), which will approach \(-\infty\). So, the given integral becomes \(\lim_{t \to -\infty} \int_{t}^{0} e^{2 x} d x\).

Apply the Fundamental Theorem of Calculus

Next, compute the integral within the limit. According to the Fundamental Theorem of Calculus, the antiderivative of \(e^{2x}\) is \(\frac{1}{2}e^{2x}\). Applying this from \(t\) to \(0\), the expression becomes \(\lim_{t \to -\infty} [\frac{1}{2} e^{2(0)} - \frac{1}{2}e^{2t}]\). Calculate this to get \(\lim_{t \to -\infty} [\frac{1}{2} - \frac{1}{2}e^{2t}]\).

Evaluate the limit

As \(t\) approaches \(-\infty\), \(e^{2t}\) approaches 0. This is because any exponential function with a negative exponent approaches 0 as the exponent approaches \(-\infty\). Therefore, the limit is \(\frac{1}{2} - 0 = \frac{1}{2}\).

Key concepts

Convergence and Divergence

Improper integrals can be a bit tricky to understand at first because they involve infinite limits, like our given integral \( \int_{-\infty}^{0} e^{2x} \, dx \). An integral is considered improper if it involves infinite limits or an undefined point in its range. In our case, the lower limit is \(-\infty\), which makes it improper.
To determine if such an integral converges or diverges, you substitute the infinite limit with a variable that approaches infinity. This turns our integral into a limit problem. In simpler terms:

• If the limit exists and equals a finite number, the integral converges.
• If the limit does not exist or is infinite, the integral diverges.

For \( \int_{-\infty}^{0} e^{2x} \), after proper substitution and calculation, the limit results in a finite value (\( \frac{1}{2} \)), indicating convergence. The integral therefore converges to \( \frac{1}{2} \). This helps in understanding whether our calculations in the realms of infinity make sense within a finite answer.

Exponential Functions

Exponential functions are one of the founding blocks in calculus, especially when integrating or deriving them. The function \( e^{2x} \) is an exponential function.
For exponential functions, growth or decay is determined by the exponent's sign. A positive exponent signifies growth, while a negative exponent signifies decay. In our improper integral example, the exponent \( 2x \) grows negatively as \( x \) approaches \(-\infty\). This makes the overall value of \( e^{2x} \) approach zero, which is crucial for evaluating limits.
When integrating, it is vital to recognize these behaviors, as they dictate the integral's convergence or divergence. With the exponential decay as \( x \to -\infty \), understanding this natural logarithmic behavior influences the resultant finite value from the integral. Knowing how to maneuver and manipulate exponential expressions is essential for evaluating improper integrals like this one.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration—two key aspects of calculus. It comes in two parts; for our exercise, the second part is most relevant. This theorem tells us that if we have a continuous function over a closed interval, the integral of a function's derivative can be evaluated using the antiderivatives at the boundary points.
In this problem, we calculate \( \int e^{2x} \, dx \) by finding its antiderivative, which is \( \frac{1}{2}e^{2x} \). The Fundamental Theorem allows us to plug this expression into our new limits to evaluate the integral.

• First, substitute the upper and lower bounds into the antiderivative.
• Then, compute the difference of these values to get the result.

For our case, the result after applying the theorem and limits was \( \frac{1}{2} - \frac{1}{2}e^{2t} \). As \( t \to -\infty \), the term \( e^{2t} \to 0 \). Hence, the integral evaluates to \( \frac{1}{2} \)—a simple but powerful demonstration of the theorem's capability in evaluating improper integrals.