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Magazine Chapter 7: Problem 11
Evaluate the integrals that converge. $$\int_{-\infty}^{0} e^{3 x} d x$$
Short Answer
Expert verified
The integral converges to \(\frac{1}{3}\).
Step by step solution
01
Set Up the Integral
The integral we need to evaluate is \int_{-\infty}^{0} e^{3x} \, dx . We start by recognizing it as a definite improper integral, because it extends from \(-\infty\) to 0.
02
Rewrite the Improper Integral
Because the integral is improper, we rewrite it with a limit: \[ \int_{-\infty}^{0} e^{3x} \, dx = \lim_{a \to -\infty} \int_{a}^{0} e^{3x} \, dx \] This enables us to evaluate the integral over a finite range \([a, 0]\) and then examine the behavior as \(a\) approaches \(-\infty\).
03
Evaluate the Indefinite Integral
Find the antiderivative of the integrand \(e^{3x}\). The antiderivative is \[ \frac{1}{3} e^{3x} + C \]where \(C\) is the constant of integration.
04
Apply the Limits of Integration
Apply the Fundamental Theorem of Calculus to evaluate \(\int_{a}^{0} e^{3x} \, dx\):\[ \begin{align*}\int_{a}^{0} e^{3x} \, dx &= \left[ \frac{1}{3} e^{3x} \right]_{a}^{0} \&= \frac{1}{3} e^{3 \times 0} - \frac{1}{3} e^{3a} \&= \frac{1}{3} - \frac{1}{3} e^{3a}.\end{align*} \]
05
Take the Limit as \(a\) Approaches \(-\infty\)
Now, take the limit:\[ \lim_{a \to -\infty} \left( \frac{1}{3} - \frac{1}{3} e^{3a} \right). \]As \(a\) approaches \(-\infty\), \(e^{3a}\) goes to zero because exponential functions decay to zero as their exponent approaches \(-\infty\). So,\[ \lim_{a \to -\infty} \left( \frac{1}{3} - \frac{1}{3} e^{3a} \right) = \frac{1}{3}. \]
06
Conclusion
Since the limit exists and equals \(\frac{1}{3}\), the integral converges and \int_{-\infty}^{0} e^{3x} \, dx = \frac{1}{3}.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral is a way to calculate the total accumulation of a quantity, such as area under a curve, over a specific interval. In the context of our original exercise, we deal with an improper definite integral:
- This definite integral is improper as it spans from \(-\infty\) to 0, making it necessary to evaluate the limit as \(-\infty\) is approached.
- We transition from an improper integral to a form with upper and lower bounds, \([a, 0]\), for practical evaluation, treating the lower bound as a variable that tends to \(-\infty\).
- The evaluated value of a definite integral gives a constant, representing the accumulation over the interval.
Antiderivative
An antiderivative of a function is another function whose derivative gives back the original function. Antiderivatives play a crucial role in solving integrals:
- In our case, the integrand is \(e^{3x}\), and its antiderivative is \(\frac{1}{3} e^{3x} + C\).
- The constant \(C\) usually appears in indefinite integrals; however, it cancels out in definite integrals, simplifying our calculations.
- Finding an antiderivative is akin to reversing differentiation, which is essential for solving integrals.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a powerful bridge between differentiation and integration.
- This theorem consists of two parts鈥搊ne relates the definite integral of a function to its antiderivative, while the other outlines how differentiating an integral function retrieves the original function.
- It simplifies the evaluation of definite integrals by allowing us to consider only the antiderivative at the boundaries of integration, \(\left[ F(b) - F(a) \right]\).
- In our exercise, we applied this theorem to evaluate the integral from \(a\) to 0, leading us from \(\frac{1}{3}e^{3 \times 0} - \frac{1}{3}e^{3a}\) to ultimately compute a convergent value of \(\frac{1}{3}\).
