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Chapter 7: Problem 19

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_{-\infty}^0 ze^{2z}\ dz \)

Short Answer

Expert verified
The integral is convergent with value of -1/4.

Step by step solution

01

Analyze the Improper Integral

The given integral is \( \int_{-\infty}^0 ze^{2z}\, dz \) which is an improper integral because it involves infinity as one of its limits. We need to determine its convergence or divergence by evaluating the limit of a proper integral with finite bounds, approaching the given infinite bound.
02

Set Up the Limit for Evaluation

Express the integral with a limit:\[\lim_{a \to -\infty} \int_{a}^0 ze^{2z}\, dz.\] This step converts the improper integral into a limit problem.
03

Determine the Antiderivative

Find the antiderivative of the integrand \( ze^{2z} \). Use integration by parts:\[\begin{align*}u & = z, & du & = dz, \du & = e^{2z} \cdot 2\, dz, & u & = \frac{1}{2}e^{2z}.\end{align*}\]Thus, \[ \int ze^{2z} \ dz = \frac{1}{2}ze^{2z} - \frac{1}{4}e^{2z} + C.\]
04

Evaluate the Definite Integral

Substitute the bounds into the antiderivative: \[\lim_{a \to -\infty} \left( \frac{1}{2}[0 \cdot e^{0}] - \frac{1}{4}e^{0} \right) - \left( \frac{1}{2}[a e^{2a}] - \frac{1}{4}e^{2a} \right).\] This simplifies to:\[\lim_{a \to -\infty} \left( -\frac{1}{4} + \frac{1}{2}ae^{2a} - \frac{1}{4}e^{2a} \right).\]
05

Simplify the Expression and Find Convergence

Observe as \( a \to -\infty \):- The term \( \frac{1}{2}ae^{2a} \) grows towards zero (products of exponential function with negative exponents go to zero).- Also, \( \frac{1}{4}e^{2a} \) tends to zero.Thus: \[\lim_{a \to -\infty} \left( -\frac{1}{4} + 0 - 0 \right) = -\frac{1}{4}.\]
06

Conclude Convergence or Divergence

The evaluated limit exists and is finite. Thus, the integral \( \int_{-\infty}^0 ze^{2z}\, dz \) is convergent with a result of \(-\frac{1}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by parts is a technique used to find the integral of a product of two functions. It is anchored in the integration rule derived from the product rule for differentiation. The formula for integration by parts is \[ \int u\, dv = uv - \int v\, du \], where:
  • u and dv are parts of the original integrand.
  • du is the differential of u.
  • v is the integral of dv.
In our example, the integrand is \( ze^{2z} \). We choose:
  • u = z (so du = dz)
  • dv = \( e^{2z} \, dz \) (leading to \( v = \frac{1}{2}e^{2z} \))
Following the integration by parts formula:\[ \int ze^{2z} \, dz = \frac{1}{2}ze^{2z} - \frac{1}{4}e^{2z} + C \].This step is crucial as it helps us to express the integral in a more manageable form for evaluation.
Convergence and Divergence
In the context of improper integrals, convergence and divergence refer to whether the integral approaches a finite number or not. It is important to first express the improper integral using a limit due to its infinity or finite discontinuity. In the problem of \( \int_{-\infty}^0 ze^{2z}\, dz \), we perceived it symmetrically as:
  • \( \lim_{{a \to -\infty}} \int_a^0 ze^{2z} \, dz \)
We then substitute the bounds into the antiderivative and simplify.
The convergence of an integral ensures we get a specific, finite result as \( a \to -\infty \). If the evaluated integral returns a real number, we determine it is convergent. In this exercise, we reached \(-\frac{1}{4}\), indicating convergence.
Limit Evaluation
Evaluating limits is an essential step when working with improper integrals. Here, it assists in transitioning from an expression that involves infinity into a more manageable finite mathematical computation.After applying integration by parts, we put the limits of the evaluated antiderivative, which formed:
  • \( \lim_{{a \to -\infty}} \left( -\frac{1}{4} + \frac{1}{2}ae^{2a} - \frac{1}{4}e^{2a} \right) \)
Simplifying this expression involves recognizing behaviors of expressions involving limits. As \( a \to -\infty \):
  • The term \( \frac{1}{2}ae^{2a} \) approaches zero.
  • Similarly, \( \frac{1}{4}e^{2a} \) also trends towards zero.
Hence, these zero contributions simplify the expression to \[ -\frac{1}{4} + 0 - 0 = -\frac{1}{4} \].This conclusion not only helps us validate convergence but also employs logical limit behavior to ascertain a specific answer.

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Most popular questions from this chapter

(a) If \( g(x) = \frac{(\sin^2 x)}{x^2} \), use your calculator or computer to make a table of approximate values of \( \displaystyle \int_1^t g(x)\ dx \) for \( t = 2, 5, 10, 100, 1000 \), and \( 10,000 \). Does it appear that \( \displaystyle \int_1^\infty g(x)\ dx \) is convergent? (b) Use the Comparison Theorem with \( f(x) = \frac{1}{x^2} \) to show that \( \displaystyle \int_1^\infty g(x)\ dx \) is convergent. (c) Illustrate part (b) by graphing \( f \) and \( g \) on the same screen for \( 1 \le x \le 10 \). Use your graph to explain intuitively why \( \displaystyle \int_1^\infty g(x)\ dx \) is convergent.

Sketch the region and find its area (if the area is finite). \( S = \\{ (x, y) \mid x \ge 1, 0 \le y \le \frac{1}{(x^3 + x)} \\} \)

The region bounded by the curve \( y = \frac{1}{(1 + e^{-x})} \), the x and y-axes, and the line \( x = 10 \) is rotated about the x-axis. Use Simpson's Rule with \( n = 10 \) to estimate the volume of the resulting solid.

If \( f(t) \) is continuous for \( t \ge 0 \), the Laplace transform of \( f \) is the function \( F \) defined by $$ F(s) = \int_0^\infty f(t) e^{-st}\ dt $$ and the domain of \( F \) is the set consisting of all numbers s for which the integral converges. Find the Laplace transforms of the following functions. (a) \( f(t) = 1 \) (b) \( f(t) = e^t \) (c) \( f(t) = t \)

Find the area under the curve \( y = \frac{1}{x^3} \) from \( x = 1 \) to \( x = t \) and evaluate it for \( t = 10, 100 \), and \( 1000 \). Then find the total area under this curve for \( x \ge 1 \).
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