Problem 49 Which of the following improper ... [FREE SOLUTION]

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AP Calculus with 8 Practice Tests 15th

Dennis Donovan

$Math Studyset 91影视 Explanations$ Math

15 Edition

Chapter 11: Problem 49

Which of the following improper integrals diverges? (A) $\int_{-\infty}^{0} e^{x} d x$ (B) $\int_{0}^{1} \frac{d x}{x}$ (C) $\int_{0}^{\infty} e^{-x} d x$ (D) $\int_{0}^{1} \frac{d x}{\sqrt{x}}$

Short Answer

Expert verified

Integral (B) diverges due to the divergence of $ \ln |x| $ at $ x = 0 $.

Step by step solution

Evaluate Integral (A)

Consider the integral $ \int_{-\infty}^{0} e^{x} \, dx $. Since $ e^{x} $ quickly approaches zero as $ x \to -\infty $, we need to find the limit: \[ \lim_{t \to -\infty} \int_{t}^{0} e^{x} \, dx. \] The antiderivative of $ e^{x} $ is itself $ e^{x} $. Thus, the integral becomes $ \lim_{t \to -\infty} [e^{x}]_{t}^{0} = 1 - \lim_{t \to -\infty} e^{t}. $ Since $ e^{t} \to 0 $ as $ t \to -\infty $, the integral converges to 1.

Evaluate Integral (B)

Consider $ \int_{0}^{1} \frac{dx}{x} $. This is an improper integral at $ x = 0 $. Evaluate it as $ \lim_{t \to 0^{+}} \int_{t}^{1} \frac{dx}{x} $. The antiderivative of $ \frac{1}{x} $ is $ \ln |x| $, so the integral is $ \lim_{t \to 0^{+}} [\ln |x|]_{t}^{1} = 0 - \ln |t|. $ As $ t \to 0^{+} $, $ \ln |t| \to -\infty $, indicating divergence.

Evaluate Integral (C)

Consider $ \int_{0}^{\infty} e^{-x} \, dx $. Since $ e^{-x} \to 0 $ as $ x \to \infty $, use $ \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx $. The antiderivative is $ -e^{-x} $, so the integral is $ \lim_{b \to \infty} [-e^{-x}]_{0}^{b} = \lim_{b \to \infty} (-(0) + 1) = 1. $ Thus, the integral converges.

Evaluate Integral (D)

Consider $ \int_{0}^{1} \frac{dx}{\sqrt{x}} $. This is improper at $ x = 0 $. Evaluate it as $ \lim_{t \to 0^{+}} \int_{t}^{1} x^{-1/2} \cdot dx $. The antiderivative of $ x^{-1/2} $ is $ 2\sqrt{x} $, yielding $ \lim_{t \to 0^{+}} [2 \sqrt{x}]_{t}^{1} = 2 - \lim_{t \to 0^{+}} 2\sqrt{t} $. As $ t \to 0^{+}, \sqrt{t} \to 0 $, so the integral converges to 2.

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Key Concepts

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

Convergence

Convergence in improper integrals is all about whether or not a specific integral results in a finite number. Imagine pouring water into a container; if it fills up without spilling, the process is finite, much like a convergent integral.

For improper integrals, this assessment involves limits. If the limit exists and results in a finite number, the integral converges. Look at the integral: \[ \int_{-\infty}^{0} e^{x} \, dx\]In this case, since the function $ e^{x} $ approaches zero rapidly as $ x \to -\infty $, the limit converges to 1:

The convergence was validated by evaluating the limit: $ \lim_{t \to -\infty} [e^{x}]_{t}^{0} = 1 - 0 = 1 $

Divergence

Divergence occurs when an improper integral does not approach a finite value. Think of a container with a hole; if water keeps leaking, it never fills "completely" 鈥� this mirrors a divergent integral.

Take the integral: \[ \int_{0}^{1} \frac{dx}{x}\] As you try to evaluate the limit as $ t \to 0^{+} $, like near an undefined point at zero, we find:

$ \ln|t| $ approaching $ -\infty $

This indicates divergence because the integral's value continuously moves towards infinity and has no finite bounds.

Antiderivative

Finding an antiderivative is like unraveling a sweater back to its yarn, returning the function to its roots. It鈥檚 re-engineering to see what function has the given derivative.

For example, the function $ e^{x} $ serves as its own antiderivative, which makes evaluating integrals like $ \int e^{x} \, dx $ straightforward:

The antiderivative of $ e^x $ is simply $ e^x $ itself.

Similarly, for $ x^{-1/2} $, we find:

The antiderivative is $ 2\sqrt{x} $, which illustrates how the functions behind integral convergence are often straightforward to deduce.

Limit

A limit in calculus is a way to understand what value a function approaches as the input gets closer and closer to a specific point. Imagine driving towards a stop sign; as you get nearer, your speed decreases to zero, representing the idea of convergence or divergence in integrals.

For improper integrals,

Limits help define behavior at infinity or points of discontinuity.

Consider $ \int_{0}^{\infty} e^{-x} dx $:

Here, $ \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx $ assesses the integral's convergence as $ x \to \infty $ and finds the limit resolves to 1.

A limit essentially helps check if an area under a graph exists finitely between certain boundaries.

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91影视

Which of the following improper integrals diverges? (A) \(\int_{-\infty}^{0} e^{x} d x\) (B) \(\int_{0}^{1} \frac{d x}{x}\) (C) \(\int_{0}^{\infty} e^{-x} d x\) (D) \(\int_{0}^{1} \frac{d x}{\sqrt{x}}\)

Short Answer

Step by step solution

Evaluate Integral (A)

Evaluate Integral (B)

Evaluate Integral (C)

Evaluate Integral (D)

Key Concepts

Convergence

Divergence

Antiderivative

Limit

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