Chapter 5: Q. 31 (page 478)

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{1}^{\infty} \frac{1}{{(x - 2)}^{2}} d x$

Short Answer

Expert verified

The improper integral $\int_{1}^{\infty} \frac{1}{{(x - 2)}^{2}} d x$ diverges.

Step by step solution

Step 1. Given information.

The given integral is the following.

$\int_{1}^{\infty} \frac{1}{{(x - 2)}^{2}} d x$

Step 2. Value of integral.

The function $f (x) = \frac{1}{{(x - 2)}^{2}}$ has a vertical asymptote at $x = 2 .$

solve this improper integral by splitting at $x = 2$

role="math" localid="1648878472203" $\int_{1}^{\infty} \frac{1}{{(x - 2)}^{2}} d x = \int_{1}^{2} \frac{1}{{(x - 2)}^{2}} d x + \int_{2}^{\infty} \frac{1}{{(x - 2)}^{2}} d x$

Take role="math" localid="1648878379284" $\int_{1}^{2} \frac{1}{{(x - 2)}^{2}} d x$

role="math" localid="1648878372798" $\int_{0}^{2} \frac{1}{{(x - 2)}^{2}} d x = \lim_{b \to 0} \int_{b}^{2} \frac{1}{{(x - 2)}^{2}} d x = \lim_{b \to 0} {[\frac{1}{(x - 2)}]}_{b}^{2} = \lim_{b \to 0} (1 / 0 - \frac{1}{(b - 2)}) = \frac{1}{0} + \frac{1}{2} = undefined$

Integral $\int_{1}^{2} \frac{1}{{(x - 2)}^{2}} d x$ is undefined and diverges.

So integral $\int_{1}^{\infty} \frac{1}{{(x - 2)}^{2}} d x$ is also diverged.

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Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{1}^{\infty} \frac{1}{{(x - 2)}^{2}} d x$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of integral.

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Use limits of definite integrals to calculate each of the improper integrals in Exercises.∫1∞1x-22dx

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of integral.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Logic and Functions

Applied Mathematics

Statistics

Calculus

Study anywhere. Anytime. Across all devices.

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{1}^{\infty} \frac{1}{{(x - 2)}^{2}} d x$