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3 Look at the boxed result on page 417. Now determine which of the following improper integrals converges. Circle them. dx dx x² -1/2 dx dx x√√x 1/2 dx

3 Look at the boxed result on page 417. Now determine which of the following improper integrals converges. Circle them. dx dx x² -1/2 dx dx x√√x 1/2 dx

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.5: Other Types Of Equations
Problem 54E
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y FIGURE 6 0 a FIGURE 7 0 y = f(x) area = T tb X x=b X Since 1/(1+x²) > 0, the given improper integral can be interpreted as the area of the infinite region that lies under the curve y = 1/(1+x²) and above the x-axis (see Figure 6). EXAMPLE 4 For what values of p is the integral 1 S₁2=5 dx XP convergent? SOLUTION We know from Example 1 that if p = 1, then the integral is divergent, so let's assume that p 1. Then ST XP 2 dx = lim f'x Sind SECTION 5.10 IMPROPER INTEGRALS 417 = lim X=P+1 -p+1 1 1-1-p = lim x™P dx 7x-1 =1 If p 1, then p-10, so as t→∞, P→∞ and 1/t10. Therefore 1²-=-=dx=p²-1 if p > 1 P- 1 *P-1 and so the integral converges. But if p < 1, then p - 1 <0 and so 1 =1¹-P-0 and the integral diverges. We summarize the result of Example 4 for future reference: as 100 dx is convergent if p > 1 and divergent if p < 1. Type 2: Discontinuous Integrands Suppose that f is a positive continuous function defined on a finite interval [a, b) but has a vertical asymptote at b. Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a horizontal direction. Here the region is infinite in a vertical direction.) The area of the part of S between a and t (the shaded region in Figure 7) is A(t) = f' f(x) dx If it happens that A(t) approaches a definite number A as t→b, then we say that the area of the region S is A and we write f"f(x) dx = lim f(x) dx We use this equation to define an improper integral of Type 2 even when f is not a posi- tive function, no matter what type of discontinuity f has at b.
Transcribed Image Text:y FIGURE 6 0 a FIGURE 7 0 y = f(x) area = T tb X x=b X Since 1/(1+x²) > 0, the given improper integral can be interpreted as the area of the infinite region that lies under the curve y = 1/(1+x²) and above the x-axis (see Figure 6). EXAMPLE 4 For what values of p is the integral 1 S₁2=5 dx XP convergent? SOLUTION We know from Example 1 that if p = 1, then the integral is divergent, so let's assume that p 1. Then ST XP 2 dx = lim f'x Sind SECTION 5.10 IMPROPER INTEGRALS 417 = lim X=P+1 -p+1 1 1-1-p = lim x™P dx 7x-1 =1 If p 1, then p-10, so as t→∞, P→∞ and 1/t10. Therefore 1²-=-=dx=p²-1 if p > 1 P- 1 *P-1 and so the integral converges. But if p < 1, then p - 1 <0 and so 1 =1¹-P-0 and the integral diverges. We summarize the result of Example 4 for future reference: as 100 dx is convergent if p > 1 and divergent if p < 1. Type 2: Discontinuous Integrands Suppose that f is a positive continuous function defined on a finite interval [a, b) but has a vertical asymptote at b. Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a horizontal direction. Here the region is infinite in a vertical direction.) The area of the part of S between a and t (the shaded region in Figure 7) is A(t) = f' f(x) dx If it happens that A(t) approaches a definite number A as t→b, then we say that the area of the region S is A and we write f"f(x) dx = lim f(x) dx We use this equation to define an improper integral of Type 2 even when f is not a posi- tive function, no matter what type of discontinuity f has at b.
3 Look at the boxed result on page 417. Now determine which of the following improper integrals converges. Circle them. dx 00 Ĵ 1 -1/2 dx 1 dx x√x 00 3dx 1/2 dx x²
Transcribed Image Text:3 Look at the boxed result on page 417. Now determine which of the following improper integrals converges. Circle them. dx 00 Ĵ 1 -1/2 dx 1 dx x√x 00 3dx 1/2 dx x²
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