Chapter 4: Integration
Section 4.5: Improper Integrals
Example 4.5.6
Evaluate the improper integral ∫−12x−2 ⅆx.
Solution
Mathematical Solution
The integrand has a vertical asymptote at x=0; hence, the integral must be split at the asymptote, as per 2c in Table 4.5.1.
∫−12x−2 ⅆx
=lims→0−∫−1sx−2 ⅆx+limt→0+∫t2x2 ⅆx
=lims→0−−x−1−1s+limt→0+−x−1t2
=lims→0−−1s+1+limt→0+−12+1t
= ∞+∞
= ∞
The integral diverges. Actually, it is enough to show that the integral on either side of the asymptote diverges to establish that the integral itself diverges.
Maple Solution
Evaluate the integral
Control-drag the integral.
Context Panel: Evaluate and Display Inline
∫−12x−2 ⅆx = ∞
Evaluate ∫−10x−2 ⅆx from first principles
Control-drag the integral; edit upper limit to s.
Context Panel: Simplify≻Assuming Real Range (See Figure 4.5.6(a).)
Context Panel: Limit (See Figure 4.5.6(b).)
∫−1sx−2 ⅆx→assuming real range−s+1s→limit∞
Figure 4.5.6(a) Simplify
Figure 4.5.6(b) Limit
Evaluate ∫02x−2 ⅆx from first principles
Control-drag the integral; edit lower limit to t.
Context Panel: Simplify≻Assuming Real Range (See Figure 4.5.6(c).)
Context Panel: Limit (See Figure 4.5.6(d).)
∫t2x−2 ⅆx→assuming real range−12t−2t→limit∞
Figure 4.5.6(c) Simplify
Figure 4.5.6(d) Limit
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