Chapter 4: Integration
Section 4.5: Improper Integrals
Evaluate the improper integral ∫01x−1/3 ⅆx.
The function fx=x−1/3 has a vertical asymptote at x=0, the left endpoint of the interval of integration. Hence, the integral is improper and the evaluation must take place as per item 2a in Table 4.5.1.
∫01x−1/3 ⅆx=limt→0+∫t1x−1/3 ⅆx = limt→0+x2/32/3t1 = limt→0+321−t=32
Evaluate the integral
Control-drag the integral.
Context Panel: Evaluate and Display Inline
∫01x−1/3 ⅆx = 32
Solution from first principles
Control-drag the integral; edit lower limit to t.
Context Panel: Limit (See Figure 4.5.5(a).)
Figure 4.5.5(a) Limit dialog
∫t1x−1/3 ⅆx = −32⁢t2/3+32→limit32
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