Improper Integrals

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Determine whether the area between the graph of \(f(x)=\frac{1}{x}\) and the x-axis over the interval \([1,\text{+}\infty )\) is finite or infinite.

   Atskleisti atsakymą

   We first do a quick sketch of the region in question, as shown in the following graph.

   We can see that the area of this region is given by \(A={\int }_{1}^{\infty }\frac{1}{x}dx.\) Then we have

   \[\begin{array}{lllll}A & ={\int }_{1}^{\infty }\frac{1}{x}dx & & & \\ & =\underset{t\to \text{+}\infty }{\text{lim}}{\int }_{1}^{t}\frac{1}{x}dx & & & \text{Rewrite the improper integral as a limit.} \\ & =\underset{t\to \text{+}\infty }{\text{lim}}\text{ln}|x||{}_{\begin{array}{l} \\ 1\end{array}}^{\begin{array}{l}t \\ \end{array}} & & & \text{Find the antiderivative.} \\ & =\underset{t\to \text{+}\infty }{\text{lim}}(\text{ln}|t|-\text{ln}\ 1) & & & \text{Evaluate the antiderivative.} \\ & =\text{+}\infty . & & & \text{Evaluate the limit.}\end{array}\]

   Since the improper integral diverges to \(+\infty ,\) the area of the region is infinite.

2. Find the volume of the solid obtained by revolving the region bounded by the graph of \(f(x)=\frac{1}{x}\) and the x-axis over the interval \([1,\text{+}\infty )\) about the \(x\)-axis.

   Atskleisti atsakymą

   The solid is shown in . Using the disk method, we see that the volume V is

   \[V=\pi {\int }_{1}^{+\infty }\frac{1}{{x}^{2}}dx.\]

   Then we have

   \[\begin{array}{lllll}V & =\pi {\int }_{1}^{+\infty }\frac{1}{{x}^{2}}dx & & & \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}{\int }_{1}^{t}\frac{1}{{x}^{2}}dx & & & \text{Rewrite as a limit.} \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}-\frac{1}{x}|{}_{\begin{array}{l} \\ 1\end{array}}^{\begin{array}{l}t \\ \end{array}} & & & \text{Find the antiderivative.} \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}(\text{-}\ \frac{1}{t}+1) & & & \text{Evaluate the antiderivative.} \\ & =\pi . & & & \end{array}\]

   The improper integral converges to \(\pi .\) Therefore, the volume of the solid of revolution is \(\pi .\)

3. In the chapter opener, we stated the following problem: Suppose that at a busy intersection, traffic accidents occur at an average rate of one every three months. After residents complained, changes were made to the traffic lights at the intersection. It has now been ten months since the changes were made and there have been no accidents. Were the changes effective or is the 10-month interval without an accident a result of chance?

   Atskleisti atsakymą

   Revise to: Let \(x\) represent the amount of time it takes for the next accident to occur. We want to know how likely it is that \(x>10\). Define the rate parameter \(\lambda\) to be the average number of accidents per month. According to probability theory, for \(a>0\),

   \[P\left(x>a\right)={\int }_{a}^{\infty }\lambda {e}^{-\lambda x}dx,\]

   In this example, since one accident happens every three months, on average, \(\lambda =\frac{1}{3}\). The desired probability is:

   \[\begin{array}{l}P(x>10)={\int }_{10}^{\infty }\frac{1}{3}{e}^{-\frac{1}{3}x}dx \\ =\underset{t\to \infty }{\lim }{\int }_{10}^{t}\frac{1}{3}{e}^{-\frac{1}{3}x}dx \\ =\underset{t\to \infty }{\lim }-{e}^{-\frac{1}{3}x}{|}_{10}^{t} \\ =\underset{t\to \infty }{\lim }(-{e}^{-\frac{t}{3}}+{e}^{-\frac{10}{3}}) \\ \approx 0.0357\end{array}\]

   The value \(3.8\ \times \ {10}^{-11}\) represents the probability of no accidents in 8 months under the initial conditions. Since this value is very, very small, it is reasonable to conclude the changes were effective.

4. Evaluate \({\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx.\) State whether the improper integral converges or diverges.

   Atskleisti atsakymą

   Begin by rewriting \({\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx\) as a limit using from the definition. Thus,

   \[\begin{array}{lllll}{\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx & =\underset{x\to \text{-}\infty }{\text{lim}}{\int }_{t}^{0}\frac{1}{{x}^{2}+4}dx & & & \text{Rewrite as a limit.} \\ & =\underset{t\to \text{-}\infty }{\text{lim}}\frac{1}{2}{\text{tan}}^{-1}\frac{x}{2}|{}_{\begin{array}{l} \\ t\end{array}}^{\begin{array}{l}0 \\ \end{array}} & & & \text{Find the antiderivative.} \\ & =\frac{1}{2}\underset{t\to \text{-}\infty }{\text{lim}}({\text{tan}}^{-1}0-{\text{tan}}^{-1}\frac{t}{2}) & & & \text{Evaluate the antiderivative.} \\ & =\frac{\pi }{4}. & & & \text{Evaluate the limit and simplify.}\end{array}\]

   The improper integral converges to \(\frac{\pi }{4}.\)

5. Evaluate \({\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx.\) State whether the improper integral converges or diverges.

   Atskleisti atsakymą

   Start by splitting up the integral:

   \[{\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx={\int }_{\text{-}\infty }^{0}x{e}^{x}dx+{\int }_{0}^{+\infty }x{e}^{x}dx.\]

   If either \({\int }_{\text{-}\infty }^{0}x{e}^{x}dx\) or \({\int }_{0}^{+\infty }x{e}^{x}dx\) diverges, then \({\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx\) diverges. Compute each integral separately. For the first integral,

   \[\begin{array}{lllll}{\int }_{\text{-}\infty }^{0}x{e}^{x}dx & =\underset{t\to \text{-}\infty }{\text{lim}}{\int }_{t}^{0}x{e}^{x}dx & & & \text{Rewrite as a limit.} \\ & =\underset{t\to \text{-}\infty }{\text{lim}}(x{e}^{x}-{e}^{x})|{}_{\begin{array}{l} \\ t\end{array}}^{\begin{array}{l}0 \\ \end{array}} & & & \begin{array}{l}\text{Use integration by parts to find the} \\ \text{antiderivative. (Here}\ u=x\ \text{and}\ dv={e}^{x}\text{dx}.)\end{array} \\ & =\underset{t\to \text{-}\infty }{\text{lim}}(-1-t{e}^{t}+{e}^{t}) & & & \text{Evaluate the antiderivative.} \\ & =-1. & & & \begin{array}{l}\text{Evaluate the limit. }\ \text{Note:}\ \underset{t\to \text{-}\infty }{\text{lim}}t{e}^{t}\ \text{is} \\ \text{indeterminate of the form}\ 0\cdot \infty .\ \text{Thus,} \\ \underset{t\to \text{-}\infty }{\text{lim}}t{e}^{t}=\underset{t\to \text{-}\infty }{\text{lim}}\frac{t}{{e}^{\text{-}t}}=\underset{t\to \text{-}\infty }{\text{lim}}\frac{-1}{{e}^{\text{-}t}}=\underset{t\to \text{-}\infty }{\text{lim}}-{e}^{t}=0\ \text{by} \\ \text{L’Hôpital’s Rule.}\end{array}\end{array}\]

   The first improper integral converges. For the second integral,

   \[\begin{array}{lllll}{\int }_{0}^{+\infty }x{e}^{x}dx & =\underset{t\to \text{+}\infty }{\text{lim}}{\int }_{0}^{t}x{e}^{x}dx & & & \text{Rewrite as a limit.} \\ & =\underset{t\to \text{+}\infty }{\text{lim}}(x{e}^{x}-{e}^{x})|{}_{\begin{array}{l} \\ 0\end{array}}^{\begin{array}{l}t \\ \end{array}} & & & \text{Find the antiderivative.} \\ & =\underset{t\to \text{+}\infty }{\text{lim}}(t{e}^{t}-{e}^{t}+1) & & & \text{Evaluate the antiderivative.} \\ & =\underset{t\to \text{+}\infty }{\text{lim}}((t-1){e}^{t}+1) & & & \text{Rewrite.}\ (t{e}^{t}-{e}^{t}\ \text{is indeterminate.)} \\ & =\text{+}\infty . & & & \text{Evaluate the limit.}\end{array}\]

   Thus, \({\int }_{0}^{+\infty }x{e}^{x}dx\) diverges. Since this integral diverges, \({\int }_{\text{-}\infty }^{+\infty }x{e}^{x}dx\) diverges as well.

6. Evaluate \({\int }_{-3}^{+\infty }{e}^{\text{-}x}dx.\) State whether the improper integral converges or diverges.

   Atskleisti atsakymą

   \({e}^{3},\) converges

7. Evaluate \({\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx,\) if possible. State whether the integral converges or diverges.

   Atskleisti atsakymą

   The function \(f(x)=\frac{1}{\sqrt{4-x}}\) is continuous over \([0,4)\) and discontinuous at 4. Using from the definition, rewrite \({\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx\) as a limit:

   \[\begin{array}{lllll}{\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx & =\underset{t\to {4}^{-}}{\text{lim}}{\int }_{0}^{t}\frac{1}{\sqrt{4-x}}dx & & & \text{Rewrite as a limit.} \\ & =\underset{t\to {4}^{-}}{\text{lim}}(-2\sqrt{4-x})|{}_{\begin{array}{l} \\ 0\end{array}}^{\begin{array}{l}t \\ \end{array}} & & & \text{Find the antiderivative.} \\ & =\underset{t\to {4}^{-}}{\text{lim}}(-2\sqrt{4-t}+4) & & & \text{Evaluate the antiderivative.} \\ & =4. & & & \text{Evaluate the limit.}\end{array}\]

   The improper integral converges.

8. Evaluate \({\int }_{0}^{2}x\ \text{ln}\ x\ dx.\) State whether the integral converges or diverges.

   Atskleisti atsakymą

   Since \(f(x)=x\ \text{ln}\ x\) is continuous over \((0,2]\) and is discontinuous at zero, we can rewrite the integral in limit form using :

   \[\begin{array}{lllll}{\int }_{0}^{2}x\ \text{ln}\ x\ dx & =\underset{t\to {0}^{+}}{\text{lim}}{\int }_{t}^{2}x\ \text{ln}\ x\ dx & & & \text{Rewrite as a limit.} \\ & =\underset{t\to {0}^{+}}{\text{lim}}(\frac{1}{2}{x}^{2}\text{ln}\ x-\frac{1}{4}{x}^{2})|{}_{\begin{array}{l} \\ t\end{array}}^{\begin{array}{l}2 \\ \end{array}} & & & \begin{array}{l}\text{Evaluate}\ {\int }^{\text{}}x\ \text{ln}\ x\ dx\ \text{using integration by parts} \\ \text{with}\ u=\text{ln}\ x\ \text{and}\ dv=x\text{dx}.\end{array} \\ & =\underset{t\to {0}^{+}}{\text{lim}}(2\ \text{ln}\ 2-1-\frac{1}{2}{t}^{2}\text{ln}\ t+\frac{1}{4}{t}^{2}). & & & \text{Evaluate the antiderivative.} \\ & =2\ \text{ln}\ 2-1. & & & \begin{array}{l}\text{Evaluate the limit.}\ \underset{t\to {0}^{+}}{\text{lim}}{t}^{2}\text{ln}\ t\ \text{is indeterminate.} \\ \text{To evaluate it, rewrite as a quotient and apply} \\ \text{L’Hôpital’s rule.}\end{array}\end{array}\]

   The improper integral converges.

9. Evaluate \({\int }_{-1}^{1}\frac{1}{{x}^{3}}dx.\) State whether the improper integral converges or diverges.

   Atskleisti atsakymą

   Since \(f(x)=1\text{/}{x}^{3}\) is discontinuous at zero, using , we can write

   \[{\int }_{-1}^{1}\frac{1}{{x}^{3}}dx={\int }_{-1}^{0}\frac{1}{{x}^{3}}dx+{\int }_{0}^{1}\frac{1}{{x}^{3}}dx.\]

   If either of the two integrals diverges, then the original integral diverges. Begin with \({\int }_{-1}^{0}\frac{1}{{x}^{3}}dx:\)

   \[\begin{array}{lllll}{\int }_{-1}^{0}\frac{1}{{x}^{3}}dx & =\underset{t\to {0}^{-}}{\text{lim}}{\int }_{-1}^{t}\frac{1}{{x}^{3}}dx & & & \text{Rewrite as a limit.} \\ & =\underset{t\to {0}^{-}}{\text{lim}}(-\frac{1}{2{x}^{2}})|{}_{\begin{array}{l} \\ -1\end{array}}^{\begin{array}{l}t \\ \end{array}} & & & \text{Find the antiderivative.} \\ & =\underset{t\to {0}^{-}}{\text{lim}}(-\frac{1}{2{t}^{2}}+\frac{1}{2}) & & & \text{Evaluate the antiderivative.} \\ & =\text{-}\infty . & & & \text{Evaluate the limit.}\end{array}\]

   Therefore, \({\int }_{-1}^{0}\frac{1}{{x}^{3}}dx\) diverges. Since \({\int }_{-1}^{0}\frac{1}{{x}^{3}}dx\) diverges, \({\int }_{-1}^{1}\frac{1}{{x}^{3}}dx\) diverges.

10. Evaluate \({\int }_{0}^{2}\frac{1}{x}dx.\) State whether the integral converges or diverges.

    Atskleisti atsakymą

    \(+\infty ,\) diverges

11. Use a comparison to show that \({\int }_{1}^{+\infty }\frac{1}{x{e}^{x}}dx\) converges.

    Atskleisti atsakymą

    We can see that

    \[0\le \frac{1}{x{e}^{x}}\le \frac{1}{{e}^{x}}={e}^{\text{-}x},\]

    so if \({\int }_{1}^{+\infty }{e}^{\text{-}x}dx\) converges, then so does \({\int }_{1}^{+\infty }\frac{1}{x{e}^{x}}dx.\) To evaluate \({\int }_{1}^{+\infty }{e}^{\text{-}x}dx,\) first rewrite it as a limit:

    \[\begin{array}{ll}{\int }_{1}^{+\infty }{e}^{\text{-}x}dx & =\underset{t\to \text{+}\infty }{\text{lim}}{\int }_{1}^{t}{e}^{\text{-}x}dx \\ & =\underset{t\to \text{+}\infty }{\text{lim}}(\text{-}{e}^{\text{-}x})|\begin{array}{l}t \\ 1\end{array} \\ & =\underset{t\to \text{+}\infty }{\text{lim}}(\text{-}{e}^{\text{-}t}+{e}^{-1}) \\ & ={e}^{-1}.\end{array}\]

    Since \({\int }_{1}^{+\infty }{e}^{\text{-}x}dx\) converges, so does \({\int }_{1}^{+\infty }\frac{1}{x{e}^{x}}dx.\)

12. Use the comparison theorem to show that \({\int }_{1}^{+\infty }\frac{1}{{x}^{p}}dx\) diverges for all \(p<1.\)

    Atskleisti atsakymą

    For \(p<1,\) \(1\text{/}x\le 1\text{/}({x}^{p})\) over \([1,\text{+}\infty ).\) In , we showed that \({\int }_{1}^{+\infty }\frac{1}{x}dx=\text{+}\infty .\) Therefore, \({\int }_{1}^{+\infty }\frac{1}{{x}^{p}}dx\) diverges for all \(p<1.\)

13. Use a comparison to show that \({\int }_{e}^{+\infty }\frac{\text{ln}\ x}{x}dx\) diverges.

    Atskleisti atsakymą

    Since \({\int }_{e}^{+\infty }\frac{1}{x}dx=\text{+}\infty ,\) \({\int }_{e}^{+\infty }\frac{\text{ln}\ x}{x}dx\) diverges.

14. \({\int }_{2}^{4}\frac{dx}{{(x-3)}^{2}}\)

    Atskleisti atsakymą

    divergent

15. \({\int }_{0}^{\infty }\frac{1}{4+{x}^{2}}dx\)

16. \({\int }_{0}^{2}\frac{1}{\sqrt{4-{x}^{2}}}dx\)

    Atskleisti atsakymą

    \(\frac{\pi }{2}\)

17. \({\int }_{1}^{\infty }\frac{1}{x\ \text{ln}\ x}dx\)

18. \({\int }_{1}^{\infty }x{e}^{\text{-}x}dx\)

    Atskleisti atsakymą

    \(\frac{2}{e}\)

19. \({\int }_{\text{-}\infty }^{\infty }\frac{x}{{x}^{2}+1}dx\)

20. Without integrating, determine whether the integral \({\int }_{1}^{\infty }\frac{1}{\sqrt{{x}^{3}+1}}dx\) converges or diverges by comparing the function \(f(x)=\frac{1}{\sqrt{{x}^{3}+1}}\) with \(g(x)=\frac{1}{\sqrt{{x}^{3}}}.\)

    Atskleisti atsakymą

    Converges

21. Without integrating, determine whether the integral \({\int }_{1}^{\infty }\frac{1}{\sqrt{x+1}}dx\) converges or diverges.

22. \({\int }_{0}^{\infty }{e}^{\text{-}x}\text{cos}\ x\ dx\)

    Atskleisti atsakymą

    Converges to 1/2.

23. \({\int }_{1}^{\infty }\frac{\text{ln}\ x}{x}dx\)

24. \({\int }_{0}^{1}\frac{\text{ln}\ x}{\sqrt{x}}dx\)

    Atskleisti atsakymą

    −4

25. \({\int }_{0}^{1}\text{ln}\ x\ dx\)

26. \({\int }_{\text{-}\infty }^{\infty }\frac{1}{{x}^{2}+1}dx\)

    Atskleisti atsakymą

    \(\pi\)

27. \({\int }_{1}^{5}\frac{dx}{\sqrt{x-1}}\)

28. \({\int }_{-2}^{2}\frac{dx}{{(1+x)}^{2}}\)

    Atskleisti atsakymą

    diverges

29. \({\int }_{0}^{\infty }{e}^{\text{-}x}dx\)

30. \({\int }_{0}^{\infty }\text{sin}\ x\ dx\)

    Atskleisti atsakymą

    diverges

31. \({\int }_{\text{-}\infty }^{\infty }\frac{{e}^{x}}{1+{e}^{2x}}dx\)

32. \({\int }_{0}^{1}\frac{dx}{\sqrt[3]{x}}\)

    Atskleisti atsakymą

    1.5

33. \({\int }_{0}^{2}\frac{dx}{{x}^{3}}\)

34. \({\int }_{-1}^{2}\frac{dx}{{x}^{3}}\)

    Atskleisti atsakymą

    diverges

35. \({\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}\)

36. \({\int }_{0}^{3}\frac{1}{x-1}dx\)

    Atskleisti atsakymą

    diverges

37. \({\int }_{1}^{\infty }\frac{5}{{x}^{3}}dx\)

38. \({\int }_{3}^{5}\frac{5}{{(x-4)}^{2}}dx\)

    Atskleisti atsakymą

    diverges

39. \({\int }_{1}^{\infty }\frac{dx}{{x}^{2}+4x};\) compare with \({\int }_{1}^{\infty }\frac{dx}{{x}^{2}}.\)

40. \({\int }_{1}^{\infty }\frac{dx}{\sqrt{x}+1};\) compare with \({\int }_{1}^{\infty }\frac{dx}{2\sqrt{x}}.\)

    Atskleisti atsakymą

    Both integrals diverge.

Pabandyk savo pačių

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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