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Integration

1.12 Improper Integrals

27 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

For which values of bb is the integral ∫0b1x2−1 dx\displaystyle\int_0^b \frac{1}{x^2-1}~\dee{x} improper?

Hint

There are two kinds of impropreity in an integral: an infinite discontinuity in the integrand, and an infinite limit of integration.

Answer

Any real number in [1,∞)[1,\infty) or (−∞,−1](-\infty,-1], and b=±∞b = \pm \infty.

Full solution
Q2Stage 1

For which values of bb is the integral ∫0b1x2+1 dx\displaystyle\int_0^b \frac{1}{x^2+1}~\dee{x} improper?

Hint

The integrand is continuous for all xx.

Answer

b=±∞b = \pm\infty

Full solution
Q3Stage 1

Below are the graphs y=f(x)y=f(x) and y=g(x)y=g(x). Suppose ∫0∞f(x) dx\displaystyle\int_0^\infty f(x)~\dee{x} converges, and ∫0∞g(x) dx\displaystyle\int_0^\infty g(x)~\dee{x} diverges. Assuming the graphs continue on as shown as x→∞x \to \infty, which graph is f(x)f(x), and which is g(x)g(x)?

Figure from prob_s1.12, line 2

Figure from prob_s1.12, line 2

Hint

What matters is which function is bigger for large values of xx, not near the origin.

Answer

The red function is f(x)f(x), and the blue function is g(x)g(x).

Full solution
Q4Stage 1Past exam · 2015A

Decide whether the following statement is true or false. If false, provide a counterexample. If true, provide a brief justification. (Assume that f(x)f(x) and g(x)g(x) are continuous functions.)

  1. If ∫1∞f(x) dx\displaystyle\int_{1}^{\infty} f(x) \,\dee{x} converges and g(x)≥f(x)≥0g(x)\ge f(x)\ge 0 for all xx, then ∫1∞g(x) dx\displaystyle\int_{1}^{\infty} g(x) \,\dee{x} converges.

Hint

Read both the question and Theorem 1.12.17 in the

CLP-2 text very carefully.

Answer

False. For example, the functions f(x)=e−xf(x)=e^{-x} and g(x)=1g(x)=1 provide a counterexample.

Full solution
Q5Stage 1

Let f(x)=e−xf(x) = e^{-x} and g(x)=1x+1g(x)=\dfrac{1}{x+1}. Note ∫0∞f(x) dx\int_{0}^\infty f(x)~\dee{x} converges while ∫0∞g(x) dx\int_{0}^\infty g(x)~\dee{x} diverges.

For each of the functions h(x)h(x) described below, decide whether ∫012∞h(x) dx\int_{0\vphantom{\frac12}}^\infty h(x)~\dee{x} converges or diverges, or whether there isn't enough information to decide. Justify your decision.

  1. h(x)h(x), continuous and defined for all x≥0x \ge0, h(x)≤f(x)h(x) \leq f(x).

  2. h(x)h(x), continuous and defined for all x≥0x\ge 0, f(x)≤h(x)≤g(x)f(x) \leq h(x) \leq g(x).

  3. h(x)h(x), continuous and defined for all x≥0x\ge 0, −2f(x)≤h(x)≤f(x)-2f(x) \leq h(x) \leq f(x).

Hint

(a) What if h(x)h(x) is negative? What if it's not?
(b) What if h(x)h(x) is very close to f(x)f(x) or g(x)g(x), rather than right in the middle?
(c) Note ∣h(x)∣≤2f(x)|h(x)| \leq 2f(x).

Answer
  1. Not enough information to decide. For example, consider h(x)=0h(x) = 0 versus h(x)=−1h(x) = -1.

  2. Not enough information to decide. For example, consider h(x)=f(x)h(x)= f(x) versus h(x)=g(x)h(x) = g(x).

  3. ∫012∞h(x) dx\displaystyle\int_{0\vphantom{\frac12}}^{\infty}h(x)~\dee{x} converges by the comparison test, since ∣h(x)∣≤2f(x)|h(x)| \leq 2f(x) and ∫0∞2f(x) dx\displaystyle\int_0^\infty 2f(x)~\dee{x} converges.

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q6Stage 2Past exam · M105 2015A

Evaluate the integral ∫01x4x5−1 dx\displaystyle\int_0^1\frac{x^4}{x^5-1}\,\dee{x} or state that it diverges.

Hint

First: is the integrand unbounded, and if so, where?

Second: when evaluating integrals, always check to see if you can use a simple substitution before trying a complicated procedure like partial fractions.

Answer

The integral diverges.

Full solution
Q7Stage 2Past exam · 2016Q4

Determine whether the integral ∫−221(x+1)4/3 dx\displaystyle\int_{-2}^2\frac{1}{(x+1)^{4/3}}\,\dee{x} is convergent or divergent. If it is convergent, find its value.

Hint

Is the integrand bounded?

Answer

The integral diverges.

Full solution
Q8Stage 2Past exam · 1997D

Does the improper integral ∫1∞14x2−x dx\displaystyle\int_1^\infty\frac{1}{\sqrt{4x^2-x}}\,\dee{x} converge? Justify your answer.

Hint

See Example 1.12.21 in the

CLP-2 text. Rather than antidifferentiating, you can find a nice comparison.

Answer

The integral does not converge.

Full solution
Q9Stage 2Past exam · 2001A

Does the integral ∫0∞dxx2+x\displaystyle\int_0^\infty\frac{\dee{x}}{x^2+\sqrt{x}} converge or diverge? Justify your claim.

Hint

Which of the two terms in the denominator is more important when x≈0x\approx 0? Which one is more important when xx is very large?

Answer

The integral converges.

Full solution
Q10Stage 2

Does the integral ∫−∞∞cos⁡x dx\displaystyle\int_{-\infty}^\infty \cos x~\dee{x} converge or diverge? If it converges, evaluate it.

Hint

Remember to break the integral into two pieces.

Answer

The integral diverges.

Full solution
Q11Stage 2

Does the integral ∫−∞∞sin⁡x dx\displaystyle\int_{-\infty}^\infty \sin x~\dee{x} converge or diverge? If it converges, evaluate it.

Hint

Remember to break the integral into two pieces.

Answer

The integral diverges.

Full solution
Q12Stage 2

Evaluate ∫10∞x4−5x3+2x−7x5+3x+8 dx\displaystyle\int_{10}^\infty \frac{x^4-5x^3+2x-7}{x^5+3x+8} ~\dee{x}, or state that it diverges.

Hint

The easiest test in this case is limiting comparison, Theorem 1.12.22 in the CLP-2 text.

Answer

The integral diverges.

Full solution
Q13Stage 2

Evaluate ∫010x−1x2−11x+10 dx\displaystyle\int_0^{10} \frac{x-1}{x^2-11x+10} ~\dee{x}, or state that it diverges.

Hint

Not all discontinuities cause an integral to be improper–only infinite discontinuities.

Answer

The integral diverges.

Full solution
Q14Stage 2Past exam · M121 2012A

Determine (with justification!) which of the following applies to the integral ∫−∞+∞xx2+1dx\displaystyle\int_{-\infty}^{+\infty}\frac{x}{x^2+1}\dee{x}:

  1. ∫−∞+∞xx2+1dx\displaystyle\int_{-\infty}^{+\infty}\frac{x}{x^2+1}\dee{x} diverges

  2. ∫−∞+∞xx2+1dx\displaystyle\int_{-\infty}^{+\infty}\frac{x}{x^2+1}\dee{x} converges but ∫−∞+∞∣xx2+1∣dx\displaystyle\int_{-\infty}^{+\infty}\left|\frac{x}{x^2+1}\right|\dee{x} diverges

  3. ∫−∞+∞xx2+1dx\displaystyle\int_{-\infty}^{+\infty}\frac{x}{x^2+1}\dee{x} converges, as does ∫−∞+∞∣xx2+1∣dx\displaystyle\int_{-\infty}^{+\infty}\left|\frac{x}{x^2+1}\right|\dee{x}

Remark: these options, respectively, are that the integral diverges, converges conditionally, and converges absolutely. You'll see this terminology used for series in Section 3.4.1 of the CLP-2 text.

Hint

Which of the two terms in the denominator is more important when xx is very large?

Answer

The integral diverges.

Full solution
Q15Stage 2Past exam · M121 1999A

Decide whether I=∫0∞∣sin⁡x∣x3/2+x1/2 dxI=\displaystyle\int_0^\infty\frac{|\sin x|}{x^{3/2}+x^{1/2}}\ \dee{x} converges or diverges. Justify.

Hint

Which of the two terms in the denominator is more important when x≈0x\approx 0? Which one is more important when xx is very large?

Answer

The integral converges.

Full solution
Q16Stage 2Past exam · M121 2000A

Does the integral ∫0∞x+1x1/3(x2+x+1) dx\displaystyle\int_0^\infty\frac{x+1}{x^{1/3}(x^2+x+1)}\,\dee{x} converge or diverge?

Hint

What are the “problem xx's” for this integral? Get a simple approximation to the integrand near each.

Answer

The integral converges.

Full solution

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q17Stage 3

We craft a tall, vuvuzela-shaped solid by rotating the line y=1x12y = \dfrac{1}{x\vphantom{\frac{1}{2}}} from x=ax=a to x=1x=1 about the yy-axis, where aa is some constant between 0 and 1.

Figure from prob_s1.12, line 2

Figure from prob_s1.12, line 2

True or false: No matter how large a constant MM is, there is some value of aa that makes a solid with volume larger than MM.

Hint

To find the volume of the solid, cut it into horizontal slices, which are thin circular disks.

The true/false statement is equivalent to saying that the improper integral giving the volume of the solid when a=0a=0 diverges to infinity.

Answer

false

Full solution
Q18Stage 3Past exam · 2016Q4

What is the largest value of qq for which the integral ∫1∞1x5q dx\displaystyle \int_1^\infty \frac1{x^{5q}}\,\dee{x} diverges?

Hint

Review Example 1.12.8 in the

CLP-2 text. Remember the antiderivative of 1x\frac{1}{x} looks very different from the antiderivative of other powers of xx.

Answer

q=15q=\frac{1}{5}

Full solution
Q19Stage 3

For which values of pp does the integral ∫0∞x(x2+1)p dx\displaystyle\int_0^\infty \dfrac{x}{(x^2+1)^p}~\dee{x} converge?

Hint

Compare to Example 1.12.14 in the CLP-2 text. You can antidifferentiate with a uu-substitution.

Answer

p>1p>1

Full solution
Q20Stage 3

Evaluate ∫2∞1t4−1dt\displaystyle\int_2^\infty \frac{1}{t^4-1}\dee{t}, or state that it diverges.

Hint

To evaluate the integral, you can factor the denominator.
Recall lim⁡x→∞arctan⁡x=π2\displaystyle\lim_{x \to \infty}\arctan x = \frac{\pi}{2}. For the other limits, use logarithm rules, and beware of indeterminate forms.

Answer

log⁡3−π4+12arctan⁡2\dfrac{\log 3-\pi}{4} + \dfrac{1}{2}\arctan 2

Full solution
Q21Stage 3

Does the integral ∫−55(1∣x∣+1∣x−1∣+1∣x−2∣)dx\displaystyle\int_{-5}^5 \left(\frac{1}{\sqrt{|x|}} + \frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-2|}}\right)\dee{x} converge or diverge?

Hint

Break up the integral. The absolute values give you a nice even function, so you can replace ∣x−a∣|x-a| with x−ax-a if you're careful about the limits of integration.

Answer

The integral converges.

Full solution
Q22Stage 3

Evaluate ∫0∞e−xsin⁡x dx\displaystyle\int_0^\infty e^{-x}\sin x~\dee{x}, or state that it diverges.

Hint

Use integration by parts twice to find the antiderivative of e−xsin⁡xe^{-x}\sin x, as in Example 1.7.10 of the CLP-2 text. Be careful with your signs — it's easy to make a mistake with all those negatives.

If you're having a hard time taking the limit at the end, review the Squeeze Theorem, Theorem 1.4.17 in the CLP-1 text.

Answer

12\dfrac{1}{2}

Full solution
Q23Stage 3Past exam · M121 2002A

Is the integral ∫0∞sin⁡4xx2 dx\displaystyle\int_0^\infty\frac{\sin^4 x}{x^2}\, \dee{x} convergent or divergent? Explain why.

Hint

What is the limit of the integrand when x→0x\rightarrow 0?

Answer

The integral converges.

Full solution
Q24Stage 3

Does the integral ∫0∞xex+x dx\displaystyle\int_0^\infty \frac{x}{e^x+\sqrt{x}} ~\dee{x} converge or diverge?

Hint

The only “source of impropriety" is the infinite domain of integration. Don't be afraid to be a little creative to make a comparison work.

Answer

The integral converges.

Full solution
Q25Stage 3Past exam · M121 2001A

Let Mn,tM_{n,t} be the Midpoint Rule approximation for ∫0te−x1+x dx\displaystyle\int_0^t \frac{e^{-x}}{1+x}\ \dee{x} with nn equal subintervals. Find a value of tt and a value of nn such that Mn,tM_{n,t} differs from ∫0∞e−x1+x dx\int_0^\infty \frac{e^{-x}}{1+x}\ \dee{x} by at most 10−410^{-4}. Recall that the error EnE_n introduced when the Midpoint Rule is used with nn subintervals obeys

∣En∣≤M(b−a)324n2\begin{align*} |E_n|\le \frac{M(b-a)^3}{24n^2} \end{align*}

where MM is the maximum absolute value of the second derivative of the integrand and aa and bb are the end points of the interval of integration.

Hint

There are two things that contribute to your error: using tt as the upper bound instead of infinity, and using nn intervals for the approximation.

First, find a tt so that the error introduced by approximating ∫0∞e−x1+x dx\int_0^\infty \frac{e^{-x}}{1+x}\ \dee{x} by ∫0te−x1+x dx\int_0^t \frac{e^{-x}}{1+x}\ \dee{x} is at most 1210−4\frac{1}{2} 10^{-4}. Then, find your nn.

Answer

t=10t=10 and n=2042n= 2042 will do the job. There are many other correct answers.

Full solution
Q26Stage 3

Suppose f(x)f(x) is continuous for all real numbers, and ∫1∞f(x) dx\displaystyle\int_1^\infty f(x)~\dee{x} converges.

  1. If f(x)f(x) is odd, does ∫−∞12−1f(x) dx\displaystyle\int_{-\infty\vphantom{\frac12}}^{-1} f(x)~\dee{x} converge or diverge, or is there not enough information to decide?

  2. If f(x)f(x) is even, does ∫−∞∞f(x) dx\displaystyle\int_{-\infty}^\infty f(x)~\dee{x} converge or diverge, or is there not enough information to decide?

Hint

Look for a place to use Theorem 1.12.20 of the CLP-2 text.

Examples 1.2.9 and 1.2.10 in the CLP-2 text have nice results about the area under an even/odd curve.

Answer

(a) The integral converges. (b) The interval converges.

Full solution
Q27Stage 3

True or false: There is some real number xx, with x≥1x \geq 1, such that ∫0x1et dt=1\displaystyle\int_0^x \frac{1}{e^t}~\dee{t} = 1.

Hint

xx should be a real number

Answer

false

Full solution

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.