∫Calc Practice
Home›Calculus 2›Improper integrals

Improper integrals practice problems

Infinite intervals and infinite integrands: replace the bad bound with a limit. 28 problems with worked solutions; in 28 of them every equation is proved by a computer algebra system.

9 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Evaluate \( \displaystyle \int_{0}^{\infty} e^{- 4 x} \, dx \), or show that it diverges.
  1. The integral is improper at infinity; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} \left(- \frac{e^{- 4 x}}{4}\right) = e^{- 4 x} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ 1 \cdot \frac{1}{4} - \frac{e^{- 4 T}}{4} = \frac{1}{4} - \frac{e^{- 4 T}}{4} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → \infty, this approaches 1/4.
    Reviewed
  5. \[ \lim_{T \to \infty}\left(\frac{1}{4} - \frac{e^{- 4 T}}{4}\right) = \frac{1}{4} \]
    The limit.✓ Proved
Answer \( \frac{1}{4} \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{0}^{4} \frac{1}{x^{\frac{2}{3}}} \, dx \), or show that it diverges.
  1. The integral is improper at x = 0; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} 3 \sqrt[3]{x} = \frac{1}{x^{\frac{2}{3}}} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ - 3 T^{1 \cdot \frac{1}{3}} + 3 \cdot 2^{\frac{2}{3}} = - 3 \sqrt[3]{T} + 3 \cdot 2^{\frac{2}{3}} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → 0, this approaches 3*2**(2/3).
    Reviewed
  5. \[ \lim_{T \to 0^+}\left(- 3 \sqrt[3]{T} + 3 \cdot 2^{\frac{2}{3}}\right) = 3 \cdot 2^{\frac{2}{3}} \]
    The limit.✓ Proved
Answer \( 3 \cdot 2^{\frac{2}{3}} \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{0}^{\infty} e^{- x} \, dx \), or show that it diverges.
  1. The integral is improper at infinity; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} \left(- e^{- x}\right) = e^{- x} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ 1 - e^{- T} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → \infty, this approaches 1.
    Reviewed
  5. \[ \lim_{T \to \infty}\left(1 - e^{- T}\right) = 1 \]
    The limit.✓ Proved
Answer \( 1 \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{x^{\frac{5}{2}}} \, dx \), or show that it diverges.
  1. The integral is improper at infinity; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} \left(- \frac{2}{3 x^{\frac{3}{2}}}\right) = \frac{1}{x^{\frac{5}{2}}} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ \frac{2}{3} - \frac{2}{3 T^{\frac{3}{2}}} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → \infty, this approaches 2/3.
    Reviewed
  5. \[ \lim_{T \to \infty}\left(\frac{2}{3} - \frac{2}{3 T^{\frac{3}{2}}}\right) = \frac{2}{3} \]
    The limit.✓ Proved
Answer \( \frac{2}{3} \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{0}^{\infty} \frac{1}{x^{2} + 9} \, dx \), or show that it diverges.
  1. The integral is improper at infinity; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} \frac{\operatorname{atan}{\left(\frac{x}{3} \right)}}{3} = \frac{1}{x^{2} + 9} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ \frac{\operatorname{atan}{\left(\frac{T}{3} \right)}}{3} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → \infty, this approaches pi/6.
    Reviewed
  5. \[ \lim_{T \to \infty}\left(\frac{\operatorname{atan}{\left(\frac{T}{3} \right)}}{3}\right) = \frac{\pi}{6} \]
    The limit.✓ Proved
Answer \( \frac{\pi}{6} \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{0}^{8} \frac{1}{\sqrt[3]{x}} \, dx \), or show that it diverges.
  1. The integral is improper at x = 0; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} \frac{3 x^{\frac{2}{3}}}{2} = \frac{1}{\sqrt[3]{x}} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ 6 - \frac{3 T^{\frac{2}{3}}}{2} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → 0, this approaches 6.
    Reviewed
  5. \[ \lim_{T \to 0^+}\left(6 - \frac{3 T^{\frac{2}{3}}}{2}\right) = 6 \]
    The limit.✓ Proved
Answer \( 6 \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{0}^{8} \frac{1}{\sqrt{x}} \, dx \), or show that it diverges.
  1. The integral is improper at x = 0; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} 2 \sqrt{x} = \frac{1}{\sqrt{x}} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ - 2 \sqrt{T} + 4 \sqrt{2} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → 0, this approaches 4*sqrt(2).
    Reviewed
  5. \[ \lim_{T \to 0^+}\left(- 2 \sqrt{T} + 4 \sqrt{2}\right) = 4 \sqrt{2} \]
    The limit.✓ Proved
Answer \( 4 \sqrt{2} \)

✓ Nihil obstat Lines: 2 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx \), or show that it diverges.
  1. The integral is improper at infinity; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} 2 \sqrt{x} = \frac{1}{\sqrt{x}} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ 2 \sqrt{T} - 2 \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → \infty, this grows without bound, so the integral diverges.
    Reviewed
Answer \( \text{diverges} \)

✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{x} \, dx \), or show that it diverges.
  1. The integral is improper at infinity; replace that bound by T and take a limit.
    Reviewed
  2. \[ \frac{d}{d x} \ln{\left(x \right)} = \frac{1}{x} \]
    An antiderivative, checked by differentiating.✓ Proved
  3. \[ \ln{\left(T \right)} \]
    The integral with T in place of the bad bound.Reviewed
  4. As T → \infty, this grows without bound, so the integral diverges.
    Reviewed
Answer \( \text{diverges} \)

✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More improper integrals practice

Evaluate \( \displaystyle \int_{0}^{4} \frac{1}{\sqrt[3]{x}} \, dx \), or show that it diverges.
Problem 4.31medium✓ Every equation proved
Evaluate \( \displaystyle \int_{0}^{\infty} e^{- 2 x} \, dx \), or show that it diverges.
Problem 4.34medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{x^{4}} \, dx \), or show that it diverges.
Problem 4.40medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\infty} \frac{1}{x^{2} + 1} \, dx \), or show that it diverges.
Problem 4.41medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx \), or show that it diverges.
Problem 4.42medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{\infty} \frac{1}{x^{3}} \, dx \), or show that it diverges.
Problem 4.44medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\infty} e^{- 3 x} \, dx \), or show that it diverges.
Problem 4.76medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{\infty} \frac{1}{x^{2}} \, dx \), or show that it diverges.
Problem 4.77medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{\infty} \frac{1}{x^{2} + 4} \, dx \), or show that it diverges.
Problem 4.78medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{1} \frac{1}{x^{\frac{2}{3}}} \, dx \), or show that it diverges.
Problem 4.79medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{4} \frac{1}{\sqrt{x}} \, dx \), or show that it diverges.
Problem 4.80medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{1} \frac{1}{\sqrt[3]{x}} \, dx \), or show that it diverges.
Problem 4.81medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{x^{3}} \, dx \), or show that it diverges.
Problem 4.82medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{\infty} \frac{1}{x^{\frac{5}{2}}} \, dx \), or show that it diverges.
Problem 4.83medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx \), or show that it diverges.
Problem 4.84medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{8} \frac{1}{x^{\frac{2}{3}}} \, dx \), or show that it diverges.
Problem 4.85medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{2}^{\infty} \frac{1}{x^{4}} \, dx \), or show that it diverges.
Problem 4.86medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \), or show that it diverges.
Problem 4.87medium✓ Nihil obstat
Evaluate \( \displaystyle \int_{1}^{\infty} \frac{1}{x^{2}} \, dx \), or show that it diverges.
Problem 4.120medium✓ Nihil obstat