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Mathematics · University

Integral methods and improper limits

Substitution reverses a chain rule; integration by parts reverses a product rule.

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Prerequisites

Differentiation, antiderivatives and limits.

Study notes

Choose a method

Substitution reverses a chain rule; integration by parts reverses a product rule.

Endpoint limits

An improper integral is a limit of proper integrals. Check every singular endpoint separately.

Convergence test

For x ≥ 1, the integral of x^(−p) converges exactly when p > 1.

Common trap

A finite antiderivative expression does not establish convergence until the boundary limits are evaluated.

Worked examples

Evaluate the integral of 2x exp(x²) from 0 to sqrt(ln 5).

Set u=x²; the limits become 0 and ln 5. The integral is exp(ln 5)−1=4.

4

Evaluate the improper integral of 3/x² from 1 to infinity.

The antiderivative is −3/x. Its limit at infinity is 0, so the value is 0−(−3)=3.

3

Check your understanding

  1. Which integral from 1 to infinity converges?

    1. The integral of 1
    2. The integral of 1/sqrt(x)
    3. The integral of x^(−3/2)
    4. The integral of 1/x

  2. Evaluate the integral of x ln x from 1 to e. Enter the coefficient of e² in the result A e² + 1/4.

  3. Explain why the integral of 1/x from −1 to 1 is not an ordinary convergent improper integral, even though the symmetric principal value is zero.

  4. Evaluate the integral of 1/sqrt(x) from 0 to 9.

Answers and marking guidance

Answers and marking notes
  1. The integral of x^(−3/2)

    The exponent p=3/2 exceeds 1; the others have p≤1.

  2. 0.25

    Integration by parts gives x² ln x/2−x²/4. Evaluation gives e²/4+1/4.

  3. Each one-sided integral at zero diverges. Symmetric cancellation is a different limiting procedure, namely a principal value.

    Award credit for separate endpoint limits, divergence, and distinguishing a principal value from ordinary improper convergence.

  4. 6

    Use the lower limit ε>0: 2sqrt(9)−2sqrt(ε) tends to 6.

Further reading and review references

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