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Mathematics · lesson 1 of 4

Integral methods and improper limits

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

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Learn the ideas

Prerequisites: Differentiation, antiderivatives and limits.

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

Example 1

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.

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Example 2

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.

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Check your understanding

8 questions

Decimals or fractions are accepted for numeric answers. Proofs and written responses use sample answers and require human marking. This is practice, not a protected examination.

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

Show answer and explanation

4

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

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

Show answer and explanation

3

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

3 Which integral from 1 to infinity converges?

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The integral of x^(−3/2)

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

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

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0.25

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

5 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.

Sample answer and marking guidance

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.

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

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

7 Show that the integral of exp(−ax) from 0 to infinity equals 1/a when a>0.

Sample answer and marking guidance

An antiderivative is −exp(−ax)/a; its upper limit is zero and its value at zero is −1/a.

A complete solution states a>0 and evaluates the limiting endpoint, not just a formal antiderivative.

8 In substitution u=x² for a definite integral, what must be changed besides the differential?

Show answer and explanation

The integration limits or the final expression back to x

Use u-limits with a u-integrand, or return the antiderivative to x before using x-limits.

Further reading and review references

These are reference links, not copied exercises, institutional endorsement or proof of independent review.