# Integral methods and improper limits

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

Course: [University calculus and multivariable methods](https://studybrix.com/en/courses/university-calculus/index.md)
Level: university
Prerequisites: Differentiation, antiderivatives and limits.

## 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.

## Practice questions

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

Answer: 4

Explanation: 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.

Answer: 3

Explanation: 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?

A. The integral of 1
B. The integral of 1/sqrt(x)
C. The integral of x^(−3/2)
D. The integral of 1/x

Answer: The integral of x^(−3/2)

Explanation: 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.

Answer: 0.25

Explanation: 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 / marking guidance: Each one-sided integral at zero diverges. Symmetric cancellation is a different limiting procedure, namely a principal value.

Explanation: 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.

Answer: 6

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

Explanation: 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?

A. The value of the integral by a free constant
B. Nothing ever
C. The sign regardless of orientation
D. The integration limits or the final expression back to x

Answer: The integration limits or the final expression back to x

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

## Further reading

- https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/
- https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/

Original AI-assisted educational draft; not independently teacher-reviewed. References support further study, not copied exercises, affiliation or curriculum accreditation.