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
✓ 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
✓ 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
✓ 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
✓ 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
✓ 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
✓ 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
✓ 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
An antiderivative, checked by differentiating.✓ Proved
\[ 2 \sqrt{T} - 2 \]
The integral with T in place of the bad bound.Reviewed
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
An antiderivative, checked by differentiating.✓ Proved
\[ \ln{\left(T \right)} \]
The integral with T in place of the bad bound.Reviewed
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