victor0905

# victor0905 - y = ln(cos x between x = 0 and x = π 4 3[5...

This preview shows pages 1–7. Sign up to view the full content.

Math 242 Section 014: Test #2 NAME: Please circle your lab/discussion: 060 (12:00) 061 (11:00) 062 (2:30) There are 6 questions on 6 pages . Points per page: 5, 5, 5, 5, 6, 6. Show your work. A correct answer with little or no reasoning may receive a low score. (However, there is no need to be wordy.) The 7th page has some formulas you can refer to if needed. [5 points] Question 1. Determine whether the improper integral Z 0 x ( x 2 + 2) 2 dx converges or diverges. If it converges, ﬁnd its value. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
[5 points] Question 2. Using an intelligent ﬁrst approximation, use one iteration of Newton’s method to get a very good approximation to 10001. 2
[5 points] Question 3. Find the length of the curve

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: y = ln(cos x ) between x = 0 and x = π/ 4. 3 [5 points] Question 4. Determine whether the series ∞ X n =1 1 + 2 n 3 n converges, and if it converges, ﬁnd its sum. 4 [6 points] Question 5. Determine whether the series ∞ X n =2 1 n (ln n ) 3 / 2 converges or diverges. (Provide a reason.) 5 [6 points] Question 6. Determine whether the series ∞ X n =1 sin 1 n converges or diverges. (Provide a reason.) 6 If helpful, you may use any of these formulas in any of the test questions. sin 2 x = 1 2 ± 1-cos 2 x ² cos 2 x = 1 2 ± 1 + cos 2 x ² Z 1 1 + x 2 dx = ? Z sec x dx = ln | sec x + tan x | + C 7...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

victor0905 - y = ln(cos x between x = 0 and x = π 4 3[5...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online