{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# e2r - Calculus II Review Problems for Exam 2 1 Test for...

This preview shows pages 1–2. Sign up to view the full content.

Calculus II, Review Problems for Exam 2 October 27, 2007 1. Test for convergence 2 0 e cos x x 3 (2 - x ) dx . SOLUTION: Since 2 - x 2 for all x [0 , 2] we have 2 0 e cos x x 3 (2 - x ) dx 1 0 e - 1 2 x 3 / 2 which diverges. Hence the given integral diverges. 2. Give a complete argument that the harmonic series diverges (saying p -series, p = 1 is not enough!). If s n denotes the n -th partial sum, show that s n 1 + ln n . Conclude that the sum of the first billion terms is less than 22. Is the sequence of partial sums unbounded? SOLUTION: Notice that f ( x ) = 1 x is continuous, positive, and decreasing on [1 , ). Further f ( n ) = 1 n and 1 1 x dx diverges. Hence by the integral test, the harmonic series diverges. s n 1 + n 1 dx x = 1 + ln n (draw the picture!!). 3. For which α and β 0 does 0 x α 1 + x β dx converge? INTUITION: “Near 0” the integrand “looks like” x α which has convergent integral if and only if α > - 1. “Near ” the integrand “looks like” x α - β which has convergent integral if and only if α - β < - 1 SOLUTION: Since β 0, for x 1 we have x β 1 + x β x β + x β = 2 x β hence 1 x α - β 2 dx 1 x α 1 + x β dx 1 x α - β

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.

{[ snackBarMessage ]}

### Page1 / 3

e2r - Calculus II Review Problems for Exam 2 1 Test for...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online