midterm1247sol

# midterm1247sol - x 1 cos x 2 x dx = R cos u du = sin u C =...

This preview shows pages 1–3. Sign up to view the full content.

Math 247, Midterm I. 0. Write your name here: 1. Find R 3 x - 1 x dx . We have R 3 x - 1 x dx = R x 1 / 3 - 1 x dx = R ( x - 2 / 3 - x - 1 ) dx = x 1 / 3 1 / 3 - ln | x | + C = 3 3 x - ln | x | + C . Answer: R 3 x - 1 x dx = 3 3 x - ln | x | + C . 2. Approximate R 1 - 1 x 2 dx using four equal subintervals and right endpoints. We have R 1 - 1 x 2 dx 0 . 5 · ( - 0 . 5) 2 +0 . 5 · 0 2 +0 . 5 · (0 . 5) 2 +0 . 5 · 1 2 = 0 . 5 · (1 . 5) = 0 . 75. Answer: R 1 - 1 x 2 dx 0 . 75.

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

View Full Document
3. A size of population N ( t ) satisﬁes the equation dN ( t ) dt = t (1 - t ). Find the cumulative change in population size during the time interval [0 , 1]. The cumulative change is R 1 0 dN ( t ) dt dt = R 1 0 t (1 - t ) dt = R 1 0 ( t - t 2 ) dt = ( t 2 / 2 - t 3 / 3) | 1 0 = 1 2 - 1 3 = 1 6 . Answer: the cumulative change is 1 6 . 4. Find R (2 x + 1) cos( x 2 + x ) dx . We will use the substitution rule with u = x 2 + x . Then du = 2 x + 1 and R (2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x + 1) cos( x 2 + x ) dx = R cos( u ) du = sin( u ) + C = sin( x 2 + x ) + C . Answer: R (2 x + 1) cos( x 2 + x ) dx = sin( x 2 + x ) + C . 5. Find R 2 x 2 √ x 3 +1 dx . We will use the substitution rule with u = x 3 + 1. Then du = 3 x 2 dx or x 2 dx = 1 3 du . Hence R 2 x 2 √ x 3 +1 dx = R 9 1 1 3 du √ u = 1 3 R 9 1 u-1 / 2 du = 1 3 u 1 / 2 1 / 2 | 9 1 = 2 3 (9 1 / 2-1 1 / 2 ) = 4 3 . Answer: R 2 x 2 √ x 3 +1 dx = 4 3 ....
View Full Document

## This note was uploaded on 04/07/2008 for the course MATH 247 taught by Professor Ostrik during the Winter '08 term at Oregon.

### Page1 / 3

midterm1247sol - x 1 cos x 2 x dx = R cos u du = sin u C =...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online