Exam 01 - Su Yung Exam 1 Due Oct 2 2007 11:00 pm Inst...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Su, Yung – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: Shinko Harper 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Below is the graph oF a Function f . 2 4 6 8 10 12 2 4 6 8 - 2 - 4 - 6 Estimate the defnite integral I = Z 12 0 f ( x ) dx with six equal subintervals using leFt end- points. 1. I 8 2. I 14 correct 3. I 12 4. I 6 5. I 10 Explanation: Since [0 , 12] is subdivided into six equal subintervals, each oF these will have length 2 and the six corresponding rectangles are shown as the shaded areas in 2 4 6 8 10 12 2 4 6 8 - 2 - 4 - 6 The heights oF the rectangles are leFt endpoint sample values oF f that can be read o± From the graph. Thus, with leFt endpoints, I 2(4 + 2 - 5 - 1 + 2 + 5) = 14 . keywords: graph, Riemann sum, leFt end- points 002 (part 1 oF 1) 10 points IF F ( x ) = Z x 0 e 10 sin 2 θ dθ , fnd the value oF F 0 ( π/ 4). 1. F 0 ( π/ 4) = e 10 2. F 0 ( π/ 4) = 5 e 3. F 0 ( π/ 4) = 5 e 10 4. F 0 ( π/ 4) = 5 e 5 5. F 0 ( π/ 4) = e 5 correct Explanation: By the ²undamental theorem oF calculus, F 0 ( x ) = e 10 sin 2 x . At x = π/ 4, thereFore, F 0 ( π/ 4) = e 5
Background image of page 1

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

View Full DocumentRight Arrow Icon
Su, Yung – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: Shinko Harper 2 since sin( π 4 ) = 1 2 . keywords: integral, FTC 003 (part 1 of 1) 10 points Determine F 0 ( x ) when F ( x ) = Z x 2 6 cos t t dt. 1. F 0 ( x ) = 6 cos x x 2. F 0 ( x ) = - 6 sin( x ) x 3. F 0 ( x ) = - 6 sin x x 4. F 0 ( x ) = 3 sin( x ) x 5. F 0 ( x ) = - 6 sin x x 6. F 0 ( x ) = 3 cos x x 7. F 0 ( x ) = 3 cos( x ) x correct 8. F 0 ( x ) = - 3 cos( x ) x Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx Z g ( x ) a f ( t ) dt · = f ( g ( x )) g 0 ( x ) . When F ( x ) = Z x 2 6 cos t t dt, therefore, F 0 ( x ) = 6 cos( x ) x d dx x · . Consequently, F 0 ( x ) = 3 cos( x ) x , since d dx x = 1 2 x . keywords: Stewart5e, FTC, Chain Rule 004 (part 1 of 1) 10 points If w 0 ( t ) is the rate of growth of Mira’s weight (in pounds per year), what does the de±nite integral I = Z 8 4 w 0 ( t ) dt represent? 1. decrease in Mira’s weight from age 4 to 8 2. average of Mira’s weight from age 4 to 8 3. Mira’s weight at age 8 4. increase in Mira’s weight from age 4 to 8 correct 5. Mira’s weight at age 4 Explanation: By the Fundamental theorem of Calculus, Z b a w 0 ( x ) dx = w ( b ) - w ( a ) , in other words, the value of the integral is the net change, w ( b ) - w ( a ), in w over the interval [ a, b ]. Consequently, I is the increase in Mira’s weight from age 4 to 8 . keywords: integral, rate growth, FTC, weight, net change, 005 (part 1 of 1) 10 points
Background image of page 2
Su, Yung – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: Shinko Harper 3 Find the value of f (0) when f 00 ( t ) = 2(6 t - 1) and f 0 (1) = 1 , f (1) = 2 .
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

Exam 01 - Su Yung Exam 1 Due Oct 2 2007 11:00 pm Inst...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online