Midterm 1 Winter 2003 Solutions - Math 125I 8 Midterm 1...

This preview shows page 1 out of 1 page.

Math 125I Midterm 1 Solutions Winter 2003 1.(a) Z 8 1 2 x + 5 3 x 2 dx = Z 8 1 2 x 1 / 3 + 5 x - 2 / 3 dx = 3 2 x 4 / 3 + 15 x 1 / 3 8 1 = 75 2 (b) Let u = cos t so du = - sin t dt . Then Z π 0 sin t 1 + cos 2 t dt = - Z - 1 1 1 1 + u 2 du = - tan - 1 t - 1 1 = π 2 (c) Let v = y 2 - 7 so that dv = 2 y dy and y 2 = v + 7. Then Z y 3 q y 2 - 7 dy = 1 2 Z ( v + 7) v dv = 1 2 Z v 3 / 2 + 7 v 1 / 2 dv = 1 5 v 5 / 2 + 7 3 v 3 / 2 + C = 1 5 ( y 2 - 7) 5 / 2 + 7 3 ( y 2 - 7) 3 / 2 + C 2. Δ t = 1 2 and the t -coordinates of the midpoints are 5 4 , 7 4 , 9 4 , 11 4 , 13 4 , 15 4 . The function v ( t ) is positive at the first 4 values and negative at the last 2. Thus the total distance is 1 2 " v 5 4 + v 7 4 + v 9 4 + v 11 4 - v 13 4 - v 15 4 # 2 . 1784 feet . 3. By the Fundamental Theorem of Calculus, f 0 ( x ) = (2 x - 4) e x 2 - 4 x . This is defined when x 2 - 4 x 0, that is x 4 or x 0. Since e u > 0 for any u , the derivative is positive if it is defined and if 2 x - 4 > 0. Thus the function is increasing when x 4. 4. Solve x 2 = x 3 - 6 x 2 + 10 x to get

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture