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School: JMU
Course: ELEM STATISTICS
Department of Mathematics and Statistics Math 220 (Section 21 and Section 22): Spring 2014 Instructor: Dr. Hasan Hamdan Course Website: Course is on Canvas Email: hamdanhx@jmu.edu Phone: 540.568.2844 Sections: 21 & 22 Class Location: Tuesday Roop 127 and
School: JMU
School: JMU
Course: Calculus II
H OMEWORK 5 5.5.44. To solve becomes (x2 + 4)3/2 dx we let x = 2 tan so dx = 2 sec2 d. The integral 1 1 (sec )1 d = 4 4 (2 sec )3 (2 sec2 d) = = 5.5.50. To solve becomes cos d = 1 sin + C 4 1 1 x sin(tan1 (x/2) + C = + C. 4 4 x2 + 4 x2 4 2x dx we let x =
School: JMU
Course: Calculus II
H OMEWORK 8 6.3.22. We can approximate this by 1 + (e0 e1 )2 + 1 + (e1 e0 )2 + 1 + (e2 e1 )2 + 1 + (e3 e2 )2 . 6.3.32. Since f (x) = 1, the arclength is given by 1 + (1)2 dx = 3 2. 5 2 6.3.38. Since f (x) = (1/2)(1 x2 )1/2 (2x), the arclength is given by
School: JMU
Course: Calculus II
H OMEWORK 6 5.7.10. In this situation, we can rewrite the integral as b c b f (x) dx. f (x) dx + f (x) dx = c a a Each of the integrands on the right side is monotonic, so we can use LEF T (n) or RIGHT (n) sums to evaluate each integral. On the rst integr
School: JMU
Course: Calculus II
H OMEWORK 10 7.2.34. We have k2 2 2k 2 = lim = lim = 1 2 + 2k + 2 k k k 2k + 2 k 2 by LHopitals rule. Since the sequence converges, it is bounded above and below. To see this, let = 1. Then the limit statement tells us that there exists an N > 0 such that
School: JMU
Course: Calculus II
H OMEWORK 5 5.5.44. To solve becomes (x2 + 4)3/2 dx we let x = 2 tan so dx = 2 sec2 d. The integral 1 1 (sec )1 d = 4 4 (2 sec )3 (2 sec2 d) = = 5.5.50. To solve becomes cos d = 1 sin + C 4 1 1 x sin(tan1 (x/2) + C = + C. 4 4 x2 + 4 x2 4 2x dx we let x =
School: JMU
Course: Calculus II
H OMEWORK 8 6.3.22. We can approximate this by 1 + (e0 e1 )2 + 1 + (e1 e0 )2 + 1 + (e2 e1 )2 + 1 + (e3 e2 )2 . 6.3.32. Since f (x) = 1, the arclength is given by 1 + (1)2 dx = 3 2. 5 2 6.3.38. Since f (x) = (1/2)(1 x2 )1/2 (2x), the arclength is given by
School: JMU
Course: Calculus II
H OMEWORK 6 5.7.10. In this situation, we can rewrite the integral as b c b f (x) dx. f (x) dx + f (x) dx = c a a Each of the integrands on the right side is monotonic, so we can use LEF T (n) or RIGHT (n) sums to evaluate each integral. On the rst integr
School: JMU
Course: Calculus II
H OMEWORK 10 7.2.34. We have k2 2 2k 2 = lim = lim = 1 2 + 2k + 2 k k k 2k + 2 k 2 by LHopitals rule. Since the sequence converges, it is bounded above and below. To see this, let = 1. Then the limit statement tells us that there exists an N > 0 such that
School: JMU
Course: Calculus II
H OMEWORK 7 1 b 6.2.12. Fitting 0 2xx2 dx to the general formula a (radius)(height) dx, we nd that the radius of the shell is x and the height is x2 . Hence, the region is bounded by y = x2 and the x-axis, from x = 0 to x = 1, revolved around the y -axis.
School: JMU
Course: Calculus II
Name: Q UIZ 5 Please write your solutions in complete sentences, as simply as you can, and justify each step. 1. Let h and r be positive constants. Consider the region R bounded by y = h x, y = h, r and the y -axis. Use the shell method to nd the volume o
School: JMU
Course: Calculus 1
MATH235 Calculus 1 Quiz 1 Show all work to receive full credit. Carefully write down your thought process. Your solution must not contain any logical errors. Good luck! 1. Solve the following inequality : 3 2 < . x1 x+1 3 2 < 0, x1 x+1 3(x + 1) 2(x 1) < 0
School: JMU
Course: Calculus 1
MATH 235 Calculus 1 Quiz 6 Solution 1. Find the following values. 1 a. sin1 ( ) = . 2 6 1 3 b. cos1 ( ) = 4 2 Note that for both part a and part b, there is one and only one answer, since if you plug in any x-value into sin1 the function value will always
School: JMU
Course: Calculus 1
MATH 235 Calculus 1 Quiz 5 1. At time t, the position of a body moving along the s-axis is s = t3 6t2 + 9t. Time is given in seconds and distance is given in meters. a. Find the bodys acceleration each time the velocity is zero. Since v (t) = s (t) = 3t2
School: JMU
Course: Calculus 1
MATH 235 Calculus 1 Quiz 4 10/04/2010 Show all work to receive full credit. Carefully write down your thought process. Your solution must not contain any logical errors. Good luck! 1. y = ( 4 t 2 dy ) . Find . t+1 dt dy 4 t 3 d 4 t = 2( ) ( ) dt t+1 dt t
School: JMU
Course: Calculus 1
MATH235 Calculus 1 Quiz 2 Show all work to receive full credit. Carefully write down your thought process. Your solution must not contain any logical errors. Good luck! 1. Find the slope of the line tangent to the curve f (x) = x2 2x 3 at the point (2, 3)
School: JMU
Course: Calculus 1
Name : MATH 235 Calculus 1 Quiz 3 09/27/2010 Show all work to receive full credit. Carefully write down your thought process. Your solution must not contain any logical errors. Good luck! 1. By using the precise denition of limits, prove that limx1 f (x)
School: JMU
Course: Calculus 1
Worksheet 6 MATH 235 10/21/2010 1. (The Derivative Rule for Inverses.) Let f be a continuous one-to-one function dened on an interval. Suppose f is differentiable at x = a and f (a) = 0. If f (a) = b, show that (f 1 ) (b) exists and that 1 ( f 1 ) ( b) =
School: JMU
Course: Calculus 1
Worksheet 5 MATH 235 10/07/2010 Discuss the following problems with your group and write down a complete solution. Show all work. 1. Prove the Product Rule: if f and g are differentiable, then so is their product P (x) = f (x)g (x), and P ( x) = f ( x) g
School: JMU
Course: Calculus 1
Name: Worksheet 4 MATH 235 Fall, 2010. Discuss the following problems with your group and write down a complete solution. Show all work. 1. Prove that if f is differentiable at x = c, then f is continuous at x = c. Proof. To show that f is continuous at x
School: JMU
Course: Calculus II
H OMEWORK 1 4.3.26. The integrand is the upper half circle of radius r centered at 0. Here, r is a constant and x is the variable of integration since it appears in the differential dx. Hence, r r2 x2 dx r is 1 2 of the area of a circle of radius r. This
School: JMU
Course: Calculus II
H OMEWORK 2 4.7.2 (c) We cannot anti-differentiate this function directly. However, we can use the second Fundamental Theorem of Calculus to write x d 2 2 esin t dt = esin x , dx 0 x d 2 2 esin t dt = esin x , dx 1 x d 2 2 esin t dt = esin x , dx 2 x sin2
School: JMU
Course: Calculus II
H OMEWORK 3 1 1 5.2.24. If we let u = ln x and dv = x dx then we have du = x dx and v = ln x. Integration by parts gives ln x ln x dx = (ln x)2 dx. x x If we rearrange by combining the integrals, then we get ln x dx = (ln x)2 + C x 2 so ln x (ln x)2 dx =
School: JMU
Course: ELEM STATISTICS
Department of Mathematics and Statistics Math 220 (Section 21 and Section 22): Spring 2014 Instructor: Dr. Hasan Hamdan Course Website: Course is on Canvas Email: hamdanhx@jmu.edu Phone: 540.568.2844 Sections: 21 & 22 Class Location: Tuesday Roop 127 and
School: JMU
School: JMU
Course: Stats
Math 220 Syllabus Spring 2008 1. Class Information: Section 8 : M,W 10:10 AM to 11:00 AM, Burruss 141 F 10:10 AM to 11:00 AM, Roop 127 MW 11:15 AM to 12:05 PM , Roop 141 F 11:15 AM to 12:05 PM, Roop 127 Section 11: Section 16: M,W 1:25 PM to 2:1