TEST 1 FORM A (SOLUTIONS)

Transcription

1 TEST FORM A SOLUTIONS) MAT 68 Directions: Solve as many problems as well as you can in the blue eamination book, writing in pencil and showing all work. Put away any cell phones; the mere appearance will give a zero.. Give both a geometric and an algebraic definition of b f ) d. a geometric: the net area between f ) and the -ais on the interval [a, b] algebraic: f i ) where i is any sample point in the ith subinterval of n [a, b] and = b a n. Compute the following its, if they eist. Use ôpital s Rule only if necessary. a) As, we are dividing a constant by larger and larger positive numbers, to the quotient dwindles to zero, so the it is zero. ôpital s Rule does not apply. b) + As from the right, we are dividing a constant by smaller and smaller positive numbers, so the quotient grows without bound, so the it is. ôpital s Rule does not apply. sin c) sin, so ôpital s Rule applies directly. Take the derivative of numerator and denominator and we have sin cos = = =. d) ln ) ln + ) ln ), so ôpital s Rule applies directly. Take the derivative of ln + ) numerator and denominator, without forgetting the chain rule, and we have ln ) ln + ) = 4 + = 4 + ) ), Date: Fall 7.

2 e) so ôpital s Rule applies again. We have ln ) ln + ) = ) 4 ) 4 TEST FORM A SOLUTIONS) = , so ôpital s Rule, if it applies, applies indirectly. Multiply by the conjugate and we have = = = 4 4 ) + 4 = Here ôpital s Rule applies directly, and we have ) 4 = + 4 = which is a right mess, but multiply numerator and denominator by 4 and we have ) 4 4 = ). This still approaches /, and even worse, it seems to grow more complicated rather than less, so we contemplate a different approach: multiply numerator and denominator by /. That gives us ) 4 4 = ) = ,. ) = 6 = 4. f) + ) + + ) approaches the form, so ôpital s Rule, if it applies, applies + indirectly. Since the variable is in the eponent, we need to use the natural logarithm: y = + ) ln y = ln + ) ln y = ln + ) ln + ) ln y = ln + ) ln y = + + Now ôpital s Rule apples, and we have ln + ) ln y = + + = + +. = = 6.

3 This is the it of ln y, so + + ) TEST FORM A SOLUTIONS) = + y = + eln y = e + ln y = e 6.. Suppose we want to approimate by using Newton s Method to find a root of, starting at =. a) Find the first four approimations. We are given f ) = and an initial approimation =. Newton s Method computes each successive approimation i+ using so we compute f ) = and thus i+ = i f i) f i ), = f ) f ) = = 4. = f ) f ) = = f ) f ) = = 4 f 4) f 4 ) = b) Are they correct to the nearest thousandth place? Why or why not? The approimations are not guaranteed correct because they have not yet repeated up to the thousandths place. Note: after this point they actually do start repeating, so technically they are correct, but you don t actually know that yet. I would not penalize you for saying no, they re not accurate when in fact they are but you lack evidence.) 4. We want to find or approimate A = d. a) Use high school geometry to approimate the area under the curve. To be clear: I do not want anything sophisticated here. I do not epect you to find the eact area. Your grade depends on how intelligently you use ideas of high school geometry to approimate the area. As long as it makes sense, you earn full credit. The graph of looks like this: It looks as if we could use a semicircle and a triangle to approimate the area beneath it:

4 TEST FORM A SOLUTIONS) 4 The area of the semicircle would be π while the area of the triangle would be ) = π 8, 6 = 6. Because the semicircle is beneath the -ais, we subtract its area from that of the triangle, so we have 6 π b) Use three rectangles and left endpoints to approimate A. We have = ) / =. The left endpoints are thus =, =, and =, whence A [f ) + f ) + f )] = + + ) =. c) Use si rectangles and left endpoints to approimate A. We have = ) /6 = /. The left endpoints are thus =, = /, =, = /, =, and = 5 /, whence [ A f ) + f = ) + f ) + f ) 4 = 5 8 =.5. ) + f ) + f )] 5 d) Use the definition of the integral to find the eact value of A.

5 TEST FORM A SOLUTIONS) 5 We have = ) /n and, since we use right endpoints, i = + i /n = i /n. Thus d = f i ) n = n = [ i ) ] i [ i ) ] i n n n = n = = [ ] 9 i i n n [ 9 n n + ) n + ) ] n n + ) n n 6 n [ ] 7n n + ) n + ) 9n n + ) 6n n. using ôpital s Rule or some other method) = = 9. e) Our approimation in part a) wasn t so bad, after all!) Why do we epect c) to be more accurate than b), and d) to be eact? We epect c) to be more accurate because it uses more rectangles, and thus typically) leaves less room for error. We epect d) to be eact because, as the number of rectangles approaches, the error should correspondingly approach. 5. bonus) Suppose we know that A L is the approimation of b f ) d using n left a endpoints, and we also want to find A R, the approimation using n right endpoints. Use the geometry of these approimations to eplain how we can find A R by subtracting one easily-found value which one?) and adding one easily-found value which one?). We can find A R by removing the leftmost rectangle s area and adding the rightmost rectangle s area, because the left endpoint of one rectangle is the right endpoint of the rectangle to its immediate left. Hence A R = A L f ) }{{} leftmost rectangle + f n ) }{{} rightmost rectangle.

6 TEST FORM A SOLUTIONS) 6 Left endpoints: i = a + i ) Right endpoints: i = a + i Midpoints: i = a + i ) Sum shortcuts: c = cn i = i i + ) Useful formulas i = n n + ) n + ) 6 i = n n + ) 4