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FINAL 01

# FINAL 01 - Portillo Tom – Final 1 – Due 11:00 pm –...

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Unformatted text preview: Portillo, Tom – Final 1 – Due: Dec 19 2006, 11:00 pm – Inst: David Benzvi 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Estimate the area under the graph of f ( x ) = 21- x 2 on [0 , 4] by dividing [0 , 4] into four equal subintervals and using right endpoints as sam- ple points. 1. area ≈ 52 2. area ≈ 54 correct 3. area ≈ 55 4. area ≈ 56 5. area ≈ 53 Explanation: With four equal subintervals and right end- points as sample points, A ≈ n f (1) + f (2) + f (3) + f (4) o 1 , since x i = x * i = i . Consequently, A ≈ 54 . keywords: estimate area, quadratic function Riemman sum 002 (part 1 of 1) 10 points Use properties of integrals to determine the value of I = Z 4 f ( x ) dx when Z 6 f ( x ) dx = 9 , Z 6 4 f ( x ) dx = 4 . 1. I =- 8 2. I =- 13 3. I = 13 4. I =- 5 correct 5. I = 8 6. I = 5 Explanation: Since Z a b f ( x ) dx =- Z b a f ( x ) dx and Z b a f ( x ) dx + Z c b f ( x ) dx = Z c a f ( x ) dx, we see that Z a b f ( x ) dx =- ‡ Z c a f ( x ) dx- Z c b f ( x ) dx · . Consequently, I =- (9- 4) =- 5 . keywords: integral, properties of integrals 003 (part 1 of 1) 10 points Determine g ( x ) when g ( x ) = Z 5 x 2 t 2 sec t dt. 1. g ( x ) = 4 x sec 2 x 2. g ( x ) =- 2 x 2 tan x Portillo, Tom – Final 1 – Due: Dec 19 2006, 11:00 pm – Inst: David Benzvi 2 3. g ( x ) = 2 x 2 tan x 4. g ( x ) =- 4 x sec 2 x 5. g ( x ) = 4 x sec x tan x 6. g ( x ) =- 4 x sec x tan x 7. g ( x ) =- 2 x 2 sec x correct 8. g ( x ) = 2 x 2 sec x Explanation: By Properties of integrals and the Funda- mental Theorem of Calculus, d dx ‡ Z a x f ( t ) dt · = d dx ‡- Z x a f ( t ) dt · =- f ( x ) . When g ( x ) = Z 5 x f ( t ) dt, f ( t ) = 2 t 2 sec t, therefore, g ( x ) =- 2 x 2 sec x . keywords: Stewart5e, FTC, derivative form, properties of integrals, 004 (part 1 of 1) 10 points Find the volume, V , of the solid obtained by rotating the region bounded by y = 2 x , x = 2 , x = 3 , y = 0 about the x-axis. 1. V = 1 6 2. V = 1 3 π 3. V = 1 6 π 4. V = 1 3 5. V = 2 3 6. V = 2 3 π correct Explanation: The volume of the solid of revolution ob- tained by rotating the graph of y = f ( x ) on [ a, b ] about the x-axis is given by volume = π Z b a f ( x ) 2 dx. When f ( x ) = 2 x , a = 2 , b = 3 , therefore, V = π Z 3 2 4 x 2 dx. Consequently, V = π •- 4 x ‚ 3 2 = 2 3 π . keywords: volume, integral, solid of revolu- tion 005 (part 1 of 1) 10 points Evaluate the integral I = Z √ ln 3 xe x 2 √ e x 2 + 2 dx. 1. I = 1 2 ‡ √ 5- √ 3 · 2. I = 2 ‡ √ 5- √ 3 · 3. I = 1 2 ‡ √ 7- 2 · Portillo, Tom – Final 1 – Due: Dec 19 2006, 11:00 pm – Inst: David Benzvi 3 4. I = √ 7- 2 5. I = √ 5- √ 3 correct 6. I = 2 ‡ √ 7- 2 · Explanation: The presence of the term xe x 2 in the nu- merator suggests the substitution u = e x 2 + 2 , for then du = 2 xe x 2 dx....
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FINAL 01 - Portillo Tom – Final 1 – Due 11:00 pm –...

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