MATH225 - Calculus Final Exam-Grade- 97.15% 2025-2026 American Public University

MATH225 - Calculus Final ExamGrade- 97.15% 2025-2026 American Public University

Final Exam - Grade: 97.15%

Instructions: This is a CUMULATIVE final exam. Click on “Start” to begin the Test.

This Test may be printed by clicking the Print icon at the top of the Test window AFTER starting the Test. We suggest you work out the answers on the printed Test, then submit your answers online.

THIS IS A TIMED TEST. YOU HAVE 3 HOURS TO COMPLETE THE TEST ONCE YOU CLICK "START." You can start and stop the Test if you need to; however, the time will continue to elapse. You can also skip questions and go back to them as needed during the test. Use the 'skip' button to skip a question and question navigation pull-down menu to jump back to any questions you skipped.

Once you have completed the Test online, click “Submit Answers.” Your answers will be scored and the answer key with step-by-step solutions will become available.

Questions? Reach out to us at support@thinkwell.com. We’re here and happy to help.

Questions 35 Questions

Limits 180 Minutes

Attempt 1

Points 100 pts possible

No due date.

97.15% (97.15 of 100)

Completed on 04/27/25 at 08:39PM

Score for this quiz: 97.15% ( 97.15 /100) Submitted Apr 27 at 8:39pm This attempt took about 1 hour.

Question 1 : 2.85 pts Let f (x) = 3 x 2 − 1. What is the average rate of change of f over the interval (1, 1.1)?

Let f (x) = 3 x 2 − 1. What is the average rate of change of f over the interval (1, 1.1)?

2.63 6.3 0.63 0.1

Due Date

f(1.1)=2.63, f(1)=2ΔfΔx=f(1.1)−f(1)1.1−1=2.63−21.1−1=0.630.1=6.3f(1.1)=2.63, f(1)=2ΔfΔx=f(1.1)−f(1)1.1−1=2.63−21.1−1=0.630.1=6.3 2.85 / 2.85 Question 2 : 2.85 ptsSkip to question text.

What is the limit of the function in the graph at x = 4?

6 The limit does not exist.

8 2

As x approaches 4 from the left side, the function value approaches 6. As x approaches 4 from the right side, the function value approaches 8. These two values are different. Thus, the limit of the function at x = 4 does not exist.

2.85 / 2.85 Question 3 : 2.85 ptsSkip to question text.

f(x)={4x+1, x<54x−1, x>5Evaluate limx→ 5 + f(x).f(x)={4x+1, x<54x−1, x>5Evaluate limx→ 5 + f(x).

5 The limit does not exist.

19 21

f(x)={4x+1, x<54x−1, x>5limx→ 5 + f(x)=limx→ 5 + (4x−1)=4(5)−1=20−1=19f(x)={4x+1, x<54x−1, x>5limx→ 5 + f(x)=limx→ 5 + (4x−1)=4(5)−1=20−1=19 2.85 / 2.85

Question 4 : 2.85 pts What is the average rate of change of the function y = 4x 3 − 2 between x = 2 and x = 4?

What is the average rate of change of the function y = 4x 3 − 2 between x = 2 and x = 4?

112 −112 110 −110

ΔyΔx=(4(4)3−2)−(4(2)3−2)4−2=2242=112ΔyΔx=(4(4)3−2)−(4(2)3−2)4−2=2242=112 2.85 / 2.85 Question 5 : 2.85 pts Consider the function y = x 2 − 2x + 1. What is the slope of the tangent line at x = 2?

Consider the function y = x 2 − 2x + 1. What is the slope of the tangent line at x = 2?

2 −2 1 −1

f′(2)=limΔx→0f(2+Δx)−f(2)Δx=limΔx→0[(2+Δx)2−2(2+Δx)+1]−[(2)2−2(2)+1]Δx=limΔx→04+4Δx+ (Δx)2−4−2Δx+1−1Δx=limΔx→0Δx(2+Δx)Δx=limΔx→02+Δx=2f′(2)=limΔx→0f(2+Δx)−f(2)Δx=limΔx→0[(2+Δx)2−2(2+Δx) +1]−[(2)2−2(2)+1]Δx=limΔx→04+4Δx+(Δx)2−4−2Δx+1−1Δx=limΔx→0Δx(2+Δx)Δx=limΔx→02+Δx=2 2.85 / 2.85 Question 6 : 2.85 pts The position of a car at time t is given by the function p (t) = t 2 − 3t − 6. At what time will the velocity of the car be 7? Assume t ≥ 0.

The position of a car at time t is given by the function p (t) = t 2 − 3t − 6. At what time will the velocity of the car be 7? Assume t ≥ 0.

1 −5 5 −1

p′(t)=limΔt→0p(t+Δt)−p(t)Δt =limΔt→0[(t+Δt)2−3(t+Δt)−6]−(t2−3t−6)Δt =limΔt→0t2+2tΔt+(Δt)2−3t−3Δt−6−t2+3t+6Δt =li mΔt→0Δt(2t−3+Δt)Δt =limΔt→02t−3+Δt =2t−3 7=p′(t) 7=2t−3 t=5p'(t)=limΔt→0p(t+Δt)−p(t)Δt =limΔt→ 0[(t+Δt)2−3(t+Δt)−6]−(t2−3t−6)Δt =limΔt→0t2+2tΔt+(Δt)2−3t−3Δt−6−t2+3t+6Δt =limΔt→0Δt(2t−3+Δt)Δt =limΔt→02t−3+Δt =2t−3 7=p'(t) 7=2t−3 t=5 2.85 / 2.85 Question 7 : 2.85 pts

Compute the derivative of the function f (x) = (2x 2 + 3)2.

Compute the derivative of the function f (x) = (2x 2 + 3)2.

2 (2x 2 + 3) 8x 2x 2 + 3 8x (2x 2 + 3)

f ′(x) = (2x 2 + 3)′ (2x 2 + 3) + (2x 2 + 3) (2x 2 + 3)′ = (4x) (2x 2 + 3) + (2x 2 + 3) (4x) = 8x (2x 2 + 3) 2.85 / 2.85 Question 8 : 2.85 ptsSkip to question text.

Compute the derivative of the function f(x)=1x2−x+2.f(x)=1x2−x+2. −1(x2−x+2)2−1(x2−x+2)2 1(x2−x+2)21(x2−x+2)2 −2x+1(x2−x+2)2−2x+1(x2−x+2)2 −2x(x2−x+2)2−2x(x2−x+2)2

f(x)=−(x2−x+2)′(x2−x+2)2 =−(2x−1)(x2−x+2)2 (x2−x+2)2 =−2x+1(x2−x+2)2 2.85 / 2.85

Question 9 : 2.85 pts Compute the derivative of the function f(x) = (x4 − 2x2 + 7x + 4)3

Compute the derivative of the function f(x) = (x4 − 2x2 + 7x + 4)3

4(x4−2x2+7x+4)2(4x3−4x+7)4(x4−2x2+7x+4)2(4x3−4x+7) 3(x4−2x2+7x+4)23(x4−2x2+7x+4)2 3(x4−2x2+7x+4)2(4x3−4x+7)3(x4−2x2+7x+4)2(4x3−4x+7) 12(x4−2x2+7x+4)212(x4−2x2+7x+4)2 f '(x) 3(x4 − 2x2 + 7x + 4)2 ⋅ (x4 − = 2x2 + 7x + 4)' = 3(x4 − 2x2 + 7x + 4)2(4x3 − 4x +

=−2x+1(x2−x+2)2f(x)=−(x2−x+2)′(x2−x+2)2

=−(2x−1)

7) 2.85 / 2.85 Question 10 : 2.85 ptsSkip to question text.

What is the derivative of the function f(x)=sin2x+tan(x2+1)?What is the derivative of the function f(x)=sin2x+tan(x2+1)?

2sinxcosx+2xsec2(x2+1)2sinxcosx+2xsec2(x2+1) 2sinxcosx+2xsec(x2+1)2sinxcosx+2xsec(x2+1) 2sinxcosx−2xsec2(x2+1)2sinxcosx−2xsec2(x2+1) 2sinxcosx+xsec2(x2+1)2sinxcosx+xsec2(x2+1) f'(x)=2sinxcosx+sec2(x2+1)(2x)=2sinxcosx+2xsec2(x2+1)f'(x)=2sinxcosx+sec2(x2+1)(2x)=2sinxcosx+2xsec2(x2+1) 2.85 / 2.85 Question 11 : 2.85 ptsSkip to question text.

What is the derivative of y=xex+x2 ?What is the derivative of y=xex+x2 ? ex + xex + 2x 2ex + x xex + x xex + 2x

Use the sum and difference rules, followed by the product rule. y'=dydx(xex)+dydx(x2) =[xdydxex+exdydx(x)] +dydx(x2) =[x(ex)+ex(1)]+2x =xex+ex+2xUse the sum and difference rules, followed by the product rule. y'=dydx(xe x)+dydx(x2) =[xdydxex+exdydx(x)]+dydx(x2) =[x(ex)+ex(1)]+2x =xex+ex+2x 2.85 / 2.85

Question 12 : 2.85 ptsSkip to question text.

What is the derivative of f(x)=ln1x?What is the derivative of f(x)=ln1x?

1x1x −x−x −1x2−1x2 −1x−1x ln(1x)=ln(x−1)=−lnx(−lnx)′=−1xln(1x)=ln(x−1)=−lnx(−lnx)′=−1x 2.85 / 2.85 Question 13 : 2.85 ptsSkip to question text.

Given the equation3x+y2=8,find dydx.Given the equation3x+y2=8,find dydx. dydx=32ydydx=32y dydx=−32ydydx=−32y dydx=2y3dydx=2y3 dydx=−2y3dydx=−2y3

3x+y2=8ddx(3x+y2)=ddx(8)ddx(3x) +ddx(y2)=ddx(8)3+2ydydx=02ydydx=−3dydx=−32y3x+y2=8ddx(3x+y2)=ddx(8)ddx(3x) +ddx(y2)=ddx(8)3+2ydydx=02ydydx=−3dydx=−32y 2.85 / 2.85 Question 14 : 2.85 pts What is the slope of the line tangent to the curve x 3 + y 3 = 2x 2y 2 at the point (1, 1)?

What is the slope of the line tangent to the curve x 3 + y 3 = 2x 2y 2 at the point (1, 1)?

1 −1 0 None of these

First, find the derivative dy/dx. Thenreplace x and y with their values. Take the derivative of both sides of the equati on and solve for dy/ dx.3x2+3y2dydx=4xy2+4x2ydydx3y2dydx−4x2ydydx=4xy2−3x2(3y2−4x2y)dydx=4xy2−3x2dydx=4xy2−3x23y2−4x2yAt t he point (1,1), dydx∣∣∣(1,1)=4−33−4=−1.First, find the derivative dy/dx. Thenreplace x and y with their values. Take the derivative of both sides of the equation and solve for dy/ dx.3x2+3y2dydx=4xy2+4x2ydydx3y2dydx−4x2ydydx=4xy2−3x2(3y2−4x2y)dydx=4xy2−3x2dydx=4xy2−3x23y2−4x2yAt t he point (1,1), dydx|(1,1)=4−33−4=−1.

2.85 / 2.85

Question 15 : 2.85 pts A ladder 25 feet long is leaning against a house. The base of the ladder is pulled away from the wall at a rate of 2 ft / sec. How fast is the top moving down the wall when the base of the ladder is 15 feet from the wall?

A ladder 25 feet long is leaning against a house. The base of the ladder is pulled away from the wall at a rate of 2 ft / sec. How fast is the top moving down the wall when the base of the ladder is 15 feet from the wall?

−32ft/sec−32ft/sec −23ft/sec−23ft/sec −416ft/sec−416ft/sec −641ft/sec−641ft/sec

Let x be the height up the wall.Let y be the distance from the wall.dydt=2Find dxdt when y=15.x 2+y2=2522xdxdt+2y dydt=0x2+y2=252x2+152=252x2=400x=20x2+y2=2522(20)dxdt+2(15)2=0dxdt=−2(15)22(20)=−32ft/sec Let x be the height up the wall.Let y be the distance from the wall.dydt=2Find dxdt when y=15.x2+y2=2522xdxdt+2ydydt=0x2+y2=2 52x2+152=252x2=400x=20x2+y2=2522(20)dxdt+2(15)2=0dxdt=−2(15)22(20)=−32ft/sec 2.85 / 2.85

Question 16 : 2.85 ptsSkip to question text.

Given: f(x)=4x2Find:f′′(−2)Given: f(x)=4x2Find:f″(−2) 3232 −32−32 8 −8

f(x)=4x−2f′(x)=−8x−3f′′(x)=24x−4 =24x4f′′(−2)=24(−2)4 (x)=24x−4 =24x4f″(−2)=24(−2)4 =2416 =32

=2416

=32f(x)=4x−2f′(x)=−8x−3f″

0 / 2.85

Question 17 : 2.85 ptsSkip to question text.

f(x)=xlnx, Find f′′(2).f(x)=xlnx, Find f″(2).

3/2 ln 2 1/2 1+ ln 2

f′(x)=x(1x)+(lnx)(1)

=1+lnxf′′(x)=1xf′′′(2)=12 f′(x)=x(1x)+(lnx)(1)

=1+lnxf″(x)=1xf‴(2)=12

2.85 / 2.85 Question 18 : 2.85 ptsSkip to question text.

Below is a graph on which five points have been labeled. Between which of these labeled points is the derivative of the function always negative?

Between 1 and 2 Between 2 and 3 Between 3 and 4 Between 4 and 5

Between 1 and 2 the function is increasing, so the derivative there is always positive. Between 3 and 4 (and between 4 and 5), the function both increases and decreases, so the derivative is positive in some places on the interval and negative in some others. Between 2 and 3 the function is only decreasing, however, so there the derivative is always negative.

2.85 / 2.85 Question 19 : 2.85 pts The function f (x) = 2x 3 − 21x 2 + 72x − 49 has critical points at x = 3 and x = 4. Which sign chart describes these critical points?

The function f (x) = 2x 3 − 21x 2 + 72x − 49 has critical points at x = 3 and x = 4. Which sign chart describes these critical points?

The derivative is f′(x)=6x2−42x+72=6(x−3)(x−4).The derivative is f′(x)=6x2−42x+72=6(x−3)(x−4).

Plugging in points x = 0, x = 3.5 and x = 5 produces a positive number, a negative number, and a positive number, respectively. Therefore, the function is increasing on the first interval, decreasing on the second interval, and increasing on the last interval.

2.85 / 2.85

Question 20 : 2.85 ptsSkip to question text.

What are the critical points of the function f(x)=x+1x?What are the critical points of the function f(x)=x+1x? x = −1, 0, 1 x = −1, 1 x=1 x=0

The derivative is f′(x)=1−1x2.To determine the critical points, find all of the points where the derivativeequals zero, a nd all of the pointswhere the derivative does not exist,but the original function does.Set f ′(x)=0 to find the critical poi ntsx=−1, 1. The derivative does notexist at x=0, but neither does the original function. The only critical pointsare x =−1, 1.The derivative is f′(x)=1−1x2.To determine the critical points, find all of the points where the derivativeequals zer o, and all of the pointswhere the derivative does not exist,but the original function does.Set f′(x)=0 to find the critical poin tsx=−1, 1. The derivative does notexist at x=0, but neither does the original function. The only critical pointsare x=−1, 1 .

2.85 / 2.85 Question 21 : 2.85 ptsSkip to question text.

On which of the following intervals is the graph off(x)=x 2−12x+1 concave up?On which of the following intervals is the graph off (x)=x2−12x+1 concave up? (−12,∞)(−12,∞) (− ∞,−12)(− ∞,−12)

f(x) is concave up for all x.f(x) is concave up for all x. f(x) is never concave up.f(x) is never concave up. f(x) is concave up when f′′(x)>0. f(x)=x2−12x+1 f′(x)=2x(2x+1)−(x2−1)(2)(2x+1)2=2x2+2x+2(2x+1)2f′′(x)=(4x+2) (2x+1)2−(2x2+2x+2)(2)(2x+1)(2)(2x+1)4=2(2x+1)[(2x+1)(2x+1)−2(2x2+2x+2)](2x+1)4=2(−3) (2x+1)3=−6(2x+1)3Set f′′(x)=0. −6(2x+1)3=0f′′(x) does not equal 0.f′′(x) does not exist at x=−12. f(x) is concave up when f″(x)>0. f(x)=x2−12x+1 f′(x)=2x(2x+1)−(x2−1)(2)(2x+1)2=2x2+2x+2(2x+1)2f″(x)=(4x+2) (2x+1)2−(2x2+2x+2)(2)(2x+1)(2)(2x+1)4=2(2x+1)[(2x+1)(2x+1)−2(2x2+2x+2)](2x+1)4=2(−3)(2x+1)3=−6(2x+1)3Set f″(x)=0. −6(2x+1)3=0f″(x) does not equal 0.f″(x) does not exist at x=−12.

f(x) is concave up on the interval (- ∞,−12).f(x) is concave up on the interval (- ∞,−12).

2.85 / 2.85 Question 22 : 2.85 pts On which interval is f (x) = 3x 5 − 20x 3 increasing?

On which interval is f (x) = 3x 5 − 20x 3 increasing?

(0, ∞)(0, ∞) (−∞, −2) and (0, 2)(−∞, −2) and (0, 2) (−2, 2)

(−∞, −2) and (2, ∞)(−∞, −2) and (2, ∞) The function f is increasing when f ′(x)>0.f(x)=3x5−20x3f′(x)=15x4−60x2Set f′ (x)=0.15x4−60x2=015x2(x2−4)=015x2(x+2)(x−2)=0x=0, x=−2, x=2The function f is increasing when f′ (x)>0.f(x)=3x5−20x3f′(x)=15x4−60x2Set f′(x)=0.15x4−60x2=015x2(x2−4)=015x2(x+2)(x−2)=0x=0, x=−2, x=2

Therefore, f is increasing on (−∞,−2)and (2, ∞).Therefore, f is increasing on (−∞,−2)and (2, ∞).

2.85 / 2.85 Question 23 : 2.85 ptsSkip to question text.

Which of the following equations could be the equation of a horizontal asymptoteof y=x2−42+x−4x2?Which of the following equations could be the equation of a horizontal asymptoteof y=x2−42+x−4x2? y=−2y=−2 y=−14y=−14 y=12y=12 y=1y=1

The horizontal asymptote is a result of taking the limit as x approaches ∞.lim x → ∞ x2−42+x−4x2=1− 4 =−14 There fore, y=−1/4 is the horizontal asymptote.The horizontal asymptote is a result of taking the limit as x approaches ∞.limx → ∞ x2−42+x−4x2=1− 4 =−14 Therefore, y=−1/4 is the horizontal asymptote.

2.85 / 2.85 Question 24 : 2.85 pts Given the position function p (x) = e 2x, find the acceleration function.

Given the position function p (x) = e 2x, find the acceleration function. e 2x e 4x 2e 4x 4e 2x

Acceleration is the second derivative of the position function. p(x)=e2x v(x)=p′(x) =2e2x a(x)=p′′(x) =2(2e2x) =4e2x

=4e2xp(x)=e2x v(x)=p′(x)

=2e2x a(x)=p″(x)

=2(2e2x)

2.85 / 2.85 Question 25 : 2.85 pts A ball is thrown directly upward. Its height h (in feet) after t seconds is given by h (t) = 5 + 80t − 16t 2. Find the maximum height the ball reaches.

A ball is thrown directly upward. Its height h (in feet) after t seconds is given by h (t) = 5 + 80t − 16t 2. Find the maximum height the ball reaches.

95 ft 100 ft 105 ft 120 ft

Maximum height occurs when velocity is 0. Therefore, v(t)=080−32t=0 t=2.5 h(2.5)=5+80(2.5)−16(2.52) 05 feet v(t)=080−32t=0 t=2.5 h(2.5)=5+80(2.5)−16(2.52) =105 feet

=5+200−16(6.25) =5+200−16(6.25)

=205−100 =205−100

2.85 / 2.85

Question 26 : 2.85 ptsSkip to question text.

Evaluate limx→1lnxx2−1.Evaluate limx→1lnxx2−1.

0 1 2 1/2

limx→1lnxx2−1→00limx→1lnxx2−1=limx→11/x2x=limx→112x2=12limx→1lnxx2−1→00limx→1lnxx2−1=limx→11/ x2x=limx→112x2=12 2.85 / 2.85 Question 27 : 2.85 ptsSkip to question text.

A cardboard box manufacturer wishes to make open boxes from 12 in. by 12 in. pieces of cardboard by cutting equal squares from each corner and turning up the sides. What is the length of the side of the square cutout that would give the box the largest possible volume?

0 in.

2 in.

6 in.

=1

128 in.

x = side of square cutout 12 – 2x = side of base V(x)=(12−2x)2x=144x−48x2+4x3V′(x)=144−96x+12x212x2−96x+144=012(x2−8x+12)=012(x−6)(x−2)=0x=6, x=2V(x)=(12−2x)2x=144x−48x2+4x3V'(x)=144−96x+12x212x2−96x+144=012(x2−8x+12)=012(x−6)(x−2)=0x=6, x=2 Because the length of the square was 12, the value of x must be in the interval [0, 6].V(0)=0, V(2)=128, V(6)=0Therefore, the extreme value theorem guarantees that x=2 in. will maximize the volume. Because the length of the square was 12, the value of x must be in the interval [0, 6].V(0)=0, V(2)=128, V(6)=0Therefore, the extreme value theo rem guarantees that x=2 in. will maximize the volume.

2.85 / 2.85 Question 28 : 2.85 ptsSkip to question text.

In order for a rectangular solid package to be mailed, the sum of the height and the perimeter of the base cannot exceed 108 in. If the base of a package is a square, what is the length of the side of the square that would maximize the volume of the package?

0 in.

12 in.

18 in.

27 in.

x=side of square basey=height4x+y=108y=108−4xVolume=V(x)V(x)=x 2yV(x)=x2(108−4x)V(x)=108x2−4x3V′ (x)=216x−12x2216x−12x2=012x(18−x)=0x=0, x=18 x=side of square basey=height4x+y=108y=108−4xVolume=V(x)V (x)=x2yV(x)=x2(108−4x)V(x)=108x2−4x3V′(x)=216x−12x2216x−12x2=012x(18−x)=0x=0, x=18

Because the sum of the dimensions is 108, x must be in the interval [0,27].V(0)=0, V(18)=11,664, V(27)=0Therefore, the extreme value theorem guarantees that x=18 in. will maximize the volume. Because the sum of the dimensions is 1 08, x must be in the interval [0,27].V(0)=0, V(18)=11,664, V(27)=0Therefore, the extreme value theorem guarantees that x=18 in. will maximize the volume.

2.85 / 2.85

Question 29 : 2.85 ptsSkip to question text.

Evaluate limx→01−cosxx2ex.Evaluate limx→01−cosxx2ex.

0 2 −2

−12−12

None of these

limx→01−cosxx2ex→00limx→01−cosxx2ex=limx→0sinx2xex+x2exlimx→0sinx2xe+x2ex→00limx→0sinx2xe+x2ex=limx→0cosx2ex +4xex+x2ex=12limx→01−cosxx2ex→00limx→01−cosxx2ex=limx→0sinx2xex+x2exlimx→0sinx2xe+x2ex→00limx→0sinx2xe +x2ex=limx→0cosx2ex+4xex+x2ex=12 2.85 / 2.85 Question 30 : 2.85 ptsSkip to question text.

Evaluate limx→∞(x−x2+x−−−−−√).Evaluate limx→∞(x−x2+x). −1 −1/2 0 −∞

limx→∞(x−x2+x−−−−−−√)→∞−∞limx→∞(x−x2+x−−−−−−√)=limx→∞x(1−1+1/x−−−−−−√)=limx→∞1−1+1/x−−−−− −√1/x=limx→∞(1/2x2)(1+1/x) −1/2−1/x2=limx→∞−12(1+1/x)1/2=−12limx→∞(x−x2+x)→∞−∞limx→∞(x−x2+x)=limx→∞x(1−1+1/x)=limx→∞1−1+1/x1/ x=limx→∞(1/2x2)(1+1/x)−1/2−1/x2=limx→∞−12(1+1/x)1/2=−12 2.85 / 2.85 Question 31 : 2.85 ptsSkip to question text.

Evaluate the indefinite integral∫ex+44 dx.Evaluate the indefinite integral∫ex+44 dx. e2x+88+Ce2x+88+C e2x4+Ce2x4+C e2x+48+Ce2x+48+C ex+44+Cex+44+C

ex+44=e44ex∫ex+44 dx=∫e44ex dx =e44∫ex dx =e4ex4+C dx =e4ex4+C =ex+44+C

=ex+44+Cex+44=e44ex∫ex+44 dx=∫e44ex dx =e44∫ex

2.85 / 2.85 Question 32 : 2.85 ptsSkip to question text.

Evaluate the indefinite integral ∫ln 2 dx.Evaluate the indefinite integral ∫ln 2 dx. x ln 2 + C ln 2 + C

(ln2)22+C(ln2)22+C ln222+Cln222+C ∫ln2 dx=ln2∫x0dx=(ln2)x0+11=xln2+C∫ln2 dx=ln2∫x0dx=(ln2)x0+11=xln2+C 2.85 / 2.85 Question 33 : 2.85 ptsSkip to question text.

Evaluate the indefinite integral∫(t√+1) (t√−1) dt.Evaluate the indefinite integral∫(t+1) (t−1) dt. t22−t+Ct22−t+C t2 − t + C

(t√+1)22(t√−1)22(t+1)22(t−1)22 None of these

(t√+1) (t√−1)=t−1∫(t√+1) (t√−1) dt=∫(t−1) dt=t22−t+C(t+1) (t−1)=t−1∫(t+1) (t−1) dt=∫(t−1) dt=t22−t+C 2.85 / 2.85 Question 34 : 2.85 ptsSkip to question text.

What is the area beneath the curve y=5−x2x3 from x = 1 to x = 2?What is the area beneath the curve y=5−x2x3 from x = 1 to x = 2?

158−ln2158−ln2 1313 358358 −152+ln1−152+ln1

∫12⎡⎣⎢5−x2x3⎤⎦⎥dx=∫12(5x−3−1x)dx =[−52x−2−lnx]∣∣12 =20−58−ln2 =158−ln2∫12[5−x2x3]dx=∫12(5x−3−1x)dx n2−(−52−ln1) =20−58−ln2 =158−ln2 2.85 / 2.85 Question 35 : 3.1 pts What is the area beneath the curve y = −x 2 − 6x − 5 and above the x-axis?

What is the area beneath the curve y = −x 2 − 6x − 5 and above the x-axis?

323323

=−58−ln2−(−52−ln1) =[−52x−2−lnx]|12

=−58−l

x 3 + 3x 2 + 5x 11 32

−x2−6x−5=0x2+6x+5=0(x+5)(x+1)=0x=−5, −1∫ −5−1(−x2−6x−5) dx =−1(x33+3x2+5x) ∣∣∣ −1 −5 =−1(−13+3−5)+ (−1253+75−25) =13+2−1253+50 =52−1243 =323−x2−6x−5=0x2+6x+5=0(x+5)(x+1)=0x=−5, −1∫−5−1(−x2− 6x−5) dx =−1(x33+3x2+5x) | −5 −1 =−1(−13+3−5)+(−1253+75−25) =13+2−1253+50 =52−1243 =323 3.1 / 3.1

Quiz Score: 97.15 out of 100