For time t ≥ 0, the velocity of a particle moving along the x-axis is given by v(t) = x3 − 3x2 + x.

Question Answered step-by-step Question 1 For time t ≥ 0, the velocity of a particle moving along the x-axis is given by v(t) = x3 − 3x2 + x. The initial position of the particle at time t = 0 is x = 3. Which of the following gives the total distance the particle traveled from time t = 0 to time t = 4?

Question 2 What is the average value of on the closed interval [1, 3]?

Question 3 Which definite integral represents the area of the region that is bounded by 3x + 3y2 = 0 and y = 2x + 1?

Question 4 A certain mosquito population changes at the rate M(t) = 12.1(1.2)t mosquitos per day, where t represents time in days.

If the mosquito population is 649 at t = 0, then how many mosquitos are there on day 5?

Round to the nearest whole number.

99 735 748

Question 5 The water level of a certain body of water is changing at a rate of

inches per hour, where t represents hours since 12 a.m.

Part A: Using correct units, explain the meaning of in terms of the context of this problem. (1 point) Part B: Write the integral that represents the number of inches that the water level changes from 10 a.m. to 3 p.m.

Part C: Use your calculator to evaluate your integral from Part B. Explain the meaning of the answer in terms of the context.

Part D:What is the average hourly number of inches that the water level changes for this lake in one day?

Question 6 For the regions A and B shown in the graph: Part A: Discuss the limits of integration.

Part B: Set up an integral expression that represents the total area.

Part C: Calculate the total area.

Image transcription text 4 B A y = -X2+3 -1 x = (y-1)2 -2 -2 -4 -3 -2 -1 2 4 6

MATH

CALCULUS MATH 101

Answer & Explanation Solved by verified expert Rated Helpful 1) b 2) b 3)a 4) c 5 and 6 check images Step-by-step explanation

Image transcription text 14 Q 1 . Six -3x tal do will give the total distance travelled by the particle from t = 0 to t= 4. ( ( x3 - 3 x 2 + x ) du + this will give the total o displacement travelled by the particle from to to t = 4 14 3 + ( 1 25 - 3x tuldad + this will give the position of particle. Q 2 . B ) In ( 3 ) a Average value = b - a 6 3 3 1 dx 3 - 1 2 = 1 ( luxe ] 2 = 1 en3 - 1 mo(1 ) 2 = 1 en3 [ As unl = 0 ] 2 Show less

Image transcription text 3 to find shaded orca 2 X= - 4 4= 2x + 1 that WA . ( 21 17. ) IR A ( acidy - fx . dy we have to determine yikyz ( #2 , 72 ) 2) x 2 -42 - (0) 4 = 2x+ 1 - (2 ) -) putting a from @ in Q y = 2 ( 4 2 ) + 1 = 1 242 + 4 - 1 = 0 => 2 4+ 2 4 + zy - y - 1 = 0 24 ( 4 +1 ) - 1 (y + 1) = 0 = (4+1). (24 +1)=0 8 = - 1 1/2 41 = 1/2 $2 = - 1 Let fly) = x = - y * 9ly ) = x = y - 2 Area = ( fig ) dy - S gly ) dy 82 1/ 2 1/ 2 - yzdy - 5 ( 4 - 1 ) dy -1 1/2 1 ( - y z + 4 ) dy Show less

Image transcription text $ 4. M It ) = 12.1 ( 1 . 2 ) t < et P' (+ = MIt ) = 12.1 ( 1.2) t S p' ( # ) dt = P It ) = / 12. 1 ( 1. 2 ) P ( t ) = 12 .

1 ( 1 . 2) * + C In ( 1 . 2 ) at t = 0 , P10 ) = 649 P ( t ) = 12 . 1 + C = ) C = 649 - 12.1 = 582.63 in ( 1. 2 ) (n ( 1 .

2 ) Population at * = 5, = 12.1 (1.2) + 5 82. 63 = 747.77 In ( 1. 2 ) = 74 8 mosquito Show less

Image transcription text Q5, Part A : [ withat 2 it shows that the number of inches 12 the water level changes from 12 to 24 hours.

B 3 Pm 15 = [ withat = (wit ) at 10 am 10 C ! 20 find w( t) at 15 - 3 sin ( 1 - t ) at 2 3 Sin / t * - 1 ) at 10 2 1 3 5 = t/ [ 3 ws ( #- 1) ] =3 Cos ( 13 ) - COX ( 4 ) =4. 8906 Inches It represent the number of inches that water level changes from 10 am to 3 pm thatis 4. gob. day : 12 pm 24 24 Swithat = [with at = sin ( 1 - t )at 12 am 16 = 3 [ cos ( 11) - co (1)] = -1. 6076 Inches It represent that water level of lake is decreased by 1:6076 Inches per day. Show less

Image transcription text

26. Past A : - for Region A : - 4 = Natz + 1 to y = - x1 + 3 4 2=-2 to 2 2- 1 (we can see from graph) To calculate limit for Region B, we find intersection point. - * + 3 = Na+ 2 + 1 2 4- 425- 2 +2 =0 x = - 1, s = 2 x = - 1 - Ns , x = - N Required point we can see from graph DC = - 1 = 0 . 6 2 2 for Region B 4=- 2 +3 to y= Natz + 1 K X= - 1 to x = 0.6 B : Total area : - -1 4= - x +3 2 = 0. 6 4 =12+2+ 1 dy da + dy dal x = - 2 y= vx+2 + 1 Show less

Image transcription text <: Total ance = 1+ 3 2 0 .6 *+2 + 1 8 - 26 + 3 = 10 .6 1( - * +3 - N2 + 2 - 1 ) dx + ( (Jxtz + 1 + x - 3) dix 2 1 0 . 6 ( - set - 5x + 2 + 2 ) du + ( ( Jx + 2 + xx " - 2 ) du - 2 - X - ( x+2 ) 3 / 2 -1 312 D. 6 + 2 21 + (x+2 ) + 2 - 2x 3 3 / 2 3 / 2 - 1 +3+8+ 2-6-0.6-2+3-1 W / N - 1 3 3 3 + 1. 8 - 0. 216 + 2 ( 2 . 6 ) 3/ 2 3 13 - + 8 + 1. 2 - 0.216 + 2 (2.6) /2. 6 W joo wp 3 3 + 1- 2 - 0. 216 + 2.79 WIN 3 = 016 7 + 1. 2 - 0. 07 + 2. 79 Jofal = 4.59 ence Show less