[PDF] MATH 1207 R02 MIDTERM EXAM 1 SOLUTION • Write your answer

View - Download MATH 1207 R02 MIDTERM EXAM 1 SOLUTION • Write your answer

Previous PDF

Next PDF

MATH 1207 R02 MIDTERM EXAM 1 SOLUTION

SPRING 2016 - MOON

Write your answer neatly and show steps.

Except calculators, any electronic devices including laptops and cell phones are not allowed. (1)

Quick survey .

(a) (1 pt) This class is: Too easy Moderate Too difficult

1 2 3 4 5 6 7

(b) (2 pts) W riteany suggestion for impr ovingthis class. (For instance, give moreexamplesinclass, explainproofsindetail, givemorehomework, slow down the tempo, ...) (2) (5 pts) Find the average of y= sinxover the interval[0;].Z 0 sinxdx=cosx]

0=cos(cos0) = 1(1) = 2

Average=10Z

0 sinxdx=2

Writing the formula10Z

0 sinxdxfor the average: 3 pts.

Getting the answer2

: 5 pts.Date: March 7, 2015. 1

MATH 1207 Midterm Exam 1 Spring 2016 - Moon

(3) (a) (3 pts) Sketch the r egionbounded by y=x+ 3,y= 2,y=x24 andy-axis.The blue region is the region bounded by given equations.

Sketching curves: 1 pt.

Indicating the region correctly: 3 pts.

(b) (6 pts) Find the ar eaof the r egionin (a). To find the area of the blue region, it is sufficient to subtract the area of the red triangle from the area between two curvesy=x+ 3andy=x24

Area ofand=Z

2 0 x+ 3x24 dx=12 x2+ 3xx312 2 0 =42 + 6812 =103

Area of=103

12 =176 Setting up an integral describing the area: 3 pts.

Getting the answer176

: 6 pts.2

MATH 1207 Midterm Exam 1 Spring 2016 - Moon

(4) (6 pts) The gr eatPyramid of King Khufu was built ar ound2560 BC. Its base is a square with side length 756 ft, and its height is 481 ft. Find the volume of it

using integral. (Using the volume formula of a cone is not allowed.)Letxbe the height (vertical distance) measured from the apex of Pyramid.

The section perpendicular to the vertical axis is a square whose side lengths(x) is proportional tox. Therefore we have s(x)x =756481 or equivalently, s(x) =756481 x

The area of the section is

s(x)2=756481 x 2 =7562481 2x2:

Volume=Z

481
0756
2481

2x2dx=7562481

213
x3 481
0

7562481

213

4813=75624813

= 91636272

Therefore the volume of Pyramid is

91,636,272 ft3.

Finding the side length756481

xof the section: +2 pts.

Setting up the volume formula7562481

2x2dx: +2 pts.

Getting the answer91,636,272 ft3: +2 pts.

Writing the answer without an appropriate unit: -1 pt.3

MATH 1207 Midterm Exam 1 Spring 2016 - Moon

(5) (a)

(3 pts) Sketch the r egionbounded by the graph of y= 2xx2andx-axis.(b)(6 pts) Find the volume of the solid generated by r otatingthe planar r egion

in (a) aboutx-axis. The section of the solid of revolution is a circular disk whose radius is2x x 2.

Volume=Z

2 0 (2xx2)2dx =Z 2 0

4x24x3+x4dx=43

x3x4+15 x5 2 0 =(323

16 +325

) =1615

Constructing the integralZ

2 0 (2xx2)2dxdescribing the volume: 4 pts.

Getting the answer1615

: 6 pts.4

MATH 1207 Midterm Exam 1 Spring 2016 - Moon

(c) (6 pts) Find the volume of the solid generated by r otatingthe planar r egion in (a) abouty-axis.

We can apply the shell method.

Volume=Z

2 0

2x(2xx2)dx

= 2Z 2 0

2x2x3dx= 223

x314 x4 2 0 = 2163 4 =83

Giving the integralZ

2 0

2x(2xx2)dx: 4 pts.

Obtaining the answer83

: 6 pts. (6) (5 pts) Evaluate Ze

1(lnx)2x

dx: u= lnx)du=1x dx;u(1) = ln1 = 0;u(e) = lne= 1 Z e

1(lnx)2x

dx=Z 1 0 u2du=13 u3 1 0 =13

Finding an appropriate substitutionu= lnx: 2 pts.

By using substitution method, gettingZ

1 0 u2du: 4 pts.

Getting the answer13

: 5 pts.5

MATH 1207 Midterm Exam 1 Spring 2016 - Moon

(7)

Let f(x) = 3x+ sinx.

(a) (2 pts) What is the domain of f? f(x)can be found for allx. So the domain is the set of all real numbersR (or(1;1). (b) (3 pts) Show that fis one-to-one. f

0(x) = 3 + cosx

Because1cosx1,f0(x)2. Thereforefis an increasing function and one-to-one.

Finding the derivativef0(x) = 3 + cosx: 1 pt.

Getting thatfis increasing fromf0(x)>0: 3 pts.

(c) (3 pts) Find f1(3). f

1(3) =x,f(x) = 3,3x+ sinx= 3

Becausesin= 0,f() = 3+ sin= 3. Thereforef1(3) =.

(d) (3 pts) Find (f1)0(3). (f1)0(3) =1f

0()=13 + cos=131=12

Applyingtheinversefunctiontheoremandgetting(f1)0(3) =1f 0():

2 pts.

Getting the answer12

: 3 pts.6

MATH 1207 Midterm Exam 1 Spring 2016 - Moon

(8) (6 pts) After the consumption of an alcoholic beverage, the concentration of alcohol in the bloodstream (blood alcohol concentration, orBAC) surges as the alcohol is absorbed, followed by a gradual decline as the alcohol is metabolized.

The function

C(t) = 1:35te2:802t

models the BAC of a male, measured in mg/mL, subjectsthours after rapid consumption of 15 mL of ethanol (corresponding to one alcoholic drink). Find the maximum BAC during the first 3 hours, and indicate when it occurs. C

0(t) = 1:35e2:802t+ 1:35te2:802t(2:802)= 1 :35e2:802t(12:802t)

Becausee2:802t>0,

C

0(t) = 0,1:35e2:802t(12:802t) = 0,12:802t= 0,t=12:802

C(12:802) = 1:3512:802e2:80212:802= 1:3512:802e10:1772

C(0) = 1:350e0= 0

C(3) = 1:353e2:80230:0009

Therefore the maximum BAC is0:1772mg/mL, and it occurs12:8020:3569 hours after the consumption. Finding the derivativeC0(t) = 1:35e2:802t(12:802t): 3 pts.

Obtaining the critical pointt=12:802: 4 pts.

Writing the maximum BAC0.1772 mg/mL with an appr opriateunit: 6 pts.

Omitting the unit: -1 pt.7

quotesdbs_dbs47.pdfusesText_47

[PDF] 4 commerce

[PDF] 4 commerce srl

[PDF] 4 commerce voorhees

[PDF] 4 franc liberation stamp web france

[PDF] 4 gestion academy

[PDF] 4 gestion economie 2015

[PDF] 4 gestion marketing economie 2016

[PDF] 4 gestion youtube comptabilite generale s2

[PDF] 4 image 1 mot vous etes a quel niveau

[PDF] 4 juillet 2016 big celebration to america

[PDF] 4 line page for english writing pdf

[PDF] 4 notions allemand bac 2015

[PDF] 4 notions anglais bac 2015

[PDF] 4 notions anglais bac 2016

[PDF] 4 notions english bac