ID blackboard.ncat.edu Apple Dieney ESPN Course Materials-MATH132002.201810 2907
ID: 2895444 • Letter: I
Question
ID blackboard.ncat.edu Apple Dieney ESPN Course Materials-MATH132002.201810 2907410-dt-content-rid-14 Below Is The Lewis Structure Of The Name: DO NOT STAPLE. TAPE.OR FOLD Instructions: Please answer all questions in the spaces provided. Show all necessary work, and express all answers in simplest non-decimal form: points are as indicated. Please write legibly, you will grade a copy. 1. Find the area of the shaded region in the following figures - your choice of variable [4*9-36pts] y=x 1) cos x 2)Explanation / Answer
posted multiple questions. please post each question seperately
1)
1)
given curves y =x3, y= x
when curves intersect , x3=x
=>x3-x=0
=>x(x2-1)=0
=>x=-1 ,0,1
area of the shaded region =[-1 to 0](x3-x) dx +[0 to 1](x-x3) dx
area of the shaded region =[-1 to 0]((1/4)x4-(1/2)x2) +[0 to 1]((1/2)x2-(1/4)x4)
area of the shaded region =[((1/4)04-(1/2)02) -((1/4)(-1)4-(1/2)(-1)2) ] +[((1/2)12-(1/4)14) -((1/2)02-(1/4)04) ]
area of the shaded region =[0 -(-(1/4)) ] +[(1/4) -0 ]
area of the shaded region =(1/4) +(1/4)
area of the shaded region =(1/2)
area of the shaded region =0.5 square units
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2)
given curves y =(1/4)sec2x, y= 4cos2x
when curves intersect , (1/4)sec2x= 4cos2x
=>cos4x=(1/16)
=> cosx=1/2
=>x=-/3,/3
area of the shaded region =[-/3 to /3]( 4cos2x -(1/4)sec2x) dx
area of the shaded region =[-/3 to /3]( 2(1+cos2x) -(1/4)sec2x) dx
area of the shaded region =[-/3 to /3]( 2+2cos2x -(1/4)sec2x) dx
area of the shaded region =[-/3 to /3]( 2x+sin2x -(1/4)tanx)
area of the shaded region =( (2/3)+sin(2/3) -(1/4)tan(/3)) -( (-2/3)+sin(-2/3) -(1/4)tan(-/3))
area of the shaded region =( (4/3)+2sin(2/3) -(1/2)tan(/3))
area of the shaded region =( (4/3)+3 -(3/2))
area of the shaded region =(4/3)+(3/2)
area of the shaded region =(1/6)[8 +33]
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3)
given curves y =sinx , y=cosx
when curves intersect, sinx =cosx
=>x=(/4)
area of the shaded region =[0 to /4](sinx -0) dx +[/4 to /2](cosx -0) dx
area of the shaded region =[0 to /4](sinx ) dx +[/4 to /2](cosx) dx
area of the shaded region =[0 to /4](-cosx) +[/4 to /2](sinx)
area of the shaded region =(-cos(/4)-(-cos0)) +(sin(/2) -sin(/4))
area of the shaded region =(-(1/2)+1) +(1 -(1/2))
area of the shaded region =2(1 -(1/2))
area of the shaded region =2-2
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3)
given curves y=x+8 , y =8-2x
=>x=y-8 , x=4 -(y/2)
when curves intersect
y-8 =4 -(y/2)
=>2y -16 =8-y
=>3y =24
=>y =8
areaof the shaded region =[0 to 8][(4 -(y/2))-(y-8)] dy
areaof the shaded region =[0 to 8][4 -(y/2)-y+8] dy
areaof the shaded region =[0 to 8][12 -(3y/2)] dy
areaof the shaded region =[0 to 8][12y -(3/4)y2]
areaof the shaded region =[12*8 -(3/4)82] -[12*0 -(3/4)02]
areaof the shaded region =[96 -48] -0
areaof the shaded region =48 square units