PROBLEM 1:    Compute the area of the enclosed region bounded by the graphs of the equations y=x , y=2x , and x=4.

SOLUTION 1:    Compute the area of the enclosed region bounded by the graphs of the equations y=x , y=2x and x=4 . Begin by finding the points of intersection of the two graphs. From y=x and y=2x we get that

x = 2x     →

−x = 0     →

x = 0     →

Using vertical cross-sections we get that

AREA = ∫₀⁴ (top  −  bottom)  dx

= ∫₀⁴ (2x − x)  dx

= ∫₀⁴ x  dx

= [(x²)/2] |₀⁴

= [(4²)/2] − [(0²)/2]

= 8 − 0

= 8

PROBLEM 2:    Compute the area of the enclosed region bounded by the graphs of the equations y=x² and y=x+2 .

SOLUTION 2:    Compute the area of the enclosed region bounded by the graphs of the equations y=x² and y=x+2 . Begin by finding the points of intersection of the two graphs. From y=x² and y=x+2 we get that

x²=x+2     →

x²−x−2=0     →

(x−2)(x+1)=0     →

x=2  or   x=−1

Using vertical cross-sections we get that

AREA = ∫₋₁² (top  −  bottom)  dx

= ∫₋₁² ((x+2)−x²)  dx

= ( [(x²)/2]+2x−[(x³)/3] )|₋₁²

= ( [(2²)/2]+2(2)−[(2³)/3] ) − ( [((−1)²)/2]+2(−1)−[((−1)³)/3] )

= (2+4−⁸/₃) − (¹/₂−2+¹/₃)

= (6−⁸/₃) − (³/₆−[12/6]+²/₆)

= ([36/6]−[16/6]) − ([(−7)/6])

= [20/6]+⁷/₆

= [27/6] .

PROBLEM 3:    Compute the area of the enclosed region bounded by the graphs of the equations y=e^x, y=e^−x, and x=ln3 .

SOLUTION 3:    Compute the area of the enclosed region bounded by the graphs of the equations y=e^x , y=e^−x , and x=ln3 . Begin by finding the points of intersection of the two graphs. From y=e^x and y=e^−x we get that

e^x = e^−x     →

e^2x = 1     →

x = 0     →

Using vertical cross-sections we get that

AREA = ∫₀^ln3 (top  −  bottom)  dx

= ∫₀^ln3 (e^x − e^−x)  dx

= (e^x − (−e^−x)) |₀^ln3

= (e^x + e^−x) |₀^ln3

= (e^ln3 + e^−ln3) − (e⁰+ e⁻⁰)

= (3 + [1/3]) − (1 + 1)

= ([10/3]) − (2)

= [10/3] − [6/3]

= [4/3]

PROBLEM 4:    Compute the area of the enclosed region bounded by the graphs of the equations y=x² and y=−x²+4x+6 .

SOLUTION 4:    Compute the area of the enclosed region bounded by the graphs of the equations y=x² and y=−x²+4x+6 . Begin by finding the points of intersection of the two graphs. From y=x² and y=−x²+4x+6 we get that

x² = −x² + 4x + 6     →

2x² − 4x − 6 = 0     →

x² − 2x − 3 = 0     →

(x−3)(x+1) = 0     →

x = 3 or x = −1

Using vertical cross-sections we get that

AREA = ∫₋₁³ (top  −  bottom)  dx

= ∫₋₁³ ((−x²+4x+6)−(x²))  dx

= ∫₋₁³ (−x²+4x+6)  dx

= ([(−2x³)/3] + [(4x²)/2] + 6x) |₋₁³

= (−[2/3]x³ + 2x² + 6x )|₋₁³

= ( −[2/3](3)³ + 2(3)² + 6(3)) − ( −[2/3](−1)³ + 2(−1)² + 6(−1) )

= ( −18 + 18 + 18 ) − ( [2/3]+ 2 − 6 )

= ( 18 ) − ( [2/3] −[6/3] − [18/3] )

= ( 18 ) − ( − [22/3] )

= 18 + [22/3]

= [54/3] + [22/3]

= [76/3]

PROBLEM 5:    Compute the area of the enclosed region bounded by the graphs of the equations y=x³+x² and y=3x²+3x .

SOLUTION 2:    Compute the area of the enclosed region bounded by the graphs of the equations y=x³+x² and y=3x²+3x . Begin by finding the points of intersection of the two graphs. From y=x³+x² and y=3x²+3x we get that

x³ + x² = 3x² + 3x     →

x³ − 2x² − 3x = 0     →

x(x² − 2x − 3) = 0     →

x(x−3)(x+1) = 0     →

x=0 , x=3 , or x=−1

Using vertical cross-sections we get that

AREA = ∫₋₁⁰ (top  −  bottom)  dx +∫₀³ (top  −  bottom)  dx

= ∫₋₁⁰ ((x³+x²)−(3x²+3x))  dx+ ∫₀³ ((3x²+3x) −(x³+x²))  dx

= ∫₋₁⁰ (x³−2x²−3x)  dx +∫₀³ (−x³+2x²+3x)  dx

= ( [(x⁴)/4] − [(2x³)/3] −[(3x²)/2] ) |₋₁⁰ + ( −[(x⁴)/4]+ [(2x³)/3] + [(3x²)/2] ) |₀³

= ( [(0⁴)/4] − [(2(0)³)/3] −[(3(0)²)/2] ) − ( [((−1)⁴)/4] −[(2(−1)³)/3] − [(3(−1)²)/2] ) + (−[(3⁴)/4] + [(2(3)³)/3] + [(3(3)²)/2] )

    − ( −[(0⁴)/4] +[(2(0)³)/3] + [(3(0)²)/2] )

= ( 0 ) − ( [1/4] +[2/3] − [3/2] ) + ( −[81/4] + 18 +[27/2] ) − ( 0 )

= − ( [3/12] + [8/12] −[18/12] ) + ( −[81/4] + [72/4] +[54/4] )

= − ( −[7/12] ) + ([45/4] )

= [7/12] + [45/4]

= [7/12] + [135/12]

= [142/12]

= [71/6]

PROBLEM 6:    Compute the area of the enclosed region bounded by the graphs of the equations y = lnx, y=1, and x=e² .

SOLUTION 6:    Compute the area of the enclosed region bounded by the graphs of the equations y=lnx , y=1 and y=e² . Begin by finding the points of intersection of the two graphs. From y=lnx and y=1 we get that

lnx = 1     →

e^lnx = e¹

x = e

Using vertical cross-sections we get that

AREA = ∫_e^e2 (top  −  bottom)  dx

= ∫_e^e2 (lnx − 1)  dx

= ∫_e^e2 lnx  dx − ∫_e^e21  dx

Let A = ∫lnx  dx .

Use integration by parts. Let u = lnx and dv = dx such that du = [1/x]  dx and v = x .

A = x lnx − ∫([1/x])(x)  dx

= x lnx − ∫ dx

= x lnx − x + C

A − ∫_e^e2 1  dx

= ( x lnx − x − x )|_e^e2

= ( x lnx − 2x )|_e^e2

= ( e² lne² − 2e² ) − ( elne − 2e )

= ( e²(2) − 2e² ) − ( e − 2e)

= ( 0 ) − ( −e )

= e

PROBLEM 7:    Compute the area of the enclosed region bounded by the graphs of the equations y = cosx, y = sinx, and x=0 .

SOLUTION 7:    Compute the area of the enclosed region bounded by the graphs of the equations y = cosx , y = sinx and x = 0 . Begin by finding the points of intersection of the two graphs. From y=x² and y=x+2 we get that

cosx = sinx     →

[cosx/sinx] = 1     →

cotx = 1     →

x = [(π)/4]

Using vertical cross-sections we get that

AREA = ∫₀^{π/ 4} (top  −  bottom)  dx

= ∫₀^{π/ 4} ( cosx − sinx )  dx

= ( sinx − (−cosx) )|₀^{π/ 4}

= ( sinx + cosx ) |₀^{π/ 4}

= ( sin[(π)/4] + cos[(π)/4]) − ( sin0 + cos0 )

= ( [(√2)/2] + [(√2)/2]) − ( 0 + 1 )

= √2 − 1

PROBLEM 8:    Compute the area of the enclosed region bounded by the graphs of the equations y=8/x, y=2x, and y=2 .

SOLUTION 8:    Compute the area of the enclosed region bounded by the graphs of the equations y = [8/x] , y = 2x and y = 2 . Begin by finding the points of intersection of the two graphs. From y = [8/x] and y = 2x we get that

[8/x] = 2x     →

8 = 2x²     →

4 = x²     →

x = 2 or x = −2     →

y = 4 or y = −4

Using horizontal cross-sections such that x = [8/y] and x = [y/2] we get that

AREA = ∫₂⁴ (right  −  left )  dy

∫₂⁴ ( [8/y] − [y/2] )  dy

= ( 8 lny − [(y²)/4] )|₂⁴

= ( 8 ln4 − [(4²)/4] ) − (8 ln2 − [(2²)/4] )

= ( 8 ln4 − 4 ) − ( 8 ln2 − 1)

= ( 8 ln2² − 4 ) − ( 8 ln2 −1 )

= ( 16 ln2 − 4 ) − ( 8 ln2 − 1)

= 8 ln2 − 3

PROBLEM 9:    Compute the area of the enclosed region bounded by the graphs of the equations y=lnx and y = (lnx)² .

SOLUTION 9:    Compute the area of the enclosed region bounded by the graphs of the equations y=lnx and y = (lnx)² . Begin by finding the points of intersection of the two graphs. From y=lnx and y = (lnx)² we get that

lnx = (lnx)²     →

lnx − (lnx)² = 0     →

lnx (1 − lnx)     →

lnx = 0 or lnx = 1     →

x = 1 or x = e

Using vertical cross-sections we get that

AREA = ∫₁^e (top  −  bottom)  dx

= ∫₁^e (lnx − (lnx)²)  dx

= ∫₁^e lnx  dx − ∫₁^e (lnx)²  dx

Let A = ∫₁^e lnx  dx and B = ∫(lnx)²  dx .

(Recall from PROBLEM 6 that ∫lnx  dx = xln− x + C )

A = ( x lnx − x ) |₁^e

= ( e lne − 1 ln1 ) − ( e − 1)

= ( e − 0 ) − ( e − 1 )

= 1

Use integration by parts. Let u = (lnx)² and dv = dx such that du = 2 lnx ([1/x])  dx and v = x .

B = ( x(lnx)² ) |₁^e −∫₁^e 2 lnx  dx

= ( x (lnx)² ) |₁^e −2 ( x lnx − x ) |₁^e

= ( e(lne)² − 1(ln1)² ) − 2( (e lne − e) − (1 ln1 − 1) )

= ( (e − 0) ) − 2 ( (e − e) − (0 −1) )

= e − 2

A − B = 1 − (e − 2)

= 3 − e

PROBLEM 10:    Compute the area of the enclosed region bounded by the graphs of the equations y=tan² x , y=0 and x=1 .

SOLUTION 10:    Compute the area of the enclosed region bounded by the graphs of the equations y = tan² x , y = 0 , and x = 1 . Begin by finding the points of intersection of the two graphs. From y = tan² x and y = 0 we get that

tan² x = 0     →

tanx = 0     →

x = 0

Using vertical cross-sections we get that

AREA = ∫₀¹ (top  −  bottom)  dx

= ∫₀¹ tan² x  dx

= ∫₀¹ ( sec² x − 1 )  dx

= ( tanx − x ) |₀¹

= ( tan1 − 1 ) − ( tan0 − 0)

= tan1 − 1

PROBLEM 11:    Compute the area of the enclosed region bounded by the graphs of the equations y=x, y=2x, and y=6−x .

SOLUTION 11:    Compute the area of the enclosed region bounded by the graphs of the equations y=x , y=2x , and y=6−x . Begin by finding the points of intersection of the two graphs.

From y=x and y=2x we get that

x = 2x     →

−x = 0     →

x = 0

From y=x and y=6−x we get that

x = 6−x     →

2x = 6     →

x = 3

From y=2x and y=6−x we get that

2x = 6−x     →

3x = 6     →

x = 2

Using vertical cross-sections we get that

AREA = ∫₀² (top  −  bottom)  dx +∫₂³ (top  −  bottom)  dx

= ∫₀² (2x − x)  dx + ∫₂³((6−x) − x)  dx

= ∫₀² x  dx + ∫₂³ (6−2x)  dx

= [(x²)/2] |₀² + ( 6x −[(2x²)/2] ) |₂³

= [(x²)/2] |₀² + ( 6x −x² ) |₂³

= ( [(2²)/2] − [(0²)/2] )+ ( (6 ·3 − 3²) − (6 ·2 − 2²) )

= ( 2 − 0 ) + ( 9 − 8 )

= 2 + 1

= 3

PROBLEM 12:    Compute the area of the enclosed region bounded by the graphs of the equations y = sin√x , y=0 , and x = π/ 2 .

SOLUTION 12:    Compute the area of the enclosed region bounded by the graphs of the equations y = sin√x , y=0 , and x = π/ 2 . Begin by finding the points of intersection of the two graphs. From y = sin√x and y=0 we get that

sin√x = 0     →

x = 0

Using vertical cross-sections we get that

AREA = ∫₀^{π/ 2} (top  −  bottom)  dx

= ∫₀^{π/ 2} ( sin√x )  dx

Use integration by parts. Let u = sin√x and dv = dx such that du = cos√x ·[1/(2√x)]  dx and v = x .

= x sin√x |₀^{π/ 2} −∫₀^{π/ 2} cos√x ·[1/(2 √x)] ·x  dx

= ( ([(π)/2]) sin√{[(π)/2]} − (0) sin√0 ) − ∫₀^π/2 cos√x ·[1/(2 √x)] ·x  dx

= [(π)/2] sin√{[(π)/2]} −∫₀^{π/ 2} cos√x ·[1/(2 √x)] ·x  dx

Use substitution. Let

w = √x

so that

dw = [1/(2 √x)]  dx = [(√x)/2x]  dx

x  dw = [(√x)/2]  dx     →

w²  dw = [(√x)/2]  dx

= [(π)/2] sin√{[(π)/2]} −∫₀^{√{π/ 2}} cosw ·w²  dw

Use integration by parts. Let u′ = w² and dv′ = cosw  dw such that du′ = 2w  dw and v′ = sinw .

= [(π)/2] sin√{[(π)/2]} −( w² sinw |₀^{√{π/ 2}} −∫₀^{√{π/ 2}} 2w sinw  dw )

= [(π)/2] sin√{[(π)/2]} −w² sinw |₀^{√{π/ 2}} + ∫₀^{√{π/ 2}} 2w sinw  dw

= [(π)/2] sin√{[(π)/2]} −( (√{[(π)/2]})² sin√{[(π)/2]} −(0)² sin0 ) + ∫₀^{√{π/ 2}} 2w sinw  dw

= [(π)/2] sin√{[(π)/2]} −[(π)/2] sin√{[(π)/2]} + ∫₀^√{π/2} 2w sinw  dw

= ∫₀^{√{π/ 2}} 2w sinw  dw

Use integration by parts. Let u" = 2w and dv" = sinw  dw such that du" = 2  dw and v" = − cosw

= −2 w cosw |₀^{√{π/ 2}} +∫₀^{√{π/ 2}} 2 cosw  dw

= −2 ( (√{[(π)/2]}) cos√{[(π)/2]} − (0) cos0 ) + 2 sinw|₀^{√{[(π)/2]}}

= −2(√{[(π)/2]}) cos√{[(π)/2]} + 2 ( sin√{[(π)/2]} − sin0)

= −√{2 π} cos√{[(π)/2]} + 2sin√{[(π)/2]}

PROBLEM 13:    Compute the area of the enclosed region bounded by the graphs of the equations x=y² and x=4 .

SOLUTION 13:    Compute the area of the enclosed region bounded by the graphs of the equations x=y² and x=4 . Begin by finding the points of intersection of the two graphs. From x=y² and x=4 we get that

y² = 4     →

y = −2 or y = 2

Using horizontal cross-sections we get that

AREA = ∫₋₂² (right  −  left)  dy

= ∫₋₂² (4 − y²)  dy

= ( 4y − [(y³)/3] )|₋₂²

= ( 4 ·2 − [(2³)/3] ) −( 4 ·(−2) − [((−2)³)/3] )

= ( 8 − [8/3] ) − ( −8 +[8/3] )

= 16 − [16/3]

= [48/3] − [16/3]

= [32/3]

PROBLEM 14:    Compute the area of the enclosed region bounded by the graphs of the equations x=y+3 and x=y²−y .

SOLUTION 14:    Compute the area of the enclosed region bounded by the graphs of the equations x=y+3 and x=y²−y . Begin by finding the points of intersection of the two graphs. From x=y+3 and x=y²−y we get that

y+3 = y²−y     →

0 = y² − 2y − 3     →

0 = (y−3)(y+1)     →

y=3 or y=−1

Using horizontal cross-sections we get that

AREA = ∫₋₁³ (right  −  left)  dy

= ∫₋₁³ ((y+3)−(y²−y))  dy

= ∫₋₁³ (−y²+2y+3)  dy

= ( −[(y³)/3] + [(2y²)/2] +3y) |₋₁³

= ( −[(y³)/3] + y² +3y )|₋₁³

= ( −[(3³)/3] + 3² + 3(3) ) −( −[((−1)³)/3] + (−1)² + 3(−1) )

= ( −9 + 9 + 9 ) − ( [1/3] + 1− 3 )

= ( −9 + 9 + 9 ) − ( [1/3] +[3/3] − [9/3] )

= ( 9 ) − ( −[5/3] )

= [27/3] + [5/3]

= [32/3]

PROBLEM 15:    Compute the area of the enclosed region bounded by the graphs of the equations x=y³ and x=y²+2y .

SOLUTION 15:    Compute the area of the enclosed region bounded by the graphs of the equations x=y³ and x=y²+2y . Begin by finding the points of intersection of the two graphs. From x=y³ and x=y²+2y we get that

y³ = y² + 2y     →

y³ − y² − 2y = 0     →

y ( y² − y − 2 ) = 0     →

y ( y−2 ) ( y+1 ) = 0     →

y=0 , y=2 , or y=−1

Using horizontal cross-sections we get that

AREA = ∫₋₁⁰ (right  −  left)  dy +∫₀² (right  −  left)  dy

= ∫₋₁⁰ (y³ − (y²+2y))  dy +∫₀² (y²+2y − y³)  dy

= ∫₋₁⁰ (y³ − y² − 2y)  dy +∫₀² (y²+2y − y³)  dy

= ( [(y⁴)/4] − [(y³)/3] −[(2y²)/2] ) |₋₁⁰ + ( [(y³)/3]+ [(2y²)/2] − [(y⁴)/4] ) |₀²

= ( [(y⁴)/4] − [(y³)/3] −y² ) |₋₁⁰ + ( [(y³)/3] + y² −[(y⁴)/4] ) |₀²

= ( [(0⁴)/4] − [(0³)/3] −0² ) − ( [((−1)⁴)/4] − [((−1)³)/3] −(−1)² ) + ( [(2³)/3] + 2² − [(2⁴)/4]) − ( [(0³)/3] + 0² − [(0⁴)/4] )

= ( 0 ) − ( [1/4] +[1/3] − 1 ) + ( [8/3] + 4 − 4 ) − ( 0)

= −( −[7/12] ) + ( [8/3])

= [7/12] + [8/3]

= [7/12] + [32/12]

= [39/12]

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File translated from T_EX by T_TH, version 4.03.
On 13 Jan 2015, 14:25.