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8/14/2019 integral calculus presentation
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INTEGRAL CALCULUSA REVIEW IN
8/14/2019 integral calculus presentation
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INTEGRATION FORMULAS du = u + c adu = a du = au+c
du = + , 1
=ln
= =
sin = cos cos =sin = tan
sec tan =sec
= cot csc cot = tan =lnse cot =lnsin
cot =ln(sec csc =ln(csc
=arcsin
+ = arctan
= arcsec
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INTEGRATION FORMULAS
Integration by Parts = Trigonometric Substitution When an integrand involves , use x= a sin When an integrand involves , use x= a tan When an integrand involves , use x= a sec
8/14/2019 integral calculus presentation
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Evaluate:
+
a.4 ln (3x+2) c. ln (3 b. ln (3 2) d. 2 ln (3x + 2)
43 2
= 4. + = 3 2
8/14/2019 integral calculus presentation
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Evaluate the integral of x Sin2xdxa. cos2 sin2 b. cos2 sin2 c. cos2 sin2 d. 2
Using integration by parts:
u= x dv= sin2xdxdu= dx v= 2
2 =1
2 2 1
2cos2 = 2 . cos2 2 = 2 2
8/14/2019 integral calculus presentation
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8/14/2019 integral calculus presentation
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Wallis Formula
= 1 3 ..(2 1)|(2 2
Where:a= when both m and n are + even a= 1 if otherwise
8/14/2019 integral calculus presentation
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Evaluate the integral of
5 using lowand upper limit= a.0.5046 c. 0.6107
b.0.3068 d. 0.4105
5 =55(3)(1)8(6)(4)(2.2 =0.3068
8/14/2019 integral calculus presentation
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Evaluate the integral of
using lower lim0 and upper limit =
=4(2)5 3 1.1 =
815
Evaluate the integral of usinglower limit = 0 and upper li2
=3(1)(2)753(1)1=
235
8/14/2019 integral calculus presentation
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PLANE AREASCONSIDERING VERTICAL STRIP
(0,2)
(-2,0) (2,0)y
dx
=
=
8/14/2019 integral calculus presentation
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PLANE AREASCONSIDERING HORIZONTAL STRIP
X X
(0,2)
(-2,0) (2,0)
=
= dy
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AREA BETWEEN TWO CURVESUSING HORIZONTAL STRIP
=
dy
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AREA BETWEEN TWO CURVESUSING VERTICAL STRIP
=
dx
8/14/2019 integral calculus presentation
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8/14/2019 integral calculus presentation
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Problem: Find the area bounded by
= = 2 =
= 2 (2,4)
(-1,1)
Solve for the point of intersection= = 2 = = 2 2= 2 1 X=2
X=-1If x=2; y=4If x=-1; y=1
(2,4)(-1,1)
=
=2 = [( 2)
2 3]2
1
= 883 12 13
4 5
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APPLICATION OF INTEGRATION(SOLIDS OF REVOLUTION)
CIRCULAR DISK METHOD
y
dx
y
dx
= Revolved
about oxWhere:y= radiusdx= thickness
x
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CIRCULAR RING
= = x
Where:
= =
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CYLINDRICAL SHELL
y
x
xdy
x
x2 y
dy
= 2
Revolve about oxWhere:Y= radiusx- length
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Problem: Given is the area in the first quadrant bounded by
= 8, y-2=0 and the y-axis. Find the volume generatedby revolving the area about the line y-2=0.
(4,2)dx
y
22-y
= 8
y-2=0dx
= ( )
Substitute : = =
= ( ) = [ )
=
= . . = *
= . .
8/14/2019 integral calculus presentation
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Problem: Find the volume generated by revolving about they-axis, the area in the first quadrant, bounded by = 8 and the x-axis and the line x-4=0.
4-xX
X-4=0
4
8
= Where:
= = = =
= = []
=
=
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Problem. Find the volume generated by revolving about 0y,the area in the first quadrant bounded by the curve
=4
Y
4X2
X
= 2 Where =4 . = 2 4
=2 (4)
=4 = 2 = 2 (4 )( 2 ) = ( ) = ( ) [( )] =8 .
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INTEGRAL CALCULUS1. The integral of is
a. 1/9 c. b. 1/3 d.3 _
2. Evaluate the integral of a. c.
b. d. 3. What is the integral of tan xdx?a. ln sec x+c c. ln cos x+cb.ln csc x+c d. ln sin x+c
4. Evaluate the integral of
.
a. c.
b. d. 5. The integral of cos 3d is
a. 1/3 sin3 +c c. sin3 +c
b. - 1/3 sin3 +c d. sin3 + c
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6. Find the area bounded by the curve = and the line y=5a. 21.34 sq. units c.27.20 sq. unitsb. 28.10 sq. units d. 25.63 sq. units
7. Find the area bounded by the curve = and the line y=x-4a. 20.83 sq. units c. 24.30 sq. unitsb.21.17 sq. units d.23.25 sq. units
8. Find the volume generated by revolving the area bounded by =,the line x-9=0 and y=0 about the line x-9=0a. 259 cu. Units c. 270.30 cu. Unitsb. 245.10 cu. Units d. 250.35 cu. Units
9. Find the volume generated by revolving about the y-axis the areabounded by = ,the line x=6 and the x-axis
a. 216 cu. Units c. 162 cu. Units
b. 512 cu. Units d. 324 cu. Units10. Evaluate:
a. 2/35 c. 2 /25b. /35 d. 4/35