integral calculus presentation

8/14/2019 integral calculus presentation

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INTEGRAL CALCULUSA REVIEW IN


2/23

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


3/23

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


4/23

Evaluate:

+

a.4 ln (3x+2) c. ln (3 b. ln (3 2) d. 2 ln (3x + 2)

43 2

= 4. + = 3 2


5/23

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


6/23


7/23

Wallis Formula

= 1 3 ..(2 1)|(2 2

Where:a= when both m and n are + even a= 1 if otherwise


8/23

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


9/23

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


10/23

PLANE AREASCONSIDERING VERTICAL STRIP

(0,2)

(-2,0) (2,0)y

dx

=

=


11/23

PLANE AREASCONSIDERING HORIZONTAL STRIP

X X

(0,2)

(-2,0) (2,0)

=

= dy


12/23

AREA BETWEEN TWO CURVESUSING HORIZONTAL STRIP

=

dy


13/23

AREA BETWEEN TWO CURVESUSING VERTICAL STRIP

=

dx


14/23


15/23

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


16/23

APPLICATION OF INTEGRATION(SOLIDS OF REVOLUTION)

CIRCULAR DISK METHOD

y

dx

y

dx

= Revolved

about oxWhere:y= radiusdx= thickness

x


17/23

CIRCULAR RING

= = x

Where:

= =


18/23

CYLINDRICAL SHELL

y

x

xdy

x

x2 y

dy

= 2

Revolve about oxWhere:Y= radiusx- length


19/23

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 : = =

= ( ) = [ )

=

= . . = *

= . .


20/23

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:

= = = =

= = []

=

=


21/23

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

integral calculus presentation

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