2012b mat171t areanotes applicationintegrals
TRANSCRIPT
-
7/29/2019 2012b Mat171t Areanotes Applicationintegrals
1/6
Application of Integration (Areas)
Andrew CPG [email protected]
2012-10-23Copyright c 2012 Tshwane University of Technology.
1 Areas Under a CurveThe area under the curve is determined by the denite integral
Z
() = [()]=
== () () (1)
where the indenite integral for the function = () is given by
Z() = () + (2)
1.1 Area Between The Function and The -axis
1.1.1 Above The -axis (Figure 1)
a b
x
y
y = f(x)
A
Figure 1: Area Between The Function and The -axis (Above The -axis).
1
-
7/29/2019 2012b Mat171t Areanotes Applicationintegrals
2/6
The area, , of the function = () on the interval 2 [ ] is determined
by
=
Z
() = [()]=
=
= () () , where
Z() = () + (3)
EXAMPLEDetermine the area between the function = 2 + 1 and the -axison the interval 2 [21].
The graph of the function = 2 + 1. Showing the area on the interval 2 [2 1].
=
Z12
2 + 1
=
1
33 +
=1=2
=
1
3(1)3 + (1)
1
3(2)3 + (2)
= 6 units2.
2
-
7/29/2019 2012b Mat171t Areanotes Applicationintegrals
3/6
1.1.2 Below The -axis (Figure 2)
a b
y
y = f(x)
A
Figure 2: Area Between The Function and The -axis (Below The -axis)
The area, , of the function = () on the interval 2 [ ] is determinedby
=
Z
()
= [()]=
== () () , where
Z() = () + (4)
NOTE: The area between the function and the -axis can also be determinedby
=
Z
() =
Z
() (5)
notice that the upper and lower limits of the denite integral change.
3
-
7/29/2019 2012b Mat171t Areanotes Applicationintegrals
4/6
EXAMPLE
Determine the area between the function =
2
7 and the -axis on theinterval 2 [2 1].
=
Z1
2
2 7
=
1
33 7
=1=2
=
13
(1)3 7(1)
+
13
(2)3 7 (2)
= 18 units2.
4
-
7/29/2019 2012b Mat171t Areanotes Applicationintegrals
5/6
1.1.3 Above and Below The -axis (Figure 3)
a b
y
y = f(x)
A1
A2c
x
Figure 3: Area Between The Function and The -axis (Above and Below The-axis).
The area, , of the function = () on the interval 2 [ ] is determinedby
= Z
() , where = 1 + 2
1 =
Z
() and 2 =
Z
() (6)
Thus
= 1 + 2 =
Z
()
Z
()
= [()]=
= [()]
=
=
= () 2() + () , where
Z() = () + (7)
NOTE: The area between the function and the -axis (above and below the-axis) can also be determined by
=
Z
() +
Z
()
= [()]==
+ [()]==
(8)
5
-
7/29/2019 2012b Mat171t Areanotes Applicationintegrals
6/6
EXAMPLE
Determine the area between the function =
2
+ 4 and the -axis on theinterval 2 [3 3].
Figure 4:
=
Z3
3
2 + 4
=
Z2
3
2 + 4
+
Z2
2
2 + 4
Z3
2
2 + 4
=
1
33 + 4
=2=3
+
1
33 + 4
=2=2
1
33 + 4
=3=2
=7
3+
32
3+
7
3
=46
3units2.
6