(d)
lim
n
→
‘
∑
‘
k
5
1
11
}
n
k
}
2
2
?
}
1
n
}
2
5
lim
n
→
‘
}
n
(
n
1
1
6
)
n
(
3
2
n
1
1)
}
5
lim
n
→
‘
5
}
2
6
}
5
}
1
3
}
(e)
Since
E
1
0
x
2
dx
equals the limit of any Riemann sum
over the interval [0, 1] as
n
approaches
‘
, part (d)
proves that
E
1
0
x
2
dx
5
}
1
3
}
.
■
Section 5.3
Definite Integrals and
Antiderivatives
(pp. 268–276)
Exploration 1
How Long is the Average Chord
of a Circle?
1.
The chord is twice as long as the leg of the right triangle in
the first quadrant, which has length
ˇ
r
2
w
2
w
x
w
2
w
by the
Pythagorean Theorem.
2.
Average value
5
}
r
2
1
(
2
r
)
}
E
r
2
r
2
ˇ
r
2
w
2
w
x
w
2
w
dx
.
3.
Average value
5
}
2
2
r
}
E
r
2
r
ˇ
r
2
w
2
w
x
w
2
w
dx
5
}
1
r
}
?
(area of semicircle of radius
r
)
5
}
1
r
}
?
}
p
2
r
2
}
5
}
p
2
r
}
4.
Although we only computed the average length of chords
perpendicular to a particular diameter, the same
computation applies to any diameter. The average length of
a chord of a circle of radius
r
is
}
p
2
r
}
.
5.
The function
y
5
2
ˇ
r
2
w
2
w
x
w
2
w
is continuous on [
2
r
,
r
], so
the Mean Value Theorem applies and there is a
c
in [
a
,
b
]
so that
y
(
c
) is the average value
}
p
2
r
}
.
Exploration 2
Finding the Derivative of an
Integral
Pictures will vary according to the value of
x
chosen.
(Indeed, this is the point of the exploration.) We show a
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 Spring '08
 GERMAN
 Derivative, lim, length r2 x2, r2 x2 dx

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