353
Answers
501–600
Answers and Explanations
Note that the formulas k kn
i
n
=
=
∑
1
, i
n n
i
n
=
+
(
)
=
∑
1
2
1
, and i
n n
n
i
n 2
1
1 2 1
6
=
+
(
) +
(
)
=
∑
were used
to simplify the summations.
545.
8
3
To begin, you split the interval into n pieces of equal width using the formula
∆x b a
n
= − , where a is the lower limit of integration and b is the upper limit of integration. In this case, you have
∆x
n
n
= − =
3 1 2
You also want to select a point from each interval. The formula x i = a + (Δ x)i gives you
the right endpoint from each interval. Here, you have
x a
x i
i
n
i = +
= +
( )
∆
1 2
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
x
x
dx
f x
x
i
n
i
n
n
i
i
n
n
2
1
3
1
2
5
1 2
1 2
5
+ −
=
=
+
( ) + +
( ) −
∫
∑
→∞ =
→∞
∆
=
+ +
+ + −
=
=
→∞ =
→
∑
∑
2
2 1 4 4
1 2 5
1
2
2
1
n
n
i
n
i
n
i
n
i
n
n
i
n
n
lim
lim
∞ ∞ =
→∞
=
=
=
+ −
=
+
−
∑
∑
∑
2 4
6 3
2 4
6
3
2
2
1
2
2
1
1
n
i
n
i
n
n n
i
n
i
i
n
n
i
n
i
n
i
lim
1 1
2
2 4
1 2 1
6
6
1
2
n
n
n n
n n
n
n
n n
∑
=
+
( ) +
(
)
+
+
( )
−
→∞
lim
3 3
4
3
1 2 1 6
1 6
4
3
2
2
3
n
n n
n
n
n n
n
n
=
+
( ) +
(
) +
+
( ) −
=
→∞
lim
( ) + + −
=
6 6
8
3
Note that the formulas k kn
i
n
=
=
∑
1
, i
n n
i
n
=
+
(
)
=
∑
1
2
1
, and i
n n
n
i
n
2
1
1 2 1
6
=
+
( ) +
(
)
=
∑
were used to
simplify the summations.
Answers
501–600
Answers and Explanations
Note that the formulas k kn
i
n
=
=
∑
1
, i
n n
i
n
=
+
(
)
=
∑
1
2
1
, and i
n n
n
i
n 2
1
1 2 1
6
=
+
(
) +
(
)
=
∑
were used
to simplify the summations.
545.
8
3
To begin, you split the interval into n pieces of equal width using the formula
∆x b a
n
= − , where a is the lower limit of integration and b is the upper limit of integration. In this case, you have
∆x
n
n
= − =
3 1 2
You also want to select a point from each interval. The formula x i = a + (Δ x)i gives you
the right endpoint from each interval. Here, you have
x a
x i
i
n
i = +
= +
( )
∆
1 2
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
x
x
dx
f x
x
i
n
i
n
n
i
i
n
n
2
1
3
1
2
5
1 2
1 2
5
+ −
=
=
+
( ) + +
( ) −
∫
∑
→∞ =
→∞
∆
=
+ +
+ + −
=
=
→∞ =
→
∑
∑
2
2 1 4 4
1 2 5
1
2
2
1
n
n
i
n
i
n
i
n
i
n
n
i
n
n
lim
lim
∞ ∞ =
→∞
=
=
=
+ −
=
+
−
∑
∑
∑
2 4
6 3
2 4
6
3
2
2
1
2
2
1
1
n
i
n
i
n
n n
i
n
i
i
n
n
i
n
i
n
i
lim
1 1
2
2 4
1 2 1
6
6
1
2
n
n
n n
n n
n
n
n n
∑
=
+
( ) +
(
)
+
+
( )
−
→∞
lim
3 3
4
3
1 2 1 6
1 6
4
3
2
2
3
n
n n
n
n
n n
n
n
=
+
( ) +
(
) +
+
( ) −
=
→∞
lim
( ) + + −
=
6 6
8
3
Note that the formulas k kn
i
n
=
=
∑
1
, i
n n
i
n
=
+
(
)
=
∑
1
2
1
, and i
n n
n
i
n
2
1
1 2 1
6
=
+
( ) +
(
)
=
∑
were used to
simplify the summations.