Part II: The Answers
352
Answers
501–600
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
4
4 1 3
3
1
4
1
1
−
=
=
− +
( )
∫
∑
∑
→∞ =
→∞ =
x dx
f x x
i
n n
n
i
i
n
n
i
n
∆
= =
−
=
−
+
( )
→∞
=
=
→∞
∑
∑
lim
lim
n
i
n
i
n
n
n
n
i
n
n n
n n
3
3 3
3 3
3
1
2
1
1
=
−
+
( )
= −
=
→∞
lim
n
n n
n
9 9
2
1
9 9
2
9
2
2
Note that the formulas k kn
i
n
=
=
∑
1
and i
n n
i
n
=
+
( )
=
∑
1
2
1
were used to simplify the
summations.
544.
3
2
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 0 3
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
n
i = +
= +
=
( )
∆
0 3
3
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
2
4
2 3
3
4
2
0
3
1
2
x
x
dx
f x x
i
n
i
n
n
i
i
n
n
− −
=
=
( ) − ( ) −
∫
∑
→∞ =
→∞
∆
( )
=
−
−
=
=
→∞
=
=
=
∑
∑
∑ ∑
3
3 18
3
4
1
2
2
1
1
1
n
n n
i
n
i
i
n
n
i
n
i
n
i
n
lim
lim m
n
n n
n n
n
n
n n
n
→∞
+
( ) +
(
)
−
+
( )
−
=
3 18
1 2 1
6
3
1
2
4
2
l lim
( )
n
n n
n
n
n n
n
→∞
+
( ) +
(
) −
+
( ) −
=
− −
=
9
1 2 1 9
2
1 12
9 2 9
2
12
3
2
2
3
352
Answers
501–600
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
4
4 1 3
3
1
4
1
1
−
=
=
− +
( )
∫
∑
∑
→∞ =
→∞ =
x dx
f x x
i
n n
n
i
i
n
n
i
n
∆
= =
−
=
−
+
( )
→∞
=
=
→∞
∑
∑
lim
lim
n
i
n
i
n
n
n
n
i
n
n n
n n
3
3 3
3 3
3
1
2
1
1
=
−
+
( )
= −
=
→∞
lim
n
n n
n
9 9
2
1
9 9
2
9
2
2
Note that the formulas k kn
i
n
=
=
∑
1
and i
n n
i
n
=
+
( )
=
∑
1
2
1
were used to simplify the
summations.
544.
3
2
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 0 3
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
n
i = +
= +
=
( )
∆
0 3
3
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
2
4
2 3
3
4
2
0
3
1
2
x
x
dx
f x x
i
n
i
n
n
i
i
n
n
− −
=
=
( ) − ( ) −
∫
∑
→∞ =
→∞
∆
( )
=
−
−
=
=
→∞
=
=
=
∑
∑
∑ ∑
3
3 18
3
4
1
2
2
1
1
1
n
n n
i
n
i
i
n
n
i
n
i
n
i
n
lim
lim m
n
n n
n n
n
n
n n
n
→∞
+
( ) +
(
)
−
+
( )
−
=
3 18
1 2 1
6
3
1
2
4
2
l lim
( )
n
n n
n
n
n n
n
→∞
+
( ) +
(
) −
+
( ) −
=
− −
=
9
1 2 1 9
2
1 12
9 2 9
2
12
3
2
2
3