351
Answers
501–600
Answers and Explanations
542.
196
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
= − =
4 0 4
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 4
4
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
1 3
1 3 4
4
3
0
4
1
3
1
+
=
=
+ ( )
∫
∑
→∞ =
→∞ =
x dx
f x x
i
n
n
n
i
i
n
n
i
∆
n n
n
i
n
i
n
n
n n
i
n n
n n
∑
∑ ∑
=
+
=
+
( )
→∞
=
=
→∞
lim
lim
4 192
1
4 192
1
2
3
3
1
1
3
+
=
+
( ) +
=
+
=
→∞
2
2
2
4
192
1
4
192 1 4
n
n n
n
n
lim
( )
1 196
Note that the formulas k kn
i
n
=
=
∑
1
and i
n n
i
n 3
2
1
1
2
=
+
(
)
=
∑
were used to simplify the
summations.
543.
9
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
= − =
4 1 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 = +
= +
( )
∆
1 3
Answers
501–600
Answers and Explanations
542.
196
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
= − =
4 0 4
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 4
4
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
1 3
1 3 4
4
3
0
4
1
3
1
+
=
=
+ ( )
∫
∑
→∞ =
→∞ =
x dx
f x x
i
n
n
n
i
i
n
n
i
∆
n n
n
i
n
i
n
n
n n
i
n n
n n
∑
∑ ∑
=
+
=
+
( )
→∞
=
=
→∞
lim
lim
4 192
1
4 192
1
2
3
3
1
1
3
+
=
+
( ) +
=
+
=
→∞
2
2
2
4
192
1
4
192 1 4
n
n n
n
n
lim
( )
1 196
Note that the formulas k kn
i
n
=
=
∑
1
and i
n n
i
n 3
2
1
1
2
=
+
(
)
=
∑
were used to simplify the
summations.
543.
9
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
= − =
4 1 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 = +
= +
( )
∆
1 3