Part II: The Answers
350
Answers
501–600
Notice that if you consider 5i
n
, this is of the form a + (Δ x)i, where a = 0.
You now have a value for a and know the length of the interval, so you can conclude
that you’re integrating over the interval [0, 5]. To produce the function, replace each
expression of the form 5i
n
that appears in the summation with the variable x. Notice
that you can rewrite the summation as lim
lim
n
i
n
n
i
n
n
i
n
i
n
n
i
n
i
n
→∞ =
→∞ =
+
=
+ ( )
∑
∑
5 5 125
5 5
5
3
3
1
3
1
; that
means you can replace 5
5
3
i
n
i
n
+ ( ) with x x
+
3 so that f x
x x
( ) =
+
3 . Therefore, the
Riemann sum could represent the definite integral
x x dx
+
∫
3
0
5
.
541.
6
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
= − =
2 0 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
n
i = +
= + =
( )
∆
0 2
2
Substituting those values into the definition of the integral gives you the following:
(
)
lim
( )
lim
1 2
1 2 2
2
0
2
1
1
+
=
=
+ ( )

 

  ( )
∫
∑
→∞ =
→∞ =
x dx
f x x
i
n
n
n
i
i
n
n
i
∆
n n
n
i
n
n
i
n
i
n
n
i
n
n
n
i
∑
∑
∑
∑
=
+

 

 
=
+






→∞ =
→∞
=
=
lim
lim
2 1 4
2
1 4
1
1
1
= =
+
+
( )






=
+
+

 

 
= +
=
→∞
→∞
lim
lim
( )
n
n
n
n n
n n
n n
n
2
4
1
2
2 4
2 4 1
6
2
2
Note that formulas k kn
i
n
=
=
∑
1
and i
n n
i
n
=
+
(
)
=
∑
1
2
1
were used to simplify the summations.