Part II: The Answers
338
Answers
501–600
x
x
2
2
3
1
1 1 1
1
2 1 1
2 1
1 25158
1 25158
1 25158 1
= −
+ −
+
−
≈
=
−
+ −
( )
( )
( )
( )
.
.
( .
)
( ( .
)
( .
)
( .
)
.
.
( .
1 25158
1
2 1 25158 1
2 1 25158
1 22120
1 22120
1 2
2
4
+
−
≈
=
−
x
2 2120 1 1 22120
1
2 1 22120 1
2 1 22120
1 22074
1 22
2
5
)
( .
)
( .
)
( .
)
.
.
+ −
+
−
≈
=
x
0 074
1 22074 1 1 22074
1
2 1 22074 1
2 1 22074
1 2207
2
−
+ −
+
−
≈
( .
)
( .
)
( .
)
( .
)
.
4 4
The approximation is no longer changing, so the solution is 1.22074.
519.
23
4
Because the given function has values that are strictly greater than or equal to zero on
the given interval, you can interpret the Riemann sum as approximating the area that’s
underneath the curve and bounded below by the x-axis.
You want to use four rectangles of equal width to estimate the area under f
 
(x) = 2 + x
2
over the interval 0 ≤ x ≤ 2. Begin by dividing the length of the interval by 4 to find the
width of each rectangle:
∆x = − =
2 0
4
1
2
Then divide the interval [0, 2] into 4 equal pieces, each with a width of 1
2
, to get the
intervals 0 1
2
,

 

 
, 1
2
1
,

 

 
, 1 3
2
,

 

 
, and 3
2
2
,

 

 
. Using the left endpoint of each interval to