# Riemann Sum Practice Problems ∫

```Riemann Sum Practice Problems
Questions:
1. Approximate the area under the curve
f ( x) = x 2 + 2,
− 2≤ x≤ 1
with a Riemann sum, using six sub-intervals and right endpoints.
_______________________________________
2 Approximate the area under the curve
f ( x) =
x + 1,
− 1≤ x ≤ 0
with a Riemann sum, using 4 sub-intervals and left endpoints.
_______________________________________
3. a) Approximate the value of the integral:
3 3
∫1 x
− 3 dx
with a Riemann sum, using three sub-intervals and right endpoints.
b) using five sub-intervals and left endpoints.
c) Compute the exact value of the integral using the Fundamental Theorem of Calculus.
_______________________________________
4. A rectangular canal, 5m wide and 100m long has an uneven bottom. Depth measurements are taken
at every 20m along the length of the canal. Use these depth measurements to construct a Riemann
sum using right endpoints to estimate the volume of water in the canal.
Distance
Depth
0m
20m
40m
60m
80m
100m
2.0m
1.6m
1.8m
2.1m
2.1m
1.9m
_______________________________________
Page 1 of 3
1.
a = − 2, b = 1,
b − a 1− ( − 2) 1
=
= , f ( x) = x 2 + 2
n
6
2

f ( xi )∆ x  = ∆ x ( f (− 1.5) + f (− 1) + f (− 0.5) + f (0) + f (0.5) + f (1) )

∆x=

Area ;  ∑
 i
1
=
(− 1.5) 2 + 2 + (− 1)2 + 2 + (− 0.5)2 + (0) 2 + 2 + (0.5)2 + 2 + (1) 2 + 2
2
= 8.375
(
2.
)
b − a 0 − ( − 1) 1
=
= , f ( x) = x + 1
n
4
4

f ( xi )∆ x  = ∆ x ( f (− 1) + f (− 0.75) + f (− 0.5) + f ( − 0.25) )

a = − 1, b = 0, ∆ x =

Area ;  ∑
 i
= 0.25
3. a)
a = 1, b = 3,

Area ;  ∑
 i
(
0+
0.25 +
0.5 +
3− 1 2
= ,
3
3
 2
f ( xi )∆ x  =  f
 3
∆x=
As before, a = 1, b = 3,
)
; 0.5183

 5
 7
  + f   + f ( 3) 
 3
 3


3 +


(
( 3) 3 − 3
f ( x) = x3 − 3, but ∆ x =

 2

Area ;  ∑ f ( xi )∆ x  =  f ( 1) + f 

 i
 5
  73 
2
3
=  ( 1) − 3 +    − 3  +
  5

5



; 9.120
(
)
f ( x) = x3 − 3
3
   73
2   5
=     − 3 +    −
   3
3    3 
 

b)
0.75
)

 ; 23.556


3− 1 2
=
5
5
7
 9
 11 
 13  
+ f  + + f  + f  
5
 5
 5
 5 
  93

−
  5 

   11  3
3 +    −
  5 
 
   13  3
3 +    −
  5 
 

3 


Page 2 of 3
c)
4.
∫
3 3
x
1
3
− 3 dx =
1 4
 81
x − 3x =  −
4
1  4
 1
9 −  −
 4

3  = 20 − 6 = 14

Volume = (width)(cross section area)


; (5m)  ∑ f ( xi )∆ x  = (5m)∆ x ( f (20) + f (40) + f (60) + f (80) + f (100) )
 i

∆ x here is the distance between our measurements, 20m, so we get:
Volume ; 100 ( f (20) + f (40) + f (60) + f (80) + f (100) )
= 100(1.6 + 1.8 + 2.1 + 2.1 + 1.9)
= 950m3
Page 3 of 3
```