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Rectangle Approximations for the Definite Integrals Worksheet

Rectangle Approximations for the Definite Integrals Worksheet
  • Page 1
 1.  
Find the approximate value of 13x3 dx by using 4 rectangles of equal width.
a.
25.8125
b.
55
c.
24.125
d.
27.5
e.
20.00


Solution:

As the interval is divided into 4 rectangles of equal width, width of each rectangle = 3-14 = 0.5


The height of each rectangle is determined by the value of f(x) of each rectangle from the graph as shown.

The table represents the values of f(x) for the function f(x) = x3.

x11.522.53
f(x)13.375815.62527


So, the approximate value of 13x3 dx
= (0.5)f(1) + (0.5)f(1.5) + (0.5)f(2) + (0.5)f(2.5) + (0.5)f(3)

= (0.5)(1) + (0.5)(3.375) + (0.5)(8) + (0.5)(15.625) + (0.5)(27)

= 27.5


Correct answer : (4)
 2.  
Find the approximate value of 13 ex dx by using 4 rectangles of equal width.
a.
17.366
b.
23.398
c.
44.077
d.
46.796
e.
22.036


Solution:

As the interval is divided into 4 rectangles of equal width, width of each rectangle = 3-14 = 0.5


The corresponding height of each rectangle is determined by the value of f(x) for each rectangle from the graph as shown.

The table represents the values of f(x) for the function f(x) = ex.

x11.522.53
f(x)2.71834.4817.38912.12320.085


So, the approximate value of 13ex dx
= (0.5)f(1) + (0.5)f(1.5) + (0.5)f(2) + (0.5)f(2.5) + (0.5)f(3)

= (0.5)(2.7183) + (0.5)(4.481) + (0.5)(7.389) + (0.5)(12.123) + (0.5)(20.085)

= 23.398


Correct answer : (2)
 3.  
Find the approximate value of 14ln (x + 2) dx using 6 rectangles of equal width.
a.
9.824
b.
5.172
c.
10.344
d.
8.215
e.
6.963


Solution:

As the interval is divided into 6 rectangles of equal width, width of each rectangle = 4-16 = 0.5


The corresponding height of each rectangle is determined by the value of f(x) for each rectangle from the graph as shown.

The table represents the values of f(x) for the function f(x) = ln (x + 2).

x11.522.533.54
f(x)1.0981.2521.3861.5041.6091.7041.791


So, the approximate value of 14ln (x + 2) dx
= (0.5)f(1) + (0.5)f(1.5) + (0.5)f(2) + (0.5)f(2.5) + (0.5)f(3) + (0.5)f(3.5) + (0.5)f(4)

= (0.5)(1.098) + (0.5)(1.252) + (0.5)(1.386) + (0.5)(1.504) + (0.5)(1.609) + (0.5)(1.704) + (0.5)(1.791)

= 5.172


Correct answer : (2)
 4.  
Find the approximate value of 13 1x using 4 rectangles of equal width.
a.
1.098
b.
1.25
c.
1.0835
d.
2.90
e.
1.45


Solution:

As the interval is divided into 4 rectangles of equal width, width of each rectangle = 3-14 = 0.5


The corresponding height of each rectangle is determined by the value of f(x) for each rectangle from the graph as shown.

The table represents the value of f(x) for the function = 1x
x11.522.53
f(x)10.6670.50.40.333


So, the approximate value of 131x dx = (0.5)f(1)+0.5f(1.5) + (0.5)f(2) + (0.5)f(2.5) + (0.5)f(3)

= (0.5)(1) + (0.5)(0.667) + (0.5)(0.5) + (0.5)(0.4) + (0.5)(0.333)

= 1.45


Correct answer : (5)
 5.  
Find the approximate value of 14(x2 - 3x + 6) dx using 6 rectangles of equal width.
a.
20.525
b.
20.125
c.
37.25
d.
18.625
e.
16.50


Solution:

As the interval is divided into 6 rectangles of equal width, width of each rectangle = 4-16 = 0.5


The corresponding height of each rectangle is determined by the value of f(x) for each rectangle from the graph as shown.

The table represents the values of f(x) for the function f(x) = x2 - 3x + 6
x11.522.533.54
f(x)43.7544.7567.7510


The approximate value of 14(x2 - 3x + 6) dx = (0.5)f(1) + (0.5)f(1.5) + (0.5)f(2) + (0.5)f(2.5) + (0.5)f(3) + (0.5)f(3.5) + (0.5)f(4)

= (0.5)(4) + (0.5)(3.75) + (0.5)(4)+ (0.5)(4.75) + (0.5)(6) + (0.5)(7.75) + (0.5)(10)

= 20.125


Correct answer : (2)
 6.  
Find the approximate value of -22 ln(x2 + 2) dx by using 4 rectangles of equal width.
a.
5.373
b.
3.582
c.
6.471
d.
5.778
e.
4.680


Solution:

As the interval is divided into 4 rectangles of equal width, width of each rectangle = 2+24 = 1


The corresponding height of each rectangle is determined by the value of f(x) for each rectangle from the graph as shown.

The table represents the values of f(x) for the function f(x) = ln(x2 + 2)
x- 2- 1012
f(x)1.7911.0980.6931.0981.791


So, the approximate value of -22 ln(x2 + 2) dx = (1)f(- 2) + (1)f(- 1) + (1)f(0) + (1)f(1) + (1)f(2)

= (1)(1.791) + (1)(1.098) + (1)(0.693) + (1)(1.098) + (1)(1.791)

= 6.471


Correct answer : (3)
 7.  
Find the approximate value of -3-1 e(2x + 4) dx by using 4 rectangles of equal width.
a.
5.894
b.
8.077
c.
3.141
d.
6.394
e.
11.7893


Solution:

As the interval is divided into 4 rectangles of equal width, width of each rectangle = -1+34 = 0.5


The corresponding height of each rectangle is determined by the value of f(x) for each rectangle from the graph as shown.

The table represents the values of f(x) for the function f(x) = e2x + 4
x- 3- 2.5- 2- 1.5- 1
f(x)0.1350.36712.71837.389


So, the approximate value of -3-1 e(2x + 4) dx = (0.5)f(- 3) + (0.5)f(- 2.5) + (0.5)f(- 2) + (0.5)f(- 1.5) + (0.5)f(- 1)

= (0.5)(0.135) + (0.5)(0.367) + (0.5)(1) + (0.5)(2.7183) + (0.5)(7.389)

= 5.894


Correct answer : (1)
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