﻿ Rectangle Approximations for the Definite Integrals Worksheet | Problems & Solutions
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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 ${\int }_{1}^{3}$$x$3 $\mathrm{dx}$ by using 4 rectangles of equal width.
 a. 25.8125 b. 55 c. 24.125 d. 27.5 e. 20

#### 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.

 x 1 1.5 2 2.5 3 f(x) 1 3.375 8 15.625 27

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 ${\int }_{1}^{3}$ $e$$x$ $\mathrm{dx}$ 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.

 x 1 1.5 2 2.5 3 f(x) 2.7183 4.481 7.389 12.123 20.085

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 ${\int }_{1}^{4}$ln ($x$ + 2) $\mathrm{dx}$ 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).

 x 1 1.5 2 2.5 3 3.5 4 f(x) 1.098 1.252 1.386 1.504 1.609 1.704 1.791

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 ${\int }_{1}^{3}$ $\frac{1}{x}$ using 4 rectangles of equal width.
 a. 1.098 b. 1.25 c. 1.0835 d. 2.9 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
 x 1 1.5 2 2.5 3 f(x) 1 0.667 0.5 0.4 0.333

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 ${\int }_{1}^{4}$($x$2 - 3$x$ + 6) $\mathrm{dx}$ using 6 rectangles of equal width.
 a. 20.525 b. 20.125 c. 37.25 d. 18.625 e. 16.5

#### 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
 x 1 1.5 2 2.5 3 3.5 4 f(x) 4 3.75 4 4.75 6 7.75 10

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 ${\int }_{-2}^{2}$ ln($x$2 + 2) $\mathrm{dx}$ by using 4 rectangles of equal width.
 a. 5.373 b. 3.582 c. 6.471 d. 5.778 e. 4.68

#### 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 - 1 0 1 2 f(x) 1.791 1.098 0.693 1.098 1.791

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 ${\int }_{-3}^{-1}$ $e$(2$x$ + 4) $\mathrm{dx}$ 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) 0.135 0.367 1 2.7183 7.389

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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