﻿ Rectangle Approximations for the Definite Integrals Worksheet | Problems & Solutions

Rectangle Approximations for the Definite Integrals Worksheet

Rectangle Approximations for the Definite Integrals Worksheet
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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

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

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

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

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

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

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