Q. 22 a function f is defined over an ... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 14: Q. 22 (page 904)

a function f is defined over an interval $[a, b]$
(a) Graph f, indicating the area A under f from a to b.
(b) Approximate the area A by partitioning $[a, b]$
into four subintervals of equal length and choosing u as the left
endpoint of each subinterval.
(c) Approximate the area A by partitioning $[a, b]$
into eight
subintervals of equal length and choosing u as the left
endpoint of each subinterval.
(d) Express the area A as an integral.
(e) Use a graphing utility to approximate the integral.
localid="1647152220951" $f (x) = \cos x [0, \frac{蟺}{2}]$

Short Answer

Expert verified

(a)

(b) The four subinterval are $[0, \frac{蟺}{8}], [\frac{蟺}{8}, \frac{蟺}{4}], [\frac{蟺}{4}, \frac{3 蟺}{8}], [\frac{3 蟺}{8}, \frac{蟺}{2}]$ and the area is $1.183$

(c) The eight subinterval are $[0, \frac{蟺}{16}], [\frac{蟺}{16}, \frac{蟺}{8}], [\frac{蟺}{8}, \frac{3 蟺}{16}], [\frac{3 蟺}{16}, \frac{蟺}{4}], [\frac{蟺}{4}, \frac{5 蟺}{16}], [\frac{5 蟺}{16}, \frac{3 蟺}{8}], [\frac{3 蟺}{8}, \frac{7 蟺}{16}], [\frac{7 蟺}{16}, \frac{蟺}{2}]$ and the area is $1.095$

(d) ${鈭�}_{0}^{\frac{蟺}{2}} \cos x d x$

e) Using graphing utility the area found is 1.

Step by step solution

Part (a) Step 1. Given

$f (x) = \cos x [0, \frac{蟺}{2}]$

Part (a) Step 2. Graph

Part (b) Step 1. Calculation

The area under the curve can be found using

$A = (\frac{b - a}{n}) (f (u_{1}) + f (u_{2}) + f (u_{3}) + f (u_{4})) w h e r e u_{1}, u_{2}, u_{3}, u_{4} a r e 4 e q u a l i n t e r v a l . n o w i n t e r v a l w i l l b e d e c i d e d b y (\frac{b - a}{n}) = \frac{\frac{蟺}{2} - 0}{4} = \frac{蟺}{8} T h e r e f o r e [0, \frac{蟺}{8}], [\frac{蟺}{8}, \frac{蟺}{4}], [\frac{蟺}{4}, \frac{3 蟺}{8}], [\frac{3 蟺}{8}, \frac{蟺}{2}] a r e t h e r e s p e c t i v e i n t e r v a l s .$

$A p p l y i n g t h e f o r m u l a f o r a r e a w e g e t A = (\frac{b - a}{n}) (f (u_{1}) + f (u_{2}) + f (u_{3}) + f (u_{4})) = (\frac{蟺}{8}) (1 + \cos (\frac{蟺}{8}) + \frac{\sqrt{2}}{2} + \cos (\frac{3 蟺}{8})) = 1.183$

Part (c) Step 1. Calculation

$A = (\frac{b - a}{n}) (f (u_{1}) + f (u_{2}) + f (u_{3}) + f (u_{4}) + f (u_{5}) + f (u_{6}) + f (u_{7}) + f (u_{8})) w h e r e u_{1}, u_{2}, u_{3}, u_{4}, u_{5}, u_{6}, u_{7}, u_{8} a r e 8 e q u a l i n t e r v a l . n o w i n t e r v a l w i l l b e d e c i d e d b y (\frac{b - a}{n}) = \frac{\frac{蟺}{2} - 0}{8} = \frac{蟺}{16} T h e r e f o r e [0, \frac{蟺}{16}], [\frac{蟺}{16}, \frac{蟺}{8}], [\frac{蟺}{8}, \frac{3 蟺}{16}], [\frac{3 蟺}{16}, \frac{蟺}{4}], [\frac{蟺}{4}, \frac{5 蟺}{16}], [\frac{5 蟺}{16}, \frac{3 蟺}{8}], [\frac{3 蟺}{8}, \frac{7 蟺}{16}], [\frac{7 蟺}{16}, \frac{蟺}{2}] a r e t h e r e s p e c t i v e i n t e r v a l s .$

$A = (\frac{b - a}{n}) (f (u_{1}) + f (u_{2}) + f (u_{3}) + f (u_{4}) + f (u_{5}) + f (u_{6}) + f (u_{7}) + f (u_{8})) = (\frac{蟺}{16}) (1 + \cos (\frac{蟺}{16}) + \cos (\frac{蟺}{8}) + \cos (\frac{3 蟺}{16}) + \frac{\sqrt{2}}{2} + \cos (\frac{5 蟺}{16}) + \cos (\frac{3 蟺}{8}) + \cos (\frac{7 蟺}{16})) = 1.095$

Part (d) Step 1. Area in integral form

$f (x) = \cos x [a, b] = [0, \frac{蟺}{2}] {鈭�}_{a}^{b} f (x) d x = {鈭�}_{0}^{\frac{蟺}{2}} \cos x d x$

Part (e) Step 1. Area using a graphing utility

The area comes out to be ${鈭�}_{0}^{\frac{蟺}{2}} \cos x d x = 1$

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Short Answer

Step by step solution

Part (a) Step 1. Given

Part (a) Step 2. Graph

Part (b) Step 1. Calculation

Part (c) Step 1. Calculation

Part (d) Step 1. Area in integral form

Part (e) Step 1. Area using a graphing utility

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