Explanation Theorem used: If f and g are continuous and f ( x ) ≥ g ( x ) over the interval [ a , b ] , then the area bounded by y = f ( x ) and y = g ( x ) for a ≤ x ≤ b is given exactly by A = ∫ a b [ f ( x ) − g ( x ) ] d x . Calculation: The given functions are y = x 3 − 3 x 2 + x + 2 and y = − x 2 + 4 x + 2 . Use the online graphing calculator and sketch the graph of the given functions y = x 3 − 3 x 2 + x + 2 and y = − x 2 + 4 x + 2 is as shown in below Figure 1. From the Figure 1, it is observed that the functions are y = x 3 − 3 x 2 + x + 2 and y = − x 2 + 4 x + 2 intersects at the points ( − 1 , 3 ) , ( 0 , 2 ) , and ( 3 , 5 ) . It is noted that the function y = x 3 − 3 x 2 + x + 2 is greater than the function y = − x 2 + 4 x + 2 over the interval [ − 1 , 0 ] and the function y = − x 2 + 4 x + 2 is greater than the function y = x 3 − 3 x 2 + x + 2 over the interval [ 0 , 3 ] . By the above mentioned theorem, the area of the shaded region is computed as follows. A = ∫ − 1 0 ( x 3 − 3 x 2 + x + 2 − ( − x 2 + 4 x + 2 ) ) d x + ∫ 0 3 ( ( − x 2 + 4 x + 2 ) − ( x 3 − 3 x 2 + x + 2 ) ) d x = ∫ − 1 0 (

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

See similar textbooks

Question

Chapter 6.1, Problem 68E

To determine

To find: The area bounded by the graphs of y=x3−3x2+x+2 and y=−x2+4x+2.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6.1, Problem 67E

Chapter 6.1, Problem 69E

Students have asked these similar questions

nt/Ray Skew Lines/ J K # H L 艹 G C D E F Diagrams m Three Points th a Protractor Answer Attempt 3 out of 3 el 1 is congruent to Submit Answer 103 Log Out REE Young the → C # $

4:54 PM Thu Jan 16 cdn.assess.prod.mheducation.com Question 3 The angle bisectors of APQR are PZ, QZ, and RZ. They meet at a single point Z. (In other words, Z is the incenter of APQR.) Suppose YZ = 22, QZ = 23, mz WPY 38°, and mzXQZ = 54°. Find the following measures. Note that the figure is not drawn to scale. P W Z X R Y mzXQW WZ = = 0 mz XRZ = 0°

a C d 2 1 -1 0 1 2 3 -1 Graph of f'(x) (5) The graph of f'(x), the derivative of f(x), is shown in the figure above. The line tangent to the graph of f'(x) at x=0 is vertical and f'(x) is not differentiable at x = 1. Which of the following statements is true? (a) f'(x) does not exist at x = 0. (b) f(x) has a point of inflection at x = 1. (c) f(x) has a local maximum at x = 0. (d) f(x) has a local maximum at x = 1.

Chapter 6 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

Ch. 6.1 - Matched Problem 1 Find the area bounded by f(x) =...Ch. 6.1 - Matched Problem 2 Find the area between the graph...Ch. 6.1 - Prob. 3MP Ch. 6.1 - Matched Problem 4Find the area bounded by f(x) =6 ...Ch. 6.1 - Matched Problem 5Find the area bounded by f(x)=2x2...Ch. 6.1 - Matched Problem 6Find the area (to three decimal...Ch. 6.1 - Prob. 7MP Ch. 6.1 - Prob. 8MP Ch. 6.1 - Prob. 1ED Ch. 6.1 - In Problems 18, use geometric formulas to find the...

Knowledge Booster

The area bounded by the graphs of y = x 3 − 3 x 2 + x + 2 and y = − x 2 + 4 x + 2 .

Solution Summary: The area of the shaded region is 11.833 Sq. units.

To find: The area bounded by the graphs of y=x3−3x2+x+2 and y=−x2+4x+2.

Want to see the full answer?

Chapter 6 Solutions