   Chapter 5.5, Problem 76E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use a graph to give a rough estimate of the area of the region that lies under the given curve. Then find the exact area.y = 2 sin x − sin 2x, 0 ≤ x ≤ π

To determine

To find: area of the region that lies under the given curve using a graph.

To find: the exact area for the function y=2sinxsin2x,0xπ.

Explanation

Given:

The function is y=2sinxsin2x,0xπ.

The region lies between 0 to π.

Calculation:

Show the equation as below:

y=2sinxsin2x (1)

Plot a graph for the equation y=2sinxsin2x,0xπ using the calculation as follows:

Calculate y value using Equation (1).

Substitute 0 for x in Equation (1).

y=2sin0sin(2×0)=0sin0=0

Hence, the co-ordinate of (x,y) is (0,0).

Calculate y value using Equation (1).

Substitute π for x in Equation (1).

y=2sinπsin(2π)=(2×0)0=0

Hence, the co-ordinate of (x,y) is (π,0)

Calculate y value using Equation (1).

Substitute 2π3 for x in Equation (1).

y=2sin(2π3)sin(2×2π3)=2sin(2π3)sin(4π3)=2.598

Hence, the co-ordinate of (x,y) is (2π3,2.598)

Draw the graph for the equation f(x)=2sinxsin2x using calculated (x,y) co-ordinates as in Figure 1.

Refer to Figure (1)

The shaded region is considered as a triangle for rough estimation of area

Calculate the area of shaded region.

A=12bh

Substitute π for b, 2.598 for h.

A=12×π×2.598=4

Hence, The area of the region lies under the given curve is 4_.

The integral function is,

A=0πydx

Substitute (2sinxsin2x) for y.

A=0π(2sinxsin2x)dx=0π2sinxdx0πsin2xdx=20πsinxdx0πsin2xdx (2)

The region lies between x=0 and x=π

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