   Chapter 13.1, Problem 32E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Use rectangles to find the area between y   =   4 x − x 2   and the x-axis from x = 0 to x = 4. Divide the interval [0,4] into n equal subintervals so that each subinterval has length 4 /n.

To determine

To calculate: The area between the curve y=4xx2 and the x-axis from x=0 to x=4 when the interval [0,4] is divided into n equal subintervals so that the length of each subinterval is 4n.

Explanation

Given Information:

The provided curve is y=4xx2 from x=0 to x=4 and the interval [0,4] is divided into n equal subintervals so that the length of each subinterval is 4n.

Formula used:

The base of the rectangles to approximate the area is ban where the interval [a,b] on which function is defined is divided into n subintervals.

The height of the rectangles is the value of the function calculated at the right-hand end point of the interval containing the base.

The area of a rectangle is base×height.

The approximated area under the curve is sum of the areas of each rectangle.

The value of the sum k=1nk2=n(n+1)(2n+1)6.

The value of the sum k=1nk=n(n+1)2.

For functions f(x) and g(x),

limn[f(x)+g(x)]=limnf(x)+limng(x).

The value of 1n when n,

limn1n=0

Calculation:

Consider the provided curve y=4xx2 from x=0 to x=4 and divide the interval [0,4] into n equal subintervals so that the length of each subinterval is 4n.

The base of the rectangles to approximate the area is ban where the interval [a,b] on which function is defined is divided into n subintervals.

Since, the function is defined from x=0 to x=4. So, a=0 and b=4.

Thus,

Base of each rectangle=40n=4n

Thus, the n subintervals, each of length 4n, are [0,4n], [4n,8n],,[4n4n,4].

Since, there are n subintervals, the number of rectangles is n.

Recall that the height of the rectangles is the value of the function calculated at the right-hand end point of the interval containing the base.

Since, the right-hand end point of the first interval [0,4n] is 4n.

Thus, the height of the first rectangle is,

y=4xx2=4(4n)(4n)2

Recall that the area of a rectangle is base×height.

Thus, the area of the first rectangle is,

Area=base×height=4n×(4(4n)(4n)2)=4n(4(4n)(4n)2)

Do a similar calculation to find the area of all n rectangles and record these values in a table.

 Rectangle Base Right endpoint Height Area=base×height 1 4n x1=4n y=4(4n)−(4n)2 4n(4(4n)−(4n)2) 2 4n x2=8n y=4(8n)−(8n)2 4n(4(8n)−(8n)2) 3 4n x3=12n y=4(12n)−(12n)2 4n(4(12n)−(12n)2) ⋮ ⋮ ⋮ ⋮ ⋮ n 4n xn=4nn=4 y=4(4nn)−(4nn)2 4n(4(4nn)−(4nn)2)

Recall that the approximated area under the curve is sum of the areas of each rectangle.

Thus, the area under the curve is 4n(4(4n)(4n)2)+4n(4(8n)(8n)2)++4n(4(4nn)(4nn)2).

Simplify 4n(4(4n)(4n)2)+4n(4(8n)(8n)2)++4n(4(4nn)(4nn)2)

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