Chapter 14.2, Problem 3E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Textbook Problem

# Approximation In Exercises 3-6, approximate the integral ∫ R ∫ f ( x , y )   d A by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2), and (0, 2) into eight equal squares and finding the sum ∑ i = 1 8 f ( x i , y i ) Δ A i , where ( x i , y i ) the center of the ith square. Evaluate the iterated integral and compare it with the approximation. ∫ 0 4 ∫ 0 2 ( x + y ) d y   d x

To determine

To calculate: The approximate value of the given integral, that is 0402(x+y)dydx, by dividing the rectangle R, which has the points (0,0),(4,0)(4,2) and (0,2) for vertices, into eight equal squares and by calculating the sum i=18(xi+yi)ΔAi, where (xi,yi) represents the centre of the ith square. Also, compare this value with that of the integrated integral.

Explanation

Given:

The integral, âˆ«04âˆ«02(x+y)dydx and the rectangle R, which has the points (0,0),(4,0)(4,2)Â andÂ (0,2) for its vertices, divided into eight equal squares with the center of each square at the point (xi,yi).

Formula used:

If f is defined on a closed, bounded region R in the xy-plane, then the approximate value of the integral of f over R is

âˆ«Râˆ«f(x,y)Â dA=limâ€–Î”â€–â†’0âˆ‘i=1nf(xi,yi)Î”Ai â€¦â€¦ (1)

Provided the limit exists.

Calculations:

The rectangle R, which has the points (0,0),(4,0)(4,2)Â andÂ (0,2), divided into eight equal squares and labelled from 1 to 8 is shown on the Cartesian plane below:

The centers of the squares that are labelled from 1 to 8 are:

(12,12),(32,12),(52,12),(72,12),(12,32),(32,32),(52,32)Â andÂ (72,32)

From the graph above, the area of the square marked 1 is Î”A1=1 and it is stated in the question that the squares are equal so Î”Ai=1.

Now, substitute the value of integral and the values of the variables fÂ andÂ n=8 in equation (1) as

âˆ«04âˆ«02(x+y)dydx=âˆ‘i=18(xi+yi)Î”Ai

Now, substitute xi,yiÂ andÂ Î”Ai with the corresponding values of i such that,

âˆ«04âˆ«02(x+y)dydx=(x1+y1)Î”A1+(x2+y2)Î”A2+(x3+y3)Î”A3+(x4+y4)Î”A4+(x5+y5)Î”A5+(x6+y6)Î”A6+(x7+y7)Î”A7+(x8+y8)Î”A8</

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