Explanation Given: The given function is: f ( x , y ) = { 1 5 x y , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 5 0 , elsewhere P ( 0 ≤ x ≤ 1 , 0 ≤ y ≤ 2 ) Formula used: Double integral formula is: ∫ − ∞ ∞ ∫ − ∞ ∞ f ( x , y ) d A = 1 Calculation: A function f ( x , y ) is a joint density function if it satisfies the following properties: f ( x , y ) ≥ 0 for all ( x , y ) . ∫ − ∞ ∞ ∫ − ∞ ∞ f ( x , y ) d A = 1 P [ ( x , y ) ∈ R ] = ∬ R f ( x , y ) d A Now, The provided function is f ( x , y ) = { 1 5 x y , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 5 0 , elsewhere Here 0 ≤ x ≤ 2 , 0 ≤ y ≤ 5 So, f ( x , y ) ≥ 0 for all ( x , y ) Since, f ( x , y ) = 0 except 0 ≤ x ≤ 2 , 0 ≤ y ≤ 5 ∫ − ∞ ∞ ∫ − ∞ ∞ f ( x , y ) d A = ∫ 0 5 ∫ 0 2 1 5 x y d x d y = 1 5

Calculus: Early Transcendental Functions

7th Edition

ISBN: 9781337552516

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Question

Chapter 14.2, Problem 62E

To determine

To calculate: The required probability for the function

f(x,y)={15xy,0≤x≤2,0≤y≤50, elsewhereP(0≤x≤1,0≤y≤2),

If the provided function is a joint density function.

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Chapter 14.2, Problem 61E

Chapter 14.2, Problem 63E

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Chapter 14 Solutions

Calculus: Early Transcendental Functions

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - CONCEPT CHECK Region of Integration Sketch the...Ch. 14.1 - Evaluate the integral: 0x(2xy)dy Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy Ch. 14.1 - Evaluate the integral: y2sin3xcosydx

Knowledge Booster

The required probability for the function f ( x , y ) = { 1 5 x y , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 5 0 , elsewhere P ( 0 ≤ x ≤ 1 , 0 ≤ y ≤ 2 ) , If the provided function is a joint density function.

Solution Summary: The author explains that the function f(x,y) is a joint density function if it satisfies the following properties.

To calculate: The required probability for the function

f(x,y)={15xy,0≤x≤2,0≤y≤50, elsewhereP(0≤x≤1,0≤y≤2),

If the provided function is a joint density function.

Want to see the full answer?

Chapter 14 Solutions

The required probability for the function f ( x , y ) = { 1 5 x y , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 5 0 , elsewhere P ( 0 ≤ x ≤ 1 , 0 ≤ y ≤ 2 ) , If the provided function is a joint density function.

Solution Summary: The author explains that the function f(x,y) is a joint density function if it satisfies the following properties.

To calculate: The required probability for the function f(x,y)={15xy,0≤x≤2,0≤y≤50, elsewhereP(0≤x≤1,0≤y≤2), If the provided function is a joint density function.

Want to see the full answer?

Chapter 14 Solutions

To calculate: The required probability for the function

f(x,y)={15xy,0≤x≤2,0≤y≤50, elsewhereP(0≤x≤1,0≤y≤2),

If the provided function is a joint density function.