   Chapter 7.3, Problem 11E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Evaluating Functions of Several Variables In Exercises 1–14, find and simplify the function values. See Example 1. f ( x , y ) = ∫ x y ( 2 t − 3 ) d t (a) f ( 1 , 2 ) (b) f ( 1 , 4 )

(a)

To determine

To calculate: The value of two variable function f(x,y)=xy(2t3)dt at x=1 and y=2.

Explanation

Given Information:

The provided two variable function is f(x,y)=xy(2t3)dt.

Formula used:

The integration formula:

tndt=tn+1n+1,n1

Calculation:

Consider two variable function,

f(x,y)=xy(2t3)dt

Now substitute the value of x=1 and y=2 in f(x,y)=xy(2t3)dt

Therefore,

f(1,2)=12(2t3)dt

Apply, tndt=tn+1n+1

(b)

To determine

To calculate: The value of two variable function f(x,y)=xy(2t3)dt at x=1 and y=4.

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