This preview shows page 1. Sign up to view the full content.
Unformatted text preview: + y ) ln x dy dx,
1 0 1
xy + y 2 l n x
2 Thus
(e2x − 1) dx = 3(x + y ) ln x dxdy.
A integrating ﬁrst with respect to y . After integration the inner integral becomes 0 3 1
30 The integral with respect to y can be carried
out immediately, whereas the integral with respect to x would require integration by parts,
so this suggests that I should be written as a
repeated integral xexy dy dx. Now after integration the inner integral becomes
2
= e2x − 1 .
exy I= 19
4 Explanation:
Since the area of the rectangle A is 30, the
average value of f over A is given by xexy dxdy I= 36
19
ln 6 +
5
4 6
0 = 6x ln x + 18 ln x. Now is the time to integrate by parts, for then
6 6x ln x + 18 ln x dx and so
I=
012 16
e −7
2 1 . 10.0 points 3
= 3x2 ln x − x2 + 18x ln x − 18x
2 6
1 . Thus Find the average value, I , of the function over the rectangle
A = { (x, y ) : 1 ≤ x ≤ 6, 0 ≤ y ≤ 6 }.
1. I = 3 19...
View
Full
Document
 Summer '10
 KUSHNER
 Multivariable Calculus

Click to edit the document details