# Evaluate the double integral. int_1^2 int_1^y y/x dx dy int_1^2 int_1^y y/x dx dy

>> int_1^y y/x dx

Since we have dx, treat y as a constant. Then, use the logarithmic formula of integral which is int (du)/u = ln u + C  .

>> = yint_1^y (dx)/x = y...

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int_1^2 int_1^y y/x dx dy

>> int_1^y y/x dx

Since we have dx, treat y as a constant. Then, use the logarithmic formula of integral which is int (du)/u = ln u + C  .

>> = yint_1^y (dx)/x = y lnx |_1^y = ylny - yln1 = ylny-0=ylny

Then,  evaluate the outer integral. Here, y is a variable.

> int_1^2 int_1^y y/x dx = int_1^2 yln y dy

Use integration by parts. Formula is int udv= uv -intvdu . So , let

u = ln y                       and             dv=ydy

du = 1/ydy=(dy)/y                              v=y^2/2

Substitute these to the formula.

(ln y)*y^2/2 - inty^2/2 (dy)/y = (y^2lny)/2 - 1/2intydy =(y^2lny)/2 -1/2y^2/2

So we have,

> int_1^2ylnydy = (y^2lny)/2-y^2/4 |_1^2

= ((2^2ln2)/2-2^2/4)-((1^2ln1)/2-1^2/4)

= (2ln2-1)-(0-1/4)=2ln2-3/4

Hence, int_1^2 int_1^y x/ydxdy = 2ln2-3/4 ` .

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