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

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lemjay eNotes educator| Certified Educator

`int_1^2 int_1^y y/x dx dy`

To evaluate, start with the inner integral.

>> `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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