Question

use type 2 to set up and then evaluate the integration of int_int_ xe^y dA, where D is bounded by y = 2x, y = x^2
anyone help me,please

Answer 1

@Spacelimbus guide me, please

Answer 2

Did you draw the regions? Unfortunately I don't have any drawint utilities here at the moment, but basically the region should be all you care about at the moment.

Answer 3

yes

Answer 4

that's the graph I got from the problem

Answer 5

This might take some fidgeting around with the bounds or rather with the order of integration, but I believe the first we might want to do is set y as constant and integrate with respect to x, that would mean
\[\frac{ y }{ 2 } \leq x \leq \sqrt{y}\]

Answer 6

yes, friend

Answer 7

and 0<= y<=4 . that's what i got

Answer 8

very good yes.

Answer 9

:) but I stuck at take integral. shame on me, cannot get the right answer

Answer 10

ok I will take a look at in

Answer 11

x = 0 to 1
y = x^2 to x

Answer 12

it should turn out to be

\[\frac{ 1 }{ 8 }y^2e^y-\frac{ 1 }{ 2 }ye^y\]

Answer 13

\[\int_{0}^{1}\int_{x^2}^{x}~xe^y~dydx\]
or
\[\int_{0}^{1}\int_{y}^{\sqrt{y}}~xe^y~dxdy\]

Answer 14

@amistre64 , the line is y=2x

Answer 15

... im blinder than a bat :/

Answer 16

won't change much though!

Answer 17

dividing by two here, adding 3 there :D

Answer 18

What are both you talking about? @amistre64 I don't get what you mean
@Spacelimbus I got what you got but 1/2 ye^y dy - 1/8 y^2e^ydy

Answer 19

yes @Hoa you are right, you got that right, I messed up the order when typing it into my equation editor.

Answer 20

and I stuck at the second part, don't know how to take int. the first part is ok, I got e^4 at the end up. try to solve the leftover.

Answer 21

well these are two integrals you can solve with the tabular method for integration, have you ever done that @Hoa

Answer 22

nope

Answer 23

teach me, please. it sounds new to me

Answer 24

Lets check the more complicated one.