Question

need help with this double integral

Answer 1

\[\int\limits_{0}^{1} \int\limits_{y}^{e^y} {\sqrt x} dx dy\]

Answer 2

so, lets start with the innards; what the integration of sqrt(x) ?

Answer 3

2/3 x^3/2

Answer 4

and when you apply the limits we get:

2/3 (e^3y/2 - y^3/2)

is what i think i see there, does that seem right to you?

Answer 5

then we work the dy part

Answer 6

youve got an e and a poly .... those are basic enough right?

Answer 7

@amistre64 i guess yes

Answer 8

\[\int_{a}^bx^{1/2}dx=\frac 23 (b^{3/2}-a^{3/2})\]

\[\frac23 \left(\int_{p}^{q} b^{3/2}dy-\int_{p}^{q} a^{3/2}\right)\]
\[\frac23 \left(\int_{p}^{q} e^{3y/2}dy-\int_{p}^{q} y^{3/2}\right)\]

Answer 9

i do by dy then will use this \[\int\limits_{y}^{e^2} \int\limits_{0}^{1} \sqrt x dy dx \]

Answer 10

when you change the limits around, you have to make sure you dont loose the integrity of the area that you are playing with. It is not a general rule that you can just swap the integrations like that

Answer 11

ok

Answer 12

the limits actually define a movement in a defined region, and that movement has to be kept intact if you swap it about

Answer 13

notice that if you just swap it like that, you are left with variables in the outside integral ... this is not a good sign in general since it tends to lead to an answer in variables and nothing specific

Answer 14

\[2/3 \int\limits_{0}^{1} x^{3/2} dy\]