Question

evaluate the iterated integral:
(from 0 to ln3),(from 1 to ln2)
e^(2x+4y)dydx

Answer 1

\[\int\limits_{0}^{\ln3}\int\limits_{1}^{\ln2} e^{(2x+4y)}dydx\]

Answer 2

\[\int\limits_{0}^{\ln3}\ \left( \int\limits_{1}^{\ln4}e^{(2x+4y)}dy \right)dx\]

Answer 3

inside the parentheses would be this?
\[\frac{ 1 }{ 4 }e^{2x+4(\ln2)}-\frac{ 1 }{ 4 }e^{2x+4}\]

Answer 4

I entered the wrong upper limit in the parentheses - it is supposed to be ln2

Answer 5

Looks good

Answer 6

You could also split the double integral into a product of two single integrals

Answer 7

\[\int\limits_{0}^{\ln3}\int\limits_{1}^{\ln2} e^{(2x+4y)}dydx \]

\[\int\limits_{0}^{\ln3}e^{2x}\int\limits_{1}^{\ln2} e^{4y}dydx \]

\[\left(\int\limits_{0}^{\ln3}e^{2x}dx \right)\cdot \left(\int\limits_{1}^{\ln2} e^{4y}dy\right) \]

Answer 8

that works whenever you can write the integrand as a product of form : "f(x)*g(y)"

Answer 9

That looks much easier to deal with

Answer 10

I messed up somewhere:
\[\left( \frac{ 1 }{ 2 }e^{2(\ln3)}-\frac{ 1 }{ 2 }e^{2(0)} \right) * \left( \frac{ 1 }{ 4 }e^{4(\ln2)}-\frac{ 1 }{ 4 }e^{4(1)} \right)\ \]

Answer 11

\[\left( \frac{ 1 }{ 2 }\left( 9-1 \right) \right)*\left( \frac{ 1 }{ 4 } \left( 16-e \right)\right)\]
\[\left( 4 \right)*\left( 4-\frac{ e }{ 4 } \right)\]
\[16-e\]

Answer 12

do you have any more questions on that

Answer 13

It's not the right answer. I made a mistake and can't spot it.

Answer 14

Oh oh oh I see.`

Answer 15

\[\Large\rm e^{4(1)}\]For the lower limit on your y's

Answer 16

\[\Large\rm \ne e^1\]

Answer 17

oh! so that makes the end result 16-e^4?

Answer 18

looks good!

Answer 19

Thanks for catching that!

Answer 20

Good eye