- #1

- 6

- 0

double integral (e^(y^3)) dy dx

Where dy is evaluated from sqrt(x/3) to 1

.....and dx is evaluated from 0 to 3.

I am lost.

I don't even know how to start.

- Thread starter ChargedTaco
- Start date

- #1

- 6

- 0

double integral (e^(y^3)) dy dx

Where dy is evaluated from sqrt(x/3) to 1

.....and dx is evaluated from 0 to 3.

I am lost.

I don't even know how to start.

- #2

mjsd

Homework Helper

- 726

- 3

double integral is just normal integrals in disguise, so when ou are doing the "y"-integral, you treat all "x"s in the integrand as just a constant, and while doing the "x"-integral you treat all "y"s as constant. Then, everything would then be exactly the same as a normal single integral. In many cases, whether you do the x or y integral first does not matter.

in your case i think you should do y first

- #3

- 13,024

- 579

The "y" integration doesn't look pretty. It's basically

[tex] \int_{a}^{b} e^{y^3}{}dy [/tex]

[tex] \int_{a}^{b} e^{y^3}{}dy [/tex]

- #4

- 6

- 0

If I integrate x first then I'll end up with having to

integrate 3e^(y^3)dy....and I again I don't know how to do this.

Thanks.

- #5

- 6

- 0

Here it is.

[tex]\int_{0}^{3} \int_{sqrt(x/3)}^{1} e^{y^3} {}dy} {}dx [/tex]

Yay!! I got the latex code....however this is the only thing I know about this problem.

[tex]\int_{0}^{3} \int_{sqrt(x/3)}^{1} e^{y^3} {}dy} {}dx [/tex]

Yay!! I got the latex code....however this is the only thing I know about this problem.

Last edited:

- #6

- 6

- 0

Nevermind....I got it. Thanks guys.

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