Iterated integrals

1. Mar 6, 2012

Kork

1. The problem statement, all variables and given/known data

Calculate the given iterated integrals ∫02 dy ∫0yy2 * exy dx

My attempt:

20dy[exy*y]y0

= ∫20 ey*y*y - ex*0*0

= ∫20ey2*y dx

= [ey^2]*y]20 = 2e4

Is this correct?

Last edited: Mar 6, 2012
2. Mar 6, 2012

Staff: Mentor

I think you skipped a step going from the above to the line below. Written in a more usual form, your integral would be
$$\int_{y = 0}^2~\int_{x = 0}^y y^2 e^{xy}~dx~dy$$

(You can see what I did in LaTeX by clicking the integral above.)
Show how you get from the step above to the step where you've carried out the inner integration.

No - see above. I get (1/2)e4 - 5/2

3. Mar 6, 2012

Kork

I dont understand what I have done wrong at all

4. Mar 6, 2012

Staff: Mentor

The error is here:
$$\left . ye^{xy}\right |_{x = 0}^y$$

You need to replace x by 0, not y.

5. Mar 6, 2012

Kork

Im still confused I get:

[ey^2*y-y]20 =

(e^2^2 - 2) - (e^0^2 - 0) =

2e^4 - 2 ?

Im lost

6. Mar 6, 2012

Staff: Mentor

You're skipping steps. You have the outer integrand right, but you have made a mistake when you integrated this integral.
$$\int_0^2 (ye^{y^2} - y)dy$$

Split this into two integrals and carry out the two integrations.

Last edited: Mar 6, 2012
7. Mar 6, 2012

Kork

Oh I have totally lost my comprehensive view now.....

8. Mar 6, 2012

Staff: Mentor

I don't know what you mean by this.

9. Mar 6, 2012

Kork

How do I get from:

= ∫(y = 0 to 2) (ye^(y^2) - y) dy

= (1/2)e^(y^2) - (1/2)y^2 {for y = 0 to 2}

?

10. Mar 6, 2012

Kork

y0 ye^(y^2) - y dy

to

= (1/2)e^(y^2) - (1/2)y^2

{for y = 0 to 2}

11. Mar 6, 2012

Staff: Mentor

Split the integral into two.
Use substitution to do the first integral.

12. Mar 6, 2012

Kork

This didnt get me further from start, but thanks anyway.

13. Mar 6, 2012

Staff: Mentor

When you evaluate the above at 2 and 0, what do you get?

Sorry, I misunderstood what you were asking before.