# 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.