Calculating Iterated Integrals - 2e4

  • Thread starter Kork
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In summary: You need to use substitution to get the first integral.This didnt get me further from start, but thanks anyway.
  • #1
Kork
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1. The problem statement, all variables and given/known data

Calculate the given iterated integrals ∫02 dy ∫0yy2 * exy dxMy 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:
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  • #2
Kork said:
1. The problem statement, all variables and given/known data

Calculate the given iterated integrals ∫02 dy ∫0yy2 * exy dx
I think you skipped a step going from the above to the line below. Written in a more usual form, your integral would be
[tex]\int_{y = 0}^2~\int_{x = 0}^y y^2 e^{xy}~dx~dy[/tex]

(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.


Kork said:
My attempt:

20dy[exy*y]y0

= ∫20 ey*y*y - ex*0*0

= ∫20ey2*y dx

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

Is this correct?

No - see above. I get (1/2)e4 - 5/2
 
  • #3
I don't understand what I have done wrong at all
 
  • #4
The error is here:
[tex]\left . ye^{xy}\right |_{x = 0}^y[/tex]

You need to replace x by 0, not y.
 
  • #5
Im still confused I get:

[ey^2*y-y]20 =

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

2e^4 - 2 ?

Im lost
 
  • #6
You're skipping steps. You have the outer integrand right, but you have made a mistake when you integrated this integral.
[tex]\int_0^2 (ye^{y^2} - y)dy[/tex]

Split this into two integrals and carry out the two integrations.
 
Last edited:
  • #7
Oh I have totally lost my comprehensive view now...
 
  • #8
Kork said:
Oh I have totally lost my comprehensive view now...
I don't know what you mean by this.
 
  • #9
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
y0 ye^(y^2) - y dy

to

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


{for y = 0 to 2}
 
  • #11
Split the integral into two.
Use substitution to do the first integral.
 
  • #12
This didnt get me further from start, but thanks anyway.
 
  • #13
Kork said:
y0 ye^(y^2) - y dy

to

= (1/2)e^(y^2) - (1/2)y^2
When you evaluate the above at 2 and 0, what do you get?

Sorry, I misunderstood what you were asking before.
Kork said:
{for y = 0 to 2}
 

1. What is an iterated integral?

An iterated integral is an integral that is evaluated over a specific region by breaking it up into smaller subregions and then summing the results of multiple integrals over each subregion.

2. How do you calculate an iterated integral?

To calculate an iterated integral, you first need to determine the limits of integration for each variable. Then, use the appropriate integration rules to evaluate the integral over each subregion and sum the results.

3. What is the purpose of calculating iterated integrals?

Calculating iterated integrals allows us to find the total value of a function over a specific region, which can be useful in solving real-world problems and analyzing complex functions.

4. What are some common methods for evaluating iterated integrals?

Some common methods for evaluating iterated integrals include the rectangular method, the trapezoidal method, and the Simpson's rule. These methods involve dividing the region into smaller rectangles, trapezoids, or parabolas and using their areas to approximate the integral.

5. How can I check if my calculated iterated integral is correct?

One way to check if your calculated iterated integral is correct is by using the Fundamental Theorem of Calculus. This theorem states that the integral of a function over a region can be found by evaluating its antiderivative at the limits of integration. Another way is to use numerical integration techniques, such as the trapezoidal rule, to compare your result with the calculated value.

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