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Difference between double and repeated integrals

  • Thread starter Benny
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  • #1
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Hi, I'm just having some trouble with definitions. I've googled repeated integrals but I haven't yet come across something which has answered my question. Anyway, I'd like to know what the difference is, between double and repeated integrals. For example if I had:

[tex]
\int\limits_0^2 {\int\limits_0^x {x^2 y} dydx}
[/tex]

I would evaluate it as follows.

[tex]
\int\limits_0^2 {\int\limits_0^x {x^2 y} dydx}
[/tex]

[tex]
= \int\limits_0^2 {\left[ {\frac{{x^2 y^2 }}{2}} \right]} _{y = 0}^{y = x} dx
[/tex]

[tex]
= \int\limits_0^2 {\left( {\frac{{x^4 }}{2}} \right)} dx
[/tex]

[tex]
= \left[ {\frac{{x^5 }}{5}} \right]_0^2
[/tex]

= 32/5.

Now I assume that I've just evaluated the integral as a "double integral." The question booklet I have lists the question under "repeated integrals." I'm wondering what the difference is and how I can evaluate this integral as a repeated integral. The only thing I can gather about repeated integrals(from the little bits of info I've found on google) is that the integration is done repeatedly wrt one variable but I'm not sure how that works. Can someone please explain to me how to evaluate this integral as a repeated integral? The answer is 16/5.
 

Answers and Replies

  • #2
88
0
Benny said:
Hi, I'm just having some trouble with definitions. I've googled repeated integrals but I haven't yet come across something which has answered my question. Anyway, I'd like to know what the difference is, between double and repeated integrals. For example if I had:

[tex]
\int\limits_0^2 {\int\limits_0^x {x^2 y} dydx}
[/tex]

I would evaluate it as follows.

[tex]
\int\limits_0^2 {\int\limits_0^x {x^2 y} dydx}
[/tex]

[tex]
= \int\limits_0^2 {\left[ {\frac{{x^2 y^2 }}{2}} \right]} _{y = 0}^{y = x} dx
[/tex]

[tex]
= \int\limits_0^2 {\left( {\frac{{x^4 }}{2}} \right)} dx
[/tex]

[tex]
= \color{red} \left ( \frac{1}{2} \right ) \cdot \color{black} \left[ {\frac{{x^5 }}{5}} \right]_0^2 \color{red} \ = \ \frac{16}{5}
[/tex]

= 32/5.

Now I assume that I've just evaluated the integral as a "double integral." The question booklet I have lists the question under "repeated integrals." I'm wondering what the difference is and how I can evaluate this integral as a repeated integral. The only thing I can gather about repeated integrals(from the little bits of info I've found on google) is that the integration is done repeatedly wrt one variable but I'm not sure how that works. Can someone please explain to me how to evaluate this integral as a repeated integral? The answer is 16/5.
the answer to above integral is 16/5.
you forgot the factor of (1/2) shown in RED above.
 
  • #3
88
0
Benny said:
Hi, I'm just having some trouble with definitions. I've googled repeated integrals but I haven't yet come across something which has answered my question. Anyway, I'd like to know what the difference is, between double and repeated integrals. For example if I had:

[tex]
\int\limits_0^2 {\int\limits_0^x {x^2 y} dydx}
[/tex]

I would evaluate it as follows.

[tex]
\int\limits_0^2 {\int\limits_0^x {x^2 y} dydx}
[/tex]

[tex]
= \int\limits_0^2 {\left[ {\frac{{x^2 y^2 }}{2}} \right]} _{y = 0}^{y = x} dx
[/tex]

[tex]
= \int\limits_0^2 {\left( {\frac{{x^4 }}{2}} \right)} dx
[/tex]

[tex]
= \color{red} \left ( \frac{1}{2} \right ) \cdot \color{black} \left[ {\frac{{x^5 }}{5}} \right]_0^2 \color{red} \ = \ \frac{16}{5}
[/tex]

= 32/5.

Now I assume that I've just evaluated the integral as a "double integral." The question booklet I have lists the question under "repeated integrals." I'm wondering what the difference is and how I can evaluate this integral as a repeated integral. The only thing I can gather about repeated integrals(from the little bits of info I've found on google) is that the integration is done repeatedly wrt one variable but I'm not sure how that works. Can someone please explain to me how to evaluate this integral as a repeated integral? The answer is 16/5.
there's no practical difference between a "double integral" and a "repeated integral". the term "repeated integral" generally refers to the method used to evalute a "double integral":

[tex] \mbox{Double Integral = } \int \, \int_{(2D) Region} f(x,y) \, dA \ = \ \int \, \int_{(x,y) Region} f(x,y) \, dx \, dy \ \mbox{ = Repeated Integral} [/tex]

in your solution above (except for the error shown in RED), you evaluated the "double integral" over a (2D) Region with the method of "repeated integrals" over an (x,y) Region, by first integrating wrt "y" and then integrating wrt "x".
 
Last edited:
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
What you did is a "repeated" integral because you first found the integral with respect to y and then "repeated" the process- found the derivative of the result with respect to x. A "double" integral is, as geosonel said, an integral over a two dimensional area, not necessarily saying anything about the coordinate system.
It is a basic theorem that, given a specific coordinate system, we can change a double integral into a repeated integral ("Fubini's Theorem"). Typically, we set up a problem in terms of a double integral and then change to a repeated integral to evaluate the integral.
 
  • #5
584
0
I just rechecked my working and after attempting some of the questions I did last night, again, I've founded that my initial attempts had many arithmetic errors which lead to deviations from the correct answer. Thanks for the help guys.
 

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