- #1

- 584

- 0

[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.