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I have a function $f_n:[0,2]->\mathbb R$

$\ f_n(x) = \begin{cases} n^3x^2 & 0

I need to calculate $\int_0^2 f(x)dx$ and $\int_0^2f_n(x)dx$.

So $\int_0^2f_n(x)dx=\frac{2}{3}$, but regarding $\int_0^2 f(x)dx$, isnt the integral $\int_0^2 f(x)dx=0?$

If so, why do I get 2 different answers? Im sure $\int_0^2f_n(x)dx=\frac{2}{3}$ is correct.

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    @Frank: Will you please edit your question to say what $f$ is, and clarify that you are asking why you can't pass limits through integrals? The question doesn't make sense as written.2012-02-15

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From your previous question, I'm assuming you have that $f$ is the pointwise limit of the $f_n$; so, $f$ is identically $0$ on $[0,2]$.

$\int_0^2 f_n(x)\, dx$ is indeed $2/3$ and $\int_0^2f(x)\, dx$ is indeed $0$.

Everything is ok, though. There is nothing here assuring that the integrals $\int_0^2 f_n(x)\,dx$ should converge to $\int_0^1 f(x)\,dx $.

In particular, the $f_n$ do not converge uniformly to $f $, as the answer to the aforementioned question shows. You are not guaranteed that $\int_0^2 f_n (x)\,dx \rightarrow \int_0^2 f(x)\,dx$, if you do not have uniform convergence.

I presume the purpose of this exercise is to show that the hypothesis of uniform convergence is needed in the following theorem:

If $(f_n)$ is a sequence of Riemann integrable functions over $[a,b]$ and if $(f_n)$ converges uniformly to the function $f$ on $[a,b]$, then $f$ is Riemann integrable over $[a,b]$ and $\lim\limits_{n\rightarrow\infty }\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx$.

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    Ahh okay, thanks for your insight!2012-02-15