$ \lim_{n\rightarrow \infty}\int_{x=a}^{x=b} f_n(x) dx = \int_{x=a}^{x=b} f(x) dx$ only when f is uniformly convergent? $ \lim_{n\rightarrow \infty}\int_{x=a}^{x=b} f_n(x) dx = \int_{x=a}^{x=b} f(x) dx$ .....(A)
where $ f_1(x), f_2(x),......,f_n(x) $ represent a sequence of functions such that $f(x) = \lim_{n\rightarrow \infty} f_n(x) $
Is the statement (A) true only if $f(x)$ is uniformily convergent? 
A sequence of functions f is said to be uniformly convergent in $[a,b]$ if  $\forall ~ \epsilon>0, ~ \exists ~ m_o  \in N: |f_n(x)-f(x)| < \epsilon , ~ \forall ~x \in [a,b] , ~ \forall ~ n \geq m_o$
i.e. $m_o$ is common for the convergence of $f(x_o) \forall x \in [a,b]$
So, a converging function whether uniformly convergent or NOT in the interval [a,b], will still follow the property that : $ \lim_{n\rightarrow \infty} f_n(x)$ = $f(x) ~ \forall ~ x \in [a,b] $
Hence, shouldn't $ \lim_{n\rightarrow \infty}\int_{x=a}^{x=b} f_n(x) dx = \int_{x=a}^{x=b} f(x) dx$  irrespective of whether f is uniformily convergent or not ?
 A: The statement is not true in general.
Example: Take $f_n=n$ on $[0,\frac{1}{n}]$ and $0$ on $[\frac{1}{n},1]$. Then $\int\limits_{0}^{1}f_n=1$, but $f_n \rightarrow 0$.
Also, $\int f_n \rightarrow \int f$ doesn't imply uniform convergence.
e.g.: $f_n=x^n$ on $[0,1]$
A: Convergence of integrals does not imply uniform convergence of the functions. Consider
$$
f_n(x)=4x^n(1-x^n)
$$
and
$$
f(x)=0
$$
We have that $\lim\limits_{n\to\infty}f_n(x)=f(x)$ and
$$
\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x=\int_0^1f(x)\,\mathrm{d}x
$$
yet the convergence is not uniform since $\sup\limits_{[0,1]}f_n(x)=1$.

Convergence of the functions does not imply convergence of the integrals. Consider
$$
f_n=2nx^{n-1}(1-x^n)
$$
and
$$
f(x)=0
$$
We have that $\lim\limits_{n\to\infty}f_n(x)=f(x)$ yet
$$
\int_0^1f_n(x)\,\mathrm{d}x=1
$$
and
$$
\int_0^1f(x)\,\mathrm{d}x=0
$$
A: A classic example is $n x^n$ on $(0,1)$. The function converges pointwise to $0$, so 
$$
\int_0^1 \lim_n n x^n \,dx = 0
$$
on the other hand
$$
\lim_n \int_0^1 n x^n \,dx = \lim_n \frac{n}{n+1} x^{n+1}\vert_0^1 = 1
$$
The main idea is that an integral is concerned with global behavior of the function, and pointwise-only results are about as local as it gets.
