= Uniform convergence preserves integrals on a compact interval
{title2=$\left|\int_a^b(f_n-f)\right|\leq(b-a)\|f_n-f\|_\infty$}
If <Riemann integrable> <functions> $f_n:[a,b]\to\mathbb R$ converge by <uniform convergence> to $f$, then $f$ is <Riemann integrable> and
$$
\left|\int_a^b f_n(x)\,dx-\int_a^b f(x)\,dx\right|\leq(b-a)\|f_n-f\|_\infty\longrightarrow0.
$$
This concerns one fixed <compact interval>, with finite length.<Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl/6.2 Interchange of limits>[Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl, §6.2.2, Theorem 6.2.4]{ref}
Here is a proof outline that also establishes <Riemann integrability> of the <uniform limit>. Assume $a<b$, and put $\ell=b-a$ and $E_N=\|f_N-f\|_\infty$. One bounded $f_N$ and finite $E_N$ show that $f$ is a <bounded function>. For any <partition of an interval> $P$, its <upper Darboux sum> and <lower Darboux sum> satisfy
$$
U(f,P)-L(f,P)\leq U(f_N,P)-L(f_N,P)+2\ell E_N.
$$
Given $\eta>0$, choose $N$ with $E_N<\eta/(4\ell)$, then a <partition of an interval> with $U(f_N,P)-L(f_N,P)<\eta/2$. The displayed gap is below $\eta$, proving <Riemann integrability>. Finally, $|f_n-f|\leq E_n$ and <monotonicity of the Riemann integral> give the stated error estimate. When $a=b$, every <integral> is zero.
For the <polynomial functions> $f_n(x)=x^n$ on $[0,0.8]$, direct <integration> and the uniform bound give
$$
\int_0^{0.8}x^n\,dx=\frac{0.8^{n+1}}{n+1}\leq0.8\cdot0.8^n\longrightarrow0.
$$
At $n=21$, the exact <integral> is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact <antiderivative>. On $[0,1]$, these same <integrals> still tend to zero, although there is no <uniform convergence>: sufficient conditions need not be necessary. On an unbounded <interval>, even <uniform convergence> can fail to control total area, as <uniformly vanishing functions can retain a nonzero integral> shows.
<Uniform convergence> alone also fails to preserve <derivatives>: $f_n(x)=\sin(n^2x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n'(0)=n$. The small amplitude does not control the slopes of the <sine> oscillations.
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