= Solution
Divide $[a,b]$ into $N$ equal subintervals of width $h=(b-a)/N$, with $x_j=a+jh$. Linear interpolation between adjacent values of $f$ gives a trapezoid of signed area $h(f(x_{j-1})+f(x_j))/2$. Adding these areas produces the <composite trapezoidal rule>
$$
\boxed{T_N=h\left(\frac{f(a)+f(b)}2+\sum_{j=1}^{N-1}f(x_j)\right)\approx\int_a^bf(x)\,dx}.
$$
For continuous $f$, this tends to the <integral> as $h\to0$; if $f$ has a bounded continuous second <derivative>, the error is at most $(b-a)h^2\max|f''|/12$.
The original PDF prints $n\to\infty$ while using $n$ as the bound summation index and $N$ as the upper limit. Literally this does not specify a varying upper limit. Interpreting the intended limit as \b[$N\to\infty$], and relabelling the summation index $j$, gives the <right Riemann sum>
$$
S_N=\frac1N\sum_{j=1}^N\left(\frac jN\right)^4=\frac1{N^5}\sum_{j=1}^Nj^4.
$$
For $f(x)=x^4$, $a=0$, and $b=1$, the only difference from the <composite trapezoidal rule> is the endpoint weighting: $S_N=T_N+1/(2N)$. Hence
$$
\boxed{\lim_{N\to\infty}\frac1{N^5}\sum_{j=1}^Nj^4=\int_0^1x^4\,dx=\frac15}.
$$
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