 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 ruleFor continuous $f$, this tends to the integral as $h\to 0$; if $f$ has a bounded continuous second derivative, the error is at most $(b-a)h^2\max |f^{\prime \prime }|/12$.

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 $N\to \infty$, and relabelling the summation index $j$, gives the right Riemann sumFor $f(x)=x^4$, $a=0$, and $b=1$, the only difference from the composite trapezoidal rule is the endpoint weighting: $S_N=T_N+1/(2N)$. Hence