Express area as a limit of left-endpoint Riemann sums

Since $a = 1$ and $b = 4$, the width of each subinterval is $\Delta x = \frac{4-1}{n} = \frac{3}{n}$. The left endpoint of the $i$-th subinterval, starting at $a=1$, is $x_i = 1 + (i-1)\Delta x = 1 + \frac{3(i-1)}{n}$.

Substituting into $f(x_i) = x_i^2 + 1$ gives $f(x_i) = \left(1 + \frac{3(i-1)}{n}\right)^2 + 1$. Multiplying by $\Delta x$ and summing over $i=1$ to $n$, then letting $n\to\infty$, gives the exact limit-of-Riemann-sum expression for the integral: $\int_1^4 (x^2+1)\,dx = \lim_{n\to\infty}\sum_{i=1}^n \left[\left(1+\frac{3(i-1)}{n}\right)^2+1\right]\cdot\frac{3}{n}.$

Each subinterval width equals three divided by n, which shrinks toward zero as n grows without bound. This is exactly why the Riemann sum approaches the exact area: the rectangles become infinitely thin, eliminating approximation error. The answer is that the width approaches zero.