Evaluate the limit $\lim_{n\to\infty}\sum_{i=1}^{n} f(c_i)\,\Delta x_i$ for $f(x)=\sqrt{x}$

Original question: EXAMPLE 1 A Partition with Subintervals of Unequal Widths

Consider the region bounded by the graph of $f(x)=\sqrt{x}$ and the x-axis for $0\le x\le 1$, as shown in Figure 4.17. Evaluate the limit

$\lim_{n\to\infty}\sum_{i=1}^{n} f(c_i)\,\Delta x_i$

where $c_i$ is the right endpoint of the partition given by $c_i=\frac{i^2}{n^2}$ and $\Delta x_i$ is the width of the $i$th interval.

Solution The width of the ith interval is given by

$\Delta x_i=\frac{i^2}{n^2}-\frac{(i-1)^2}{n^2}$ $=\frac{i^2-i^2+2i-1}{n^2}$ $=\frac{2i-1}{n^2}.$

So, the limit is

$\lim_{n\to\infty}\sum_{i=1}^{n} f(c_i)\,\Delta x_i=\lim_{n\to\infty}\sum_{i=1}^{n}\sqrt{\frac{i^2}{n^2}}\left(\frac{2i-1}{n^2}\right)$ $=\lim_{n\to\infty}\frac{1}{n^3}\sum_{i=1}^{n}(2i^2-i)$ $=\lim_{n\to\infty}\frac{1}{n^3}\left[2\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\right]$ $=\lim_{n\to\infty}\frac{4n^3+3n^2-n}{6n^3}$ $=\frac{2}{3}.$

From Example 7 in Section 4.2, you know that the region shown in Figure 4.18 has an area of $\frac{1}{3}$. Because the square bounded by $0\le x\le 1$ and $0\le y\le 1$ has an area of $1$, you can conclude that the area of the region shown in Figure 4.17 has an area of $\frac{2}{3}$. This agrees with the limit found in Example 1, even though that example used a partition having subintervals of unequal widths. The reason this particular partition gave the proper area is that as $n$ increases, the width of the largest subinterval approaches zero. This is a key feature of the development of definite integrals.

Figure 4.17 The subintervals do not have equal widths.

Figure 4.18 The area of the region bounded by the graph of $x=y^2$ and the y-axis for $0\le y\le 1$ is $\frac{1}{3}$. The point $(1,1)$.

$y$

$x=f(x)=\sqrt{x}$

$\frac{n-1}{n}$

$\frac{2}{n}$

$\frac{1}{n}$

$x$

$\frac{1}{n^2}$

$\frac{2^2}{n^2}$

$\frac{(n-1)^2}{n^2}$

$x=y^2$

Area $=\frac{1}{3}$

$(0,0)$

$(1,1)$