91Ó°ÊÓ

We have given,

$f\left(x\right)=\sqrt{x-1},[a,b]=[2,3]\mathrm{and}n=4$

Left endpoint Riemann sum formula:

$$\begin{gathered} {∫}_a^{b}f\left(x\right)dx ≈\mathrm{Δ}x\left(f\right(x_0)+f(x_1)+f(x_2)+...+f(x_{n-1}\left)\right) \\ Where, \\ \mathrm{Δ} x = \frac{b-a}{n} \end{gathered}$$

Right endpoint Riemann sum formula:

${∫}_a^{b}f\left(x\right)dx ≈\mathrm{Δ}x\left(f\right(x_1)+f(x_2)+f(x_3)+...+f(x_n\left)\right)$

We have given,

$f\left(x\right)=\sqrt{x-1},[a,b]=[2,3]\mathrm{and}n=4$

So length of the subintervals is,

$\mathrm{Δ} x = \frac{b-a}{n}=\frac{1}{4}$

So dividing the interval [2, 3] in to the subintervals with length $\frac{1}{4}$is,

role="math" localid="1648623565407" $\left[2,\frac{9}{4}\right],\left[\frac{9}{4},\frac{5}{2}\right],\left[\frac{5}{2},\frac{11}{4}\right],\left[\frac{11}{4},3\right]$

Left end points are:role="math" localid="1648623573506" $2,\frac{9}{4},\frac{5}{2}\mathrm{and}\frac{11}{4}$

Now, just evaluating the function at the left endpoints of the subintervals,

$f\left(2\right)=1,f\left(\frac{9}{4}\right)=\frac{\sqrt{5}}{2},f\left(\frac{5}{2}\right)=\frac{\sqrt{6}}{2}\mathrm{and}f\left(\frac{11}{4}\right)=\frac{\sqrt{7}}{2}$

Using left end point formula,

$$\begin{gathered} {∫}_2^3\sqrt{x-1}dx≈\frac{1}{4}\left[f\left(2\right)+f\left(\frac{9}{4}\right)+f\left(\frac{5}{2}\right)+f\left(\frac{11}{4}\right)\right] \\ ≈\frac{1}{4}\left[1+\frac{\sqrt{5}}{2}+\frac{\sqrt{6}}{2}+\frac{\sqrt{7}}{2}\right] \\ ≈1.17 \end{gathered}$$

Hence, using approximation method to approximate the signed area between the graph of f and the x-axis on [a, b] is,

${∫}_2^3\sqrt{x-1}dx≈1.17$

We have given,

$f\left(x\right)=\sqrt{x-1},[a,b]=[2,3]\mathrm{and}n=4$

Length of the subintervals is,

$\mathrm{Δ} x = \frac{b-a}{n}=\frac{1}{4}$

So dividing the interval [2, 3] in to the 4 subintervals with length $\frac{1}{4}$,

$\left[2,\frac{9}{4}\right],\left[\frac{9}{4},\frac{5}{2}\right],\left[\frac{5}{2},\frac{11}{4}\right],\left[\frac{11}{4},3\right]$

Right end points are:

$\frac{9}{4},\frac{5}{2},\frac{11}{4}\mathrm{and}3$

Now, just evaluating the function at the right endpoints of the subintervals,

$f\left(\frac{9}{4}\right)=\frac{\sqrt{5}}{2},f\left(\frac{5}{2}\right)=\frac{\sqrt{6}}{2},f\left(\frac{11}{4}\right)=\frac{\sqrt{7}}{2}andf\left(3\right)=\sqrt{2}$

Using the right end point Riemann sum formula,

$$\begin{gathered} {∫}_2^3\sqrt{x-1}dx≈\frac{1}{4}\left[f\left(\frac{9}{4}\right)+f\left(\frac{5}{2}\right)+f\left(\frac{11}{4}\right)+f\left(3\right)\right] \\ ≈\frac{1}{4}\left[\frac{\sqrt{5}}{2}+\sqrt{\frac{3}{2}}+\frac{\sqrt{7}}{2}+\sqrt{2}\right] \\ ≈1.27 \end{gathered}$$

Hence, using approximation method to approximate the signed area between the graph of f and the x-axis on [a, b] is,

$ {∫}_2^3\sqrt{x-1}dx≈1.27$