approximate the integral ∫₁⁸ (x² + 1)dx with both left - hand and right - hand rectangular sums, each with 4…

Explanation:

Step1: Calculate the width of each sub - interval

The interval is $[a,b]=[1,8]$ and $n = 4$. The width $\Delta x=\frac{b - a}{n}=\frac{8 - 1}{4}=\frac{7}{4}=1.75$. The sub - intervals are $[1,2.75],[2.75,4.5],[4.5,6.25],[6.25,8]$.

Step2: Calculate the left - hand Riemann sum

The left - hand endpoints are $x_0 = 1,x_1=2.75,x_2 = 4.5,x_3=6.25$. The function is $f(x)=x^{2}+1$. $f(x_0)=1^{2}+1 = 2$ $f(x_1)=2.75^{2}+1=7.5625 + 1=8.5625$ $f(x_2)=4.5^{2}+1=20.25+1 = 21.25$ $f(x_3)=6.25^{2}+1=39.0625+1=40.0625$ The left - hand Riemann sum $L_4=\Delta x\sum_{i = 0}^{3}f(x_i)=\frac{7}{4}(2 + 8.5625+21.25+40.0625)=\frac{7}{4}(71.875)=125.78125$.

Step3: Calculate the right - hand Riemann sum

The right - hand endpoints are $x_1 = 2.75,x_2=4.5,x_3 = 6.25,x_4=8$. $f(x_1)=2.75^{2}+1=8.5625$ $f(x_2)=4.5^{2}+1=21.25$ $f(x_3)=6.25^{2}+1=40.0625$ $f(x_4)=8^{2}+1=65$ The right - hand Riemann sum $R_4=\Delta x\sum_{i = 1}^{4}f(x_i)=\frac{7}{4}(8.5625+21.25+40.0625+65)=\frac{7}{4}(134.875)=236.03125$.

Answer:

Left - Hand Riemann Sum with 4 rectangles = $125.78125$ Right - Hand Riemann Sum with 4 rectangles = $236.03125$