Answer:
65 units²
Step-by-step explanation:
To approximate the area under the function f(x) = x ²+ 2 between a = 1 and b = 6 using a left-hand sum with 5 intervals, we first need to divide the interval from 'a' to 'b' into equal parts. The left-hand sum is calculated as follows:
[tex]\boxed{ \left \begin{array}{ccc} \text{\underline{Left-Hand Rectangle Approximation Method:}} \\\\ A \approx \sum_{i=1}^{n} f(x_{i-1}) \Delta x \\\\ \text{Where:} \\ \bullet \ A \ \text{is the approximate area under the curve} \\ \bullet \ n \ \text{is the number of rectangles} \\ \bullet \ f(x_{i-1}) \ \text{is the function value at the left end of the ith subinterval} \\ \bullet \ \Delta x \ \text{is the width of each subinterval} \end{array} \right}[/tex]
[tex]\hrulefill[/tex]
Determine the width of each interval:
[tex]\Delta x = \dfrac{b-a}{n}=\dfrac{6-1}{5}=\boxed{1}[/tex]
Now we can approximate the area:
[tex]L_n=\Delta x (f(x_0)+f(x_1)+\dots+f(x_{n-1}))[/tex]
[tex]\Longrightarrow L_5=\Delta x (f(1)+f(2)+f(3)+f(4)+f(5))[/tex]
[tex]\Longrightarrow L_5=(1) ([(1)^2 + 2]+[(2)^2 + 2]+[(3)^2 + 2]+[(4)^2 + 2]+[(5)^2 + 2])[/tex]
[tex]\Longrightarrow L_5=(1) (3+6+11+18+27)\\\\\\\Longrightarrow L_5=(1) (65)\\\\\\\therefore L_5 = \boxed{65 \text{ units$^2$}}[/tex]
Thus, the approximated area is 65 units².
