use finite approximation to estimate the area under the graph of f(x)=6x² and above the graph of f(x)=0 from…

use finite approximation to estimate the area under the graph of f(x)=6x² and above the graph of f(x)=0 from x₀ = 0 to xₙ = 6 using i) a lower sum with two rectangles of equal width. ii) a lower sum with four rectangles of equal width. iii) an upper sum with two rectangles of equal width. iv) an upper sum with four rectangles of equal width.

use finite approximation to estimate the area under the graph of f(x)=6x² and above the graph of f(x)=0 from x₀ = 0 to xₙ = 6 using i) a lower sum with two rectangles of equal width. ii) a lower sum with four rectangles of equal width. iii) an upper sum with two rectangles of equal width. iv) an upper sum with four rectangles of equal width.

Answer

Explanation:

Step1: Calculate width of rectangles

For (n) rectangles on ([a,b]=[0,6]), the width (\Delta x=\frac{b - a}{n}).

Step2: i) Lower - sum with 2 rectangles

(n = 2), (\Delta x=\frac{6-0}{2}=3). The sub - intervals are ([0,3]) and ([3,6]). For a lower sum, we use the minimum value of (y = f(x)=6x^{2}) on each sub - interval. On ([0,3]), (f(0)=0); on ([3,6]), (f(3)=6\times3^{2}=54). The lower sum (L_2=\sum_{i = 0}^{1}f(x_i)\Delta x=f(0)\times3 + f(3)\times3=0\times3+54\times3 = 162).

Step3: ii) Lower - sum with 4 rectangles

(n = 4), (\Delta x=\frac{6 - 0}{4}=1.5). The sub - intervals are ([0,1.5]), ([1.5,3]), ([3,4.5]), ([4.5,6]). (f(0)=0), (f(1.5)=6\times(1.5)^{2}=13.5), (f(3)=6\times3^{2}=54), (f(4.5)=6\times(4.5)^{2}=121.5). (L_4=\sum_{i = 0}^{3}f(x_i)\Delta x=f(0)\times1.5+f(1.5)\times1.5+f(3)\times1.5+f(4.5)\times1.5=(0 + 13.5+54 + 121.5)\times1.5=283.5).

Step4: iii) Upper - sum with 2 rectangles

(n = 2), (\Delta x = 3). The sub - intervals are ([0,3]) and ([3,6]). For an upper sum, we use the maximum value of (y = f(x)) on each sub - interval. On ([0,3]), (f(3)=54); on ([3,6]), (f(6)=6\times6^{2}=216). The upper sum (U_2=\sum_{i = 0}^{1}f(x_{i + 1})\Delta x=f(3)\times3+f(6)\times3=54\times3+216\times3=810).

Step5: iv) Upper - sum with 4 rectangles

(n = 4), (\Delta x=1.5). The sub - intervals are ([0,1.5]), ([1.5,3]), ([3,4.5]), ([4.5,6]). (f(1.5)=13.5), (f(3)=54), (f(4.5)=121.5), (f(6)=216). (U_4=\sum_{i = 0}^{3}f(x_{i+1})\Delta x=(13.5 + 54+121.5+216)\times1.5=594).

Answer:

i) (162) ii) (283.5) iii) (810) iv) (594)