Answer:
[tex] 55 < A < \frac{149}{2} [/tex]
Step-by-step explanation:
Given:
Function g(x) = 2x²-x-1, [2,5],
N = 6 rectangles
To find:
Two approximation (Left endpoint and Right endpoint of the area) of the area.
Solution:
Using Right endpoint approximation,
[tex]\Delta x = \frac{b - a}{N} \\ \Delta x = \frac{5 - 2}{6} \\ \Delta x = \frac{3}{6} = \frac{1}{2}[/tex]
Now,
[tex]\displaystyle\sf \: R_n = \Delta x \: \sum_{i=1}^N f(a + i \Delta x)[/tex]
Where i = 1,2,3,4......
Substituting value of N, ∆x and a in above equation,
[tex]\displaystyle\sf \: R_n = \Delta x \: \sum_{i=1}^N f(a + i \Delta x) \\ \displaystyle\sf \: R_n = \frac{1}{2} \: \sum_{i=1}^6 f(2 + i \cdot \frac{1}{2} ) \\ \displaystyle\sf \: R_n = \frac{1}{2} \: \sum_{i=1}^6 f(2 + \frac{i}{2} ) \\ \displaystyle\sf \: R_n = \frac{1}{2} \: \sum_{i=1}^6 f( \frac{4 + i}{2} ) [/tex]
[tex]\displaystyle\sf \: R_n = \frac{1}{2} \bigg( f( \frac {5}{2}) + f( 3) +f( \frac {7}{2}) + f( 4) + f( \frac {9}{2}) + f( 5) \bigg) [/tex]
Substituting the corresponding values of x in given function 2x²-x-1
[tex]\displaystyle\sf \: R_n = \frac{1}{2} \bigg(2 \times { (\frac{5}{2} )}^{2} - \frac{5}{2} - 1 ....... +2 \times { ({5} )}^{2} - {5}- 1 \bigg) [/tex]
After solving each function,
[tex]\displaystyle\sf \: R_n = \frac{1}{2} \bigg(9 + 14 +20 + 27 + 35 + 44\bigg) \\ \displaystyle\sf \: R_n \: = \frac{149}{2} [/tex]
Similarly for left endpoint approximation,
[tex]\displaystyle\sf \: L_n = \Delta x \: \sum_{i=0}^{N - 1} f(a + i \Delta x)[/tex]
Where i = 0,1,2,3......
Substituting value of N, ∆x and a in above equation,
[tex]\displaystyle\sf \: L_n = \Delta x \: \sum_{i=0}^{N } f(a + i \Delta x) \\ \displaystyle\sf \: L_n = \frac{1}{2} \: \sum_{i=0}^{N } f(2 + i \frac{1}{2} ) \\\displaystyle\sf \: L_n = \frac{1}{2} \: \sum_{i=0}^6 f( \frac{4 + i}{2} ) [/tex]
[tex]\displaystyle\sf \: L_n = \frac{1}{2} \bigg(f( 2) + f( \frac {5}{2}) + f( 3) +f( \frac {7}{2}) + f( 4) + f( \frac {9}{2}) + \bigg) [/tex]
Substituting the corresponding values of x in given function 2x²-x-1
[tex]\displaystyle\sf \: L_n = \frac{1}{2} \bigg(5+ 9 + 14 +20 + 27 + 35 \bigg) \\ \displaystyle\sf \: L_n \: = \frac{110}{2} = 55 \\ [/tex]
Right approximation 149/2
Left approximation 55
Hence the Area is bounded in,
[tex] 55 < A < \frac{149}{2} [/tex]
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