Question

PLEASE HELP! CALCULUS

Answer 1

Answer:

$55 < A < \frac{149}{2}$

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,

$\Delta x = \frac{b - a}{N} \\ \Delta x = \frac{5 - 2}{6} \\ \Delta x = \frac{3}{6} = \frac{1}{2}$

Now,

$\displaystyle\sf \: R_n = \Delta x \: \sum_{i=1}^N f(a + i \Delta x)$

Where i = 1,2,3,4......

Substituting value of N, ∆x and a in above equation,

$\displaystyle\sf \: R_n = \Delta x \: \sum_{i=1}^N f(a + i \Delta x) \\ \displaystyle\sf \: R_n = \frac{1}{2} \: \sum_{i=1}^6 f(2 + i \cdot \frac{1}{2} ) \\ \displaystyle\sf \: R_n = \frac{1}{2} \: \sum_{i=1}^6 f(2 + \frac{i}{2} ) \\ \displaystyle\sf \: R_n = \frac{1}{2} \: \sum_{i=1}^6 f( \frac{4 + i}{2} )$

$\displaystyle\sf \: R_n = \frac{1}{2} \bigg( f( \frac {5}{2}) + f( 3) +f( \frac {7}{2}) + f( 4) + f( \frac {9}{2}) + f( 5) \bigg)$

Substituting the corresponding values of x in given function 2x²-x-1

$\displaystyle\sf \: R_n = \frac{1}{2} \bigg(2 \times { (\frac{5}{2} )}^{2} - \frac{5}{2} - 1 ....... +2 \times { ({5} )}^{2} - {5}- 1 \bigg)$

After solving each function,

$\displaystyle\sf \: R_n = \frac{1}{2} \bigg(9 + 14 +20 + 27 + 35 + 44\bigg) \\ \displaystyle\sf \: R_n \: = \frac{149}{2}$

Similarly for left endpoint approximation,

$\displaystyle\sf \: L_n = \Delta x \: \sum_{i=0}^{N - 1} f(a + i \Delta x)$

Where i = 0,1,2,3......

Substituting value of N, ∆x and a in above equation,

$\displaystyle\sf \: L_n = \Delta x \: \sum_{i=0}^{N } f(a + i \Delta x) \\ \displaystyle\sf \: L_n = \frac{1}{2} \: \sum_{i=0}^{N } f(2 + i \frac{1}{2} ) \\\displaystyle\sf \: L_n = \frac{1}{2} \: \sum_{i=0}^6 f( \frac{4 + i}{2} )$

$\displaystyle\sf \: L_n = \frac{1}{2} \bigg(f( 2) + f( \frac {5}{2}) + f( 3) +f( \frac {7}{2}) + f( 4) + f( \frac {9}{2}) + \bigg)$

Substituting the corresponding values of x in given function 2x²-x-1

$\displaystyle\sf \: L_n = \frac{1}{2} \bigg(5+ 9 + 14 +20 + 27 + 35 \bigg) \\ \displaystyle\sf \: L_n \: = \frac{110}{2} = 55$

Right approximation 149/2

Left approximation 55

Hence the Area is bounded in,

$55 < A < \frac{149}{2}$

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