Judging by the model I would assume that ΔDBC is congruent to ΔDBA.
Thus, line CD should be congruent to line AD as well, which the solution should be line CD=13 units.
Ắner 3x+2y
Step-by-step explanation:
Answer:
![55 < A < \frac{149}{2}](https://tex.z-dn.net/?f=%2055%20%3C%20A%20%20%3C%20%20%5Cfrac%7B149%7D%7B2%7D%20)
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.
<em>Solution:</em>
Using Right endpoint approximation,
![\Delta x = \frac{b - a}{N} \\ \Delta x = \frac{5 - 2}{6} \\ \Delta x = \frac{3}{6} = \frac{1}{2}](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20%20%5Cfrac%7Bb%20-%20a%7D%7BN%7D%20%5C%5C%20%5CDelta%20x%20%3D%20%20%5Cfrac%7B5%20-%202%7D%7B6%7D%20%5C%5C%20%5CDelta%20x%20%3D%20%20%20%5Cfrac%7B3%7D%7B6%7D%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D)
Now,
![\displaystyle\sf \: R_n = \Delta x \: \sum_{i=1}^N f(a + i \Delta x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%5CDelta%20x%20%5C%3A%20%5Csum_%7Bi%3D1%7D%5EN%20f%28a%20%2B%20i%20%5CDelta%20x%29)
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} )](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%5CDelta%20x%20%5C%3A%20%5Csum_%7Bi%3D1%7D%5EN%20f%28a%20%2B%20i%20%5CDelta%20x%29%20%5C%5C%20%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5C%3A%20%5Csum_%7Bi%3D1%7D%5E6%20f%282%20%2B%20i%20%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%29%20%5C%5C%20%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5C%3A%20%5Csum_%7Bi%3D1%7D%5E6%20f%282%20%2B%20%5Cfrac%7Bi%7D%7B2%7D%20%29%20%5C%5C%20%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5C%3A%20%5Csum_%7Bi%3D1%7D%5E6%20f%28%20%5Cfrac%7B4%20%2B%20i%7D%7B2%7D%20%20%29%20)
![\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)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cbigg%28%20f%28%20%5Cfrac%20%7B5%7D%7B2%7D%29%20%20%2B%20f%28%203%29%20%2Bf%28%20%5Cfrac%20%7B7%7D%7B2%7D%29%20%20%2B%20f%28%204%29%20%2B%20f%28%20%5Cfrac%20%7B9%7D%7B2%7D%29%20%2B%20f%28%205%29%20%5Cbigg%29%20)
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)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cbigg%282%20%20%5Ctimes%20%20%7B%20%28%5Cfrac%7B5%7D%7B2%7D%20%29%7D%5E%7B2%7D%20%20%20-%20%20%5Cfrac%7B5%7D%7B2%7D%20-%201%20.......%20%2B2%20%20%5Ctimes%20%20%7B%20%28%7B5%7D%20%29%7D%5E%7B2%7D%20%20%20-%20%20%7B5%7D-%201%20%20%5Cbigg%29%20)
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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cbigg%289%20%2B%2014%20%2B20%20%2B%20%2027%20%2B%2035%20%2B%2044%5Cbigg%29%20%5C%5C%20%5Cdisplaystyle%5Csf%20%5C%3A%20R_n%20%5C%3A%20%20%3D%20%20%5Cfrac%7B149%7D%7B2%7D%20)
Similarly for left endpoint approximation,
![\displaystyle\sf \: L_n = \Delta x \: \sum_{i=0}^{N - 1} f(a + i \Delta x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20L_n%20%3D%20%5CDelta%20x%20%5C%3A%20%5Csum_%7Bi%3D0%7D%5E%7BN%20-%201%7D%20f%28a%20%2B%20i%20%5CDelta%20x%29)
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} )](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20L_n%20%3D%20%5CDelta%20x%20%5C%3A%20%5Csum_%7Bi%3D0%7D%5E%7BN%20%7D%20f%28a%20%2B%20i%20%5CDelta%20x%29%20%5C%5C%20%5Cdisplaystyle%5Csf%20%5C%3A%20L_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5C%3A%20%5Csum_%7Bi%3D0%7D%5E%7BN%20%7D%20f%282%20%2B%20i%20%20%5Cfrac%7B1%7D%7B2%7D%20%29%20%5C%5C%5Cdisplaystyle%5Csf%20%5C%3A%20L_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5C%3A%20%5Csum_%7Bi%3D0%7D%5E6%20f%28%20%5Cfrac%7B4%20%2B%20i%7D%7B2%7D%20%20%29%20)
![\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)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20L_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cbigg%28f%28%202%29%20%2B%20f%28%20%5Cfrac%20%7B5%7D%7B2%7D%29%20%20%2B%20f%28%203%29%20%2Bf%28%20%5Cfrac%20%7B7%7D%7B2%7D%29%20%20%2B%20f%28%204%29%20%2B%20f%28%20%5Cfrac%20%7B9%7D%7B2%7D%29%20%2B%20%20%5Cbigg%29%20)
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 \\](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csf%20%5C%3A%20L_n%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Cbigg%285%2B%209%20%2B%2014%20%2B20%20%2B%20%2027%20%2B%2035%20%20%5Cbigg%29%20%5C%5C%20%5Cdisplaystyle%5Csf%20%5C%3A%20L_n%20%5C%3A%20%20%3D%20%20%5Cfrac%7B110%7D%7B2%7D%20%20%3D%2055%20%5C%5C%20%20)
Right approximation 149/2
Left approximation 55
Hence the Area is bounded in,
![55 < A < \frac{149}{2}](https://tex.z-dn.net/?f=%2055%20%3C%20A%20%20%3C%20%20%5Cfrac%7B149%7D%7B2%7D%20)
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Since in y=8, the slope is 0 (because there is no x) and b is 8 in y=mx+b, we know that the perpendicular line is x=(something) because the perpendicular lines of y=<number> are x=<number> due to that y=<number> equations are strictly horizontal while x=<number> equations are strictly vertical. Since we know that x=<number> has to pass through (7, 8), x =7