## Finding area

1. Use left and right endpoints and the given
number of rectangles to find two approximations of the area of
the region between the graph of the function and the x-axis over
the given interval.

f(x) = 2x + 5; [0, 2]; 4 rectangles

2. Relevant equations

i = n(n+1)/2

3. The attempt at a solution
i can find the upper/right endpoint but the left endpoint is difficult.

n
Ʃ [2(2(i -1)/(n)) + 5](2/n)
i = 1

n
(2/n)Ʃ [2(2(i -1)/(n)) + 5]
i = 1

n
(2/n)Ʃ [(4(i -1)/(n)) + 5]
i = 1

n n
(2/n){(4/n)Ʃ (i -1) + Ʃ 5}
i = 1 i = 1

and then i sub the equation in for i and solve but i do not get the right answer.

btw the correct answer is 13.
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Recognitions:
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 Quote by physics=world i = n(n+1)/2
What are i and n in the context of this question?
 n Ʃ [2(2(i -1)/(n)) + 5](2/n) i = 1 n (2/n)Ʃ [2(2(i -1)/(n)) + 5] i = 1 n (2/n)Ʃ [(4(i -1)/(n)) + 5] i = 1 n n (2/n){(4/n)Ʃ (i -1) + Ʃ 5} i = 1 i = 1 and then i sub the equation in for i and solve but i do not get the right answer.
It was ok up to that point. Exactly what substitution did you make?
 n is going to equal 4. and i sub in n(n+1)/2 for i in the equation

Recognitions:
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Science Advisor

## Finding area

 Quote by physics=world n is going to equal 4. and i sub in n(n+1)/2 for i in the equation
No, it's $\sum_{i=1}^n i=n(n+1)/2$.
 Recognitions: Gold Member Science Advisor Staff Emeritus Is there a reason why you are using that general formula for this very specific problem? You are given the interval from 0 to 2 and and asked to divide it into 4 rectangles. The problem does NOT say "rectangles with the same base" but that is the simplest thing to do- each base will have length 2/4= 1/2. The endpoints of the bases of those rectangles will be 0, 1/2, 1, 3/2, and 2. For the "left endpoints", evaluate 2x+ 5 at x= 0, 1/2, 1, and 3/2. For the "right endpoints", evaluate 2x+ 5 at x= 1/2, 1, 3/2, and 2.
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