Find Area of Region w/ 4 Rectangles

[physics=world]

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

Homework Equations

i = n(n+1)/2

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.

 

[haruspex]

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?

 

[physics=world]

n is going to equal 4.

and i sub in n(n+1)/2 for i in the equation

 

[haruspex]

n is going to equal 4.

and i sub in n(n+1)/2 for i in the equation

No, it's $\sum_{i=1}^n i=n(n+1)/2$.

 

[HallsofIvy]

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.