The area under the graph of f(x) = 2 + 4x^2 from x = -1 to x = 2 has to be determined using different methods.

a) Using three rectangles and right endpoints.

The area under the graph is approximated by 3 rectangles.

The width of each of the rectangles...

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The area under the graph of f(x) = 2 + 4x^2 from x = -1 to x = 2 has to be determined using different methods.

a) Using three rectangles and right endpoints.

The area under the graph is approximated by 3 rectangles.

The width of each of the rectangles is (2 - (-1))/3 = 3/3 = 1.

Length of the three rectangles starting from right is,

R1 = f(2) = 18

R2 = f(1) = 6

R3 = f(0) = 2

The sum of their area is 1*18 + 1*6 + 1*2 = 26.

b) Using six rectangles and right endpoints,

The width of each of the rectangles is (2 - (-1))/6 = 3/6 = 1/2.

The length of each rectangle starting from the right is,

R1 = f(2) = 18

R2 = f(1.5) = 11

R3 = f(1) = 6

R4 = f(0.5) = 3

R5 = f(0) = 2

R6 = f(-0.5) = 3

The sum of their area is 0.5*18+0.5*11+0.5*6+0.5*3+0.5*2+0.5*3 = 21.5.

c) Using the left endpoints method, for 3 rectangles the width of each rectangle is equal to (2 - (-1))/3 = 1.

The length of the rectangles starting from the leftmost one is,

L1 = f(-1) = 6

L2 = f(0) = 2

L3 = f(1) = 6

Adding the area of these rectangles gives an area under the graph of 1*6 + 1*2+1*6 = 14.

d) Using the left endpoint method, using 6 rectangles the width of each of the rectangles is (2 - (-1))/6 = 0.5.

The length of the rectangles starting from the leftmost is:

L1 = f(-1) = 6

L2 = f(-0.5) = 3

L3 = f(0) = 2

L4 = f(0.5) = 3

L5 = f(1) = 6

L6 = f(1.5) = 11

Adding the area of these rectangles gives 0.5*6 + 0.5*3 + 0.5*2 + 0.5*3 + 0.5*6 + 0.5*11 = 15.5.

A) You need to evaluate the area under the graph of function `f(x)= 2 + 4x^2` and above x axis, using 3 rectangles and right endpoints.

You need to remember the area of rectangle such that:

A = l*w (length*width)

You need to find the total width of three rectangles such that:

`W = 2 - (-1) = 2+1 = 3`

You need to divide this total width by the number of rectangles such that:

`W/3 = w = 3/3 = 1`

Hence, evaluating the width of first rectangle yields:

`w_1 = -1+1 = ` 0

Evaluating the length yields:

`l_1 = f(0) = 2+4*0^2 =gt f(0)=2`

The area of first rectangle is: `A_1 = w_1*l_1`

`A_1 = 1*2 = 2`

You need to evaluate the width of the second rectangle such that:

`w_2 = 0+1 = 1`

Evaluating the length yields:

`l_2 = f(1) = 2 + 4 = 6`

`A_2 = 1*6 = 6`

You need to evaluate the width of the third rectangle such that:

`w_3 = 1+1 = 2`

Evaluating the length yields:

`w_3 = f(2) = 2+ 4*4 = 18`

`A_3 = 1*18 = 18`

`A = A_1+A_2+A_3`

`A = 2+6+18 = 26`

You need to evaluate the area under the graph of function `f(x)= 2 + 4x^2` and above x axis, using 6 rectangles and right endpoints.

You need to remember the area of rectangle such that:

A = l*w (length*width)

You need to find the total width of six rectangles such that:

`W = 2 - (-1) = 2+1 = 3`

You need to divide this total width by the number of rectangles such that:

`W/6 = w = 3/6 = 1/2`

Hence, evaluating the widths, lentgths and areas of rectangles yields:

`w_1 = -1+1/2 = -1/2`

`l_1 = f(-1/2) = 2 + 4*(1/4) = 3`

`A_1 = (1/2)*3 = 3/2`

`w_2 = -1/2 + 1/2 = 0`

`l_2 = f(0) = 2`

`A_2 = (1/2)*2 =` 1

`w_3 = 0 + 1/2 = 1/2`

`l_3 = f(1/2) = 2 + 1 = 3`

`A_3 = (1/2)*3 = 3/2`

`w_4 = 1/2 + 1/2 = 1`

`l_4 = f(1) = 2 +4 = 6`

`A_4 = (1/2)*6 = 3`

`w_5 = 1/2 + 1 = 3/2`

`l_5 = f(3/2) = 2 + 4*(9/4) = 11`

`A_5 = (1/2)*11 = 11/2`

`w_6 = 1/2 + 3/2 = 2`

`l_6 = f(2) = 2 + 16= 18`

`A_6= (1/2)*18 = 9`

Hence, evaluating the area using 6 rectangles yields:

`A = A_1+A_2+A_3+A_4+A_5+A_6`

`A = 3/2 + 1 + 3/2 + 3 + 11/2 + 9`

`A = 16 + 11/2 =gt A = 43/2 =gt A = 21.5`

B) Estimating the area using 3 rectangles and the left points yields:

`w_1 = 2-1 = 1`

`l_1 = f(1) = 2+4=6`

`A_1=1*6=6`

`w_2 = 1-1 = 0`

`l_2=f(0)=2`

`A_2=1*2=2`

`w_3=0-1=-1`

`l_3=f(-1)=2+4=6`

`A_3 = 1*6=6`

`A = 6+2+6 = 14`

Estimating the area using 6 rectangles and the left points yields:

`w_1 = 2-1/2 = 3/2`

`l_1 = f(3/2) = 2+9=11`

`A_1=(1/2)*11=11/2`

`w_2 = 3/2-1/2 = 1`

`l_2=f(1)=6`

`A_2=(1/2)*6=3`

`w_3=1-1/2=1/2`

`l_3=f(1/2)=2+1=3`

`A_3 = (1/2)*3=3/2`

`w_4=0-1/2=-1/2`

`l_4=f(-1/2)=2+1=3`

`A_4 = (1/2)*3=3/2`

`w_5=-1/2-1/2=-1`

`l_5=f(-1)=2+4=6`

`A_5 = (1/2)*6=3`

`A = 11/2+3+3/2+3/2+3=29/2`

**Hence, evaluating the area using 3 or 6 rectangles and right points or left points yields A= 26, A=21.5, A=14, A=14.5.**