Estimate the area under the graph of y=3x for x between 0 and 2 . Use a partition that consists of 4?

I'm guessing trapezium rule?

area: 0.5h{(y0 + yn) + 2(y1 + y2+ ... +y(n-1)} Where h = (b-a)/n

n = 4

So your'll have 4 trapeziums, and 5 lines:

x y

0 0

0.5 1.5

1 3

1.5 4.5

2 6

h = (2-0) / 5

so h = 0.5

0.5 x 0.5 {(0+6)+2(1.5+3+4.5)}

Simplifies to:

0.25 (6 + 18)

Simplifies to:

0.25 x 24

Which equals

6

So your answer is:

An estimate to the area under the graph of y = 3x for x between 0 and 2, using a partition consisting of 4 is equal to 6.

4 partitions will result interior the aptitude pattern factors x = 0,a million/2,a million,3/2,2. As we are employing the magnificent endpoint we will not use x = 0. A rectangle demands a length and a top. The length is given by using the area of the partition, a million/2 for each, the top is given by using the value of the function on the magnificent endpoint. the section is then approximately (a million/2)f(a million/2) + (a million/2)f(a million) + (a million/2)f(3/2) + (a million/2)f(2) (a million/2)(3/4) + (a million/2)(2) + (a million/2)(15/4) + (a million/2)(6) (3/8) + a million + (15/8) + 3 4 + (9/4) 25/4

The (very coarse Riemann mesh) rectangles have areas:

1/2 * ( 3 * 0) = 0

1/2 * ( 3 * 1/2) = 3/4

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

1/2 * ( 3 * 3/2) = 9/4

Hence the estimated area is (3/4 + 6/4 + 9/4) = 18/4 = 9/2.

(The exact area (triangle) of course equals 1/2 * 2 * 6 = 12/2 = 6 )