To determine

(a)

To find:

Estimates of

i) L6(sample points are left endpoints)

ii) R6(sample points are right endpoints)

iii) M6(sample points are mid points)

Answer

i) L6=86.6

ii) R6=70.6

iii) M6=79.4

Explanation

Given:

Graph of f from x=0 to x=12, n=6

Concept:

Formula:

i) Width of the interval [a, b] is (b-a)

ii) Width of each of strip is ∆x=b-an

iii) Area: Rn=fx1∆x+fx2∆x+…..+fxn∆x

That is, Rn=∆x(f(x1)+f(x2)+…..+f(xn))

 Ln=∆x(f(x0)+f(x1)+…..+f(xn-1))

 Mn=∆x(f(x1')+f(x2')+…..+f(xn'))  where xi' is the midpoint of each interval.

Calculation:

i) For L6

Divide the given area into 6 rectangle strips by drawing vertical lines at

x=2, x=4, x=6, x=8, x=10

Graph (i):

Approximate each strip by a rectangle that has the same base as the strip and whose height is the value of the function f(x) at the left end point of the subintervals [0, 2], [2, 4],[4, 6], [6, 8], [8, 10], [10, 12].

Width of each rectangle ∆x=12-06=2

From the graph, heights of the strips are 9, 8.8, 8.2, 7.3, 5.9, 4.1

The sum of areas of these approximating rectangles is

L6=2(f0+f2+f4+f6+f8+f10)

=29+8.8+8.2+7.3+5.9+4.1

=2(43.3)

=86.6

ii) For R6

Divide the given area into 6 rectangle strips by drawing vertical lines at

x=2, x=4, x=6, x=8, x=10

Graph (ii):

Approximate each strip by a rectangle that has the same base as the strip and whose height is the value of the function f(x) at the right end point of the subintervals [0, 2], [2, 4],[4, 6], [6, 8], [8, 10], [10, 12]

Width of each rectangle ∆x=12-06=2

From the graph, heights of the strips are 8.8, 8.2, 7.3, 5.9, 4.1, 1

The sum of areas of these approximating rectangles is

R6=2(f2+f4+f6+f8+f10+f(12))

=28.8+8.2+7.3+5.9+4.1+1

=2(35.3)

=70.6

ii) For M6

Here M means the midpoint of the left end point and right end point of the rectangular strip.

Divide the given area into 6 rectangle strips by drawing vertical lines at

x=2, x=4, x=6, x=8, x=10

Graph (iii):

Approximate each strip by a rectangle that has the same base as the strip and whose height is the value of the function f(x) at the midpoint of the subintervals

[0, 2], [2, 4], [4, 6], [6, 8], [8, 10], [10, 12],

that is, at points x=1, x=3, x=5, x=7, x=9, x=11.

Width of each rectangle ∆x=12-06=2

The sum of areas of these approximating rectangles is

M6=2(f1+f3+f5+f7+f9+f(11))

=28.9+8.5+7.8+6.6+5.1+2.8

=2(39.7)

=79.4

Conclusion:

i) L6(Sample points are left endpoints) = 86.6

ii) R6(Sample points are right endpoints) = 70.6

iii) M6(Sample points are midpoints) = 79.4

To determine

(b)

To check:

Whether L6  is an underestimate or overestimate of the true area

Answer

Overestimate

Explanation

Given:

L6=86.6

R6=70.6

Concept:

i) Overestimate means greater than the actual area.

ii) Underestimate means less than the actual area.

Calculation:

Here, the function is decreasing, and the value of lower estimate L6  is larger than the value of the upper estimate  R6, Therefore,L6>R6. This means L6 is an overestimate.

Conclusion:

L6  is an overestimate.

To determine

(c)

To check:

Whether R6 is an underestimate or overestimate of the true area

Answer

Underestimate

Explanation

Given:

L6=86.6

R6=70.6

Concept:

i) Overestimate means greater than the actual area.

ii) Underestimate means less than the actual area.

Calculation:

Here, the function is decreasing, and the value of the upper estimate R6  is smaller than the value of the lower estimate   L6. Therefore,L6>R6. This means R6 is an underestimate.

Conclusion:

R6  is an underestimate

To determine

(d)

To check:

Best estimation.

Answer

M6

Explanation

Given:

L6=86.6

R6=70.6

M6=79.4

Concept:

i) Overestimate means greater than the actual area.

ii) Underestimate means less than the actual area.

Calculation:

Here the function is decreasing and L₆ > M₆ > R₆. L₆ is an overestimate, and R₆ is an underestimate. M₆ gives the best estimate since the area of each rectangular strip appears to be closer to the true area than L₆ & R₆.

Conclusion:

M6   is the best estimate.