To determine

(a)

To estimate:

The area under the graph of fx=sin⁡x by right end points.

Answer

1.1835, Overestimate

Explanation

1) Concept:

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

ii) Width of each of the strips is ∆x=b-an

iii) Area: Rn=fx1∆x+fx2∆x+…..+fxn∆x

That is,Rn=∆x(fx1+fx2+…..+fxn)

2) Given:

fx=sin⁡x, from x=0 to x=π2, n=4

3) Calculation:

Divide the given interval into 4 rectangle strips by drawing vertical lines at

x=0, x=π8, x=π4, x=3π8,x=π2

Graph:

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, π8, π8, π4, π4, 3π8, 3π8, π2

Width of each rectangle ∆x=π2-04=π8

Each rectangle has width π8  and the heights are sin⁡π8,sin⁡π4, sin⁡3π8,sin⁡π2

i.e.  0.3827, 0.7071, 0.9238, 1

The sum of the areas of these approximating rectangles is

R4=π8sin⁡π8+sin⁡π4+ sin⁡3π8+sin⁡π2

=π80.3827+0.7071+0.9239+1

=π8(3.0137)

=(3.14159×3.0137)8

=1.1835

Since the function is increasing and when use the right end points to evaluate the area, the bars are always little bit above the graph, giving an overestimate.

Conclusion:

Area is 1.1835, and it is an overestimate.

To determine

(b)

To estimate:

The area under the graph of fx=sin⁡x by right end points.

Answer

0.7908, underestimate

Explanation

1) Concept:

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

ii) Width of each of the n strips is ∆x=b-an

iii) Area: Ln=fx0∆x+fx1∆x+…..+fxn-1∆x

i.e.Ln=∆x(fx0+fx1+…..+fxn-1)

2) Given:

fx=sin⁡x, from x=0 to x=π2, n=4

3) Calculation:

Divide the given interval into 4 rectangle strips by drawing vertical lines at

x=0, x=π8, x=π4, x=3π8,x=π2

Graph:

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, π8, π8, π4, π4, 3π8, 3π8, π2

Width of each rectangle ∆x=π2-04=π8

Each rectangle has width π8, and the heights are sin⁡0,sin⁡π8, sin⁡π4, sin⁡3π8 

i.e.  0, 0.3827, 0.7071, 0.9239

The sum of the areas of these approximating rectangles is

L4=π8sin⁡0+sin⁡π8+sin⁡π4+ sin⁡3π8

=π80+0.3827+0.7071+0.9239

=π8(2.0137)

=(3.14159×2.0137)8

=0.7908

Since the function is increasing and when use the left end points to evaluate the area, the bars are always little bit below the graph, giving an underestimate.

Conclusion:

Area is 0.7908, and it is an underestimate.