#### To determine

**(a) **

**To estimate:**

The area under the graph of f(x) by right endpoints.

#### Answer

0.6345, underestimate

#### Explanation

**1) Given:**

fx=1x, from x=1 to x=2, n=4

**2) Concept:**

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

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

iii) Area: Rn=fx1∆x+fx2∆x+…..+fxn∆x

That is,Rn=∆x(fx1+fx2+…..+fxn)

**3) Calculation:**

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

x=1, x=1.25, x=1.50, x=1.75, x=2

Graph:

Approximate each strip by 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 [1, 1.25], [1.25, 1.50],[1.50, 1.75], [1.75, 2]

Width of each rectangle ∆x=2-14=0.25

Each rectangle has width 0.25, and the heights are 11.25,11.50,11.75, 2

That is,0.8, 0.6666, 0.5714, 0.5.

The sum of the areas of these approximating rectangles is

R4=0.25(0.8+0.6666+0.5714+0.5)

=0.25(2.538)

=0.6345

Since the function is decreasing, and when we use the right end point to evaluate the area, the bars are always little bit below the graph, giving an underestimate.

**Conclusion:**

Area =0.6345, it is an underestimate.

#### To determine

**(b) **

**To estimate:**

The area under the graph of f(x) by left endpoints.

#### Answer

0.7595, overestimate

#### Explanation

**1) Given:**

fx=1x, from x=1 to x=2, n=4

**2) 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

That is,Ln=∆x(fx0+fx1+…..+fxn-1)

**3) Calculation: **

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

x=1, x=1.25, x=1.50, x=1.75, x=2

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

[1, 1.25], [1.25, 1.50], [1.50, 1.75], [1.75, 2]

Width of each rectangle ∆x=2-14=0.25

Each rectangle has width 0.25, and the heights are 1, 11.25,11.50,11.75

That is, 1, 0.8, 0.6666, 0.5714

The sum of the areas of these approximating rectangles is

L4=0.25(1+0.8+0.6666+0.5714)

=0.25(3.0338)

=0.7595

Since the graph is decreasing and when we use the left end point to evaluate the area, the bars are always little bit above the graph, giving an overestimate.

**Conclusion:**

Area =0.7595, it is an overestimate.