#### To determine

(a)

**To estimate:**

The area under the graph of fx=1+x2 by right endpoints.

#### Answer

R3=8 R6=6.875

#### Explanation

**Concept:**

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

ii) Width of each strip is ∆x=b-an

iii) Area: Rn=fx1∆x+fx2∆x+…..+fxn∆x

i.e.,Rn=∆x(fx1+fx2+…..+fxn)

Ln=∆x(fx0+fx1+…..+fxn-1)

Mn=∆x(fx1'+fx2'+…..+fxn'), where xi' is the midpoint of each interval

**Calculation:**

Divide the given area into 3 rectangle strips by drawing vertical lines at

x=-1, x=0, x=1, 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 right end point of the subintervals -1, 0, 0, 1,1, 2, width of each rectangle ∆x=3-03=1

From the graph, heights of the strips are 1, 2, 5.

The sum of areas of these approximating rectangles is

R3=1(f0+f1+f2

=11+2+5

=1(8)

=8

ii) For R6

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

x=-1. x=-0.5, x=0, x=0.5, x=1, x=1.5, 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 right end point of the subintervals [-1, -0.5], [-0.5, 0],[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]

Width of each rectangle ∆x=2-(-1)6=32=0.5

From the graph, heights of the strips are 1.25, 1, 1.25, 2, 3.25, 5.

The sum of areas of these approximating rectangles is

R6=0.5(f-0.5+f0+f0.5+f1+f1.5+f(2))

=0.51.25+1+1.25+2+3.25+5

=0.5(13.75)

R6=6.875

**Conclusion: **

R3=8 R6=6.875

#### To determine

**b)**

**To estimate:**

The area under the graph of fx=1+x2 by right endpoints.

#### Answer

L3=5, L6=5.375

#### Explanation

**Concept:**

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

ii) Width of each strip is ∆x=b-an

iii) Area: Rn=fx1∆x+fx2∆x+…..+fxn∆x

i.e.,Rn=∆x(fx1+fx2+…..+fxn)

Ln=∆x(fx0+fx1+…..+fxn-1)

Mn=∆x(fx1'+fx2'+…..+fxn'), where xi' is the midpoint of each interval

**Calculation:**

Divide the given area into 3 rectangle strips by drawing vertical lines at

x=-1, x=0, x=1

**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, 0, 0, 1,1, 2.

Width of each rectangle ∆x=2-(-1)3=1

From the graph, heights of the strips are 2, 1, 2

The sum of areas of these approximating rectangles is

L6=1(f-1+f0+f1

=12+1+2

=5

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

x=-1, x=-0.5, x=0, x=0.5, x=1, x=1.5

**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, -0.5, -0.5, 0,0, 0.5,0.5,1,1,1.5,[1.5,2]

Width of each rectangle ∆x=2-(-1)6=0.5

From the graph, heights of the strips are 2, 1.25, 1, 1.25, 2, 3.25

The sum of areas of these approximating rectangles is

L6=0.5(f-1+f-0.5+f0+f0.5+f1+f1.5)

=0.52+1.25+1+1.25+2+3.25

=5.375

**Conclusion: **

L3=5, L6=5.375

#### To determine

**c)**

**To estimate:**

The area under the graph of fx=1+x2 by right endpoints.

#### Answer

M3=5.75, M6=5.9375

#### Explanation

**Concept:**

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

ii) Width of each strip is ∆x=b-an

iii) Area: Rn=fx1∆x+fx2∆x+…..+fxn∆x

i.e.,Rn=∆x(fx1+fx2+…..+fxn)

Ln=∆x(fx0+fx1+…..+fxn-1)

Mn=∆x(fx1'+fx2'+…..+fxn'), where xi' is the midpoint of each interval

**Calculation:**

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

Divide the given area into 3 rectangle strips by drawing vertical lines at x=-0.5, x=0.5, x=6, x=1.5.

**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 midpoint of the subintervals [-1, 0], [0, 1], [1, 2] i.e. at points x=-0.5, x=0.5, x=1.5

Width of each rectangle ∆x=3-03=1

The sum of areas of these approximating rectangles is

M3=1(f-0.5+f0.5+f1.5)

=11.25+1.25+3.25

=5.75

Now divide the given area into 6 rectangle strips by drawing vertical lines at x=-0.75, x=-0.25, x=0.25, x=0.75, x=1.25, x=1.75

**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 midpoint of the subintervals -1, -0.5, -0.5, 0, 0, 0.5,0.5, 1, 1, 1.5, 1.5, 2, i.e., at points x=-0.75, x=-0.25, x=0.25, x=0.75, x=1.25, x=1.75.

Width of each rectangle ∆x=2-(-1)6=0.5

The sum of areas of these approximating rectangles is

M6=0.5(f-0.75+f-0.25+f0.25+f0.75+f1.25+f(1.75))

=0.51.5625+1.0625+1.0625+1.5625+2.5625+4.0625

=0.511.875

=5.9375

**Conclusion: **

M3=5.75, M6=5.9375

#### To determine

**d)**

**To answer:**

The best estimation among the parts (a)-(c)

#### Answer

M6=6

#### Explanation

From parts (a)-(c), the values of area approximated by right end, left end, and midpoint are

R3=8 R6=6.875 L3=5 L6=5.375 M3=5.75 M6=5.9375

Among these, the best estimation is M6=6 since from the graph, it looks more close to the actual answer.

**Conclusion:**

The best estimation is M6=6.