#### To determine

a)

**To graph:**

The function fx=11+x2, -2≤x≤2

#### Answer

#### Explanation

Use a graphing calculator to draw the graph of the given function.

Graph of fx=11+x2 -2≤x≤2

#### To determine

b)

**To estimate:**

The area under the graph of fx=11+x2 -2≤x≤2 by using 4 rectangles.

#### Answer

R4=2.2, M4=2.2154

#### Explanation

**1) 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)

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

2) **Calculation:**

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

x=-2, 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 -2, -1, -1, 0,0, 1, [1, 2]

Width of each rectangle ∆x=2-(-2)4=1

From the graph, heights of the strips are 0.5, 1, 0.5, 0.2

The sum of areas of these approximating rectangles is

R4=1(f-1+f0+f1+f(2)=112+1+12+15

=1115

R4=2.2

ii)

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

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

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

Width of each rectangle ∆x=2-(-2)4=1

The sum of areas of these approximating rectangles is

M4=1(f-1.5+f-0.5+f0.5+f1.5)

M4=1413+45+45+413

M4=14465=2.2154

**Conclusion:**

The area using right endpoints is R4=2.2, and area using mid points is M4=2.2154.

#### To determine

c)

**To estimate:**

The area under the graph of fx=11+x2 -2≤x≤2 by using 8 rectangles.

#### Answer

R8=2.2077, M8=2.2176

#### 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: Rn=fx1∆x+fx2∆x+…..+fxn∆x

i.e.,Rn=∆x(fx1+fx2+…..+fxn)

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

**2) Calculation:**

i) Divide the given area into 8 rectangle strips by drawing vertical lines at

x=-2, x=-1.5, x=-1, x=-0.5, x=0, x=0.5, x=1, x=1.5, x=2

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

Width of each rectangle ∆x=2-(-2)8=48=0.5

The sum of areas of these approximating rectangles is

R8=0.5(f-1.5+f-1+f-0.5+f0+f0.5+f1+f1.5+f2)

=12413+12+45+1+45+12+413+15

=287130

R8=2.2077

ii)

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

Divide the given area into 8 rectangle strips by drawing vertical lines.

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 -2, -1.5, -1.5, -1,-1, -0.5, -0.5, 0,0, 0.5,0.5, 1, 1, 1.5, 1.5, 2, i.e., at points x=-1.75, x=-1.25, x=-0.75, x=-0.25, x=0.25, x=0.75, x=1.25, x=1.75

Width of each rectangle ∆x=2-(-2)8=0.5

The sum of areas of these approximating rectangles is

M8=0.5(f-1.75+f-1.25+f-0.75+f-0.25+f0.25+f0.75+f1.25+f1.75)

M8=121665+1641+1625+1617+1617+1625+1641+1665

M8≈2.2176

**Conclusion:**

The area using right endpoints is R8=2.2077, and the area using midpoints is M8=2.2176.