#### To determine

**a)**

**To evaluate:**

i. gx for x=0

ii. gx for x=1

iii. gx for x=2

iv. gx for x=3

v. gx for x=4

vi. gx for x=5

vii. gx for x=6

#### Answer

i. g0=0

ii. g1=12

iii. g2=0

iv. g3=-12

v. g4=0

vi. g5=32

vii. g6= 4

#### Explanation

**Concept:**

Value of a definite integral is the same as the net area enclosed between the graph and x axis.

**Given:**

gx=∫0xf(t)dt

**Calculation:**

i. To evaluate gx for x=0

The integral of a function over a trivial interval is zero. So

g0=∫00f(t)dt =0

So

g0=0

ii. To evaluate gx for x=1

See from the graph

Geometrically, the area under the graph from 0 to 1 is g(1)

Thus it is the area of the triangle with width and height 1

g1=12·1·1

Simplify

g1=12

iii. To evaluate gx for x=2

That is to evaluate g2 subtract from g1 the area of the triangle with width and height 1

The area beneath the graph from 1 to 2 cancels out the area above the graph from 0 to 1

g2=∫02f(t)dt=∫01f(t)dt-∫12f(t)dt

g2=1-1=0

iv. To evaluate g3

The area from 0 to 2 cancels out and then there is a 1 by 1 triangle left underneath the x-axis

g3=∫03f(t)dt=g2-∫23ftdt

Simplifying

g3=0-12·1·1

g3=-12

v. To evaluate g4

The area from 0 to 2 and from 2 to 4 cancels out

g4=∫04f(t)dt=g3+∫34f(t)dt

Simplifying

g4=-12+12

g4=0

vi. To evaluate g5

The area from 0 to 4 cancels out, and then a 1 by 1 box which has area 1 and 1 by 1 triangle which has area ½ are left. So

g5=g4+∫45f(t)dt

Simplifying

g5=0+(1+12)

Simplify

g5=32

vii. To evaluate g6

The areas from 0 to 4 cancel out, and there are three 1 by 1 boxes with total area 3, and two 1 by 1 triangles under the graph each of which has area 1/2.So

g6=g5+∫56f(t)dt

g6=32+(1+1+12)

Simplifying

g6=32+52

g6=4

Therefore, g0=0,g1=12,g2=0, g3=-12,g4=0. g5=32, g6=4

#### To determine

**b)**

**To estimate:**

g(7)

#### Answer

6.2

#### Explanation

**Calculation:**

From part a) we have

g0=0,g1=12,g2=0, g3=-12,g4=0. g5=32, g6=4

To estimate g7

g7=∫07f(t)dt

This is the total area between the graph of g and x-axis from x=0 to 7

The areas that are below the x-axis are negative

From x=0 to 4 the total area above and below cancels out to 0

From x=4 to 6 there are a total of 4 whole squares

From x=6 to 7 there are roughly about 2.2 whole squares

Taking all together

g7=g6+∫67f(t)dt

g7=4+2.2

Add

g7=6.2

Therefore, g7=6.2

#### To determine

**c)**

**To find:**

Where does g have maximum value and minimum value

#### Answer

g has a maximum value at 7 and a minimum value at 3

#### Explanation

**Calculation: **From part a) and b) we see that g7=6.2 and g3=-12 are the maximum and minimum values of g.

Therefore, g has a maximum value at 7 and a minimum value at 3.

#### To determine

**d)**

**To sketch:**

A rough graph of g

#### Answer

#### Explanation

From part a) we have

g0=0,g1=12,g2=0, g3=-12,g4=0. g5=32, g6=4

With the help of the above information, the graph is