#### To determine

**a)**

**To evaluate:**

i. g0

ii. g6

#### Answer

i. 0

ii. 0

#### Explanation

**Given:**

gx=∫0xf(t)dt

**Calculation:**

i. To evaluate g0

The integral of a function over a trivial interval is zero so

g0=∫00f(t)dt=0

ii. To evaluate g6

The graph of f has rotational symmetry about the point (3, 0)

So the part from x=0 to 3 is equal in area to the part from x=3 to 6

But the area of part from x=3 to 6 has to subtracted since it is below the x-axis

So, g6=∫06f(t)dt=∫03f(t)dt-∫36f(t)dt =0

Therefore, g0=0,6=0

#### To determine

**b)**

**To estimate:**

gx for x=1, 2, 3, 4 and 5

#### Answer

g1=2.8,g2=4.7, g3=5.3,g4=4.7, g5=2.8

#### Explanation

**Concept:**

Estimate the area by counting squares and observation

**Given:**

gx=∫0xf(t)dt

**Calculation:**

To evaluate gx for x=1

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

By counting squares

g1=2+0.8

Simplifying

g1=2.8

To evaluate gx for x=2

g2=∫02f(t)dt=∫01f(t)dt+∫12f(t)dt Thus we have

g2=2.8+1.9=4.7

To evaluate g3

g3=∫03f(t)dt=g2+∫23ftdt

Thus we have

g3=4.7+0.6

g3=5.3

To evaluate g4

g4=∫04f(t)dt=g3-∫34f(t)dt

Thus we have

g4=5.3-0.6

Simplifying

g4=4.7

To evaluate g5

g5=g4-∫45f(t)dt

So

g5=4.7-1.9

Simplifying

g5=2.8

Therefore, g1=2.8,g2=4.7, g3=5.3,g4=4.7, g5=2.8

#### To determine

**c)**

**To find:**

On what interval is g increasing

#### Answer

(0, 3)

#### Explanation

More area is added to the total area as x increases from 0 to 3. And beyond 3 area shall get subtracted. So it is increasing in the interval (0,3).Alternatively, by fundamental theorem of calculus g'=f. It is positive only in [0,3] so g is increasing only (0,3).

Therefore, g is increasing on the interval (0, 3)

#### To determine

**d)**

**To find:**

Where does g have the maximum value

#### Answer

3

#### Explanation

**Concept:**

Fundamental theorem of calculus says that g'=f

**Calculation:**

By Fundamental theorem of calculus g'=f

Thus g is increasing when f is positive and g is decreasing when f is negative

So g is increasing on (0, 3) and decreasing on (3, 7)

Therefore, g reaches its maximum at x=3

#### To determine

**e)**

**To sketch:**

A rough graph of g

#### Answer

#### Explanation

From part a) we have

g0=0,g1=2.8,g2=4.7, g3=5.3,g4=4.7, g5=2.8,g6=0

With the help of the above information draw graph

#### To determine

**f)**

**To sketch:**

A rough graph of g' and compare it to f

#### Answer

The graph of g' is similar to that of f as expected from fundamental theorem of calculus.

#### Explanation

As we can see that g is maximum at x=3 and the slope of the tangent line at x=3 is 0

That is g'(3)=0

On the left hand side of x=3 the function g is increasing

Therefore g’ is positive

The rate of change of g is getting smaller from 0 to 3

Therefore g’(0) must be maximum and g’ decreases as x moves 0 to 3

On the right hand side of x=3g is decreasing

Therefore g’ is negative

The rate of change of g is getting larger from 3 to 6

Therefore g’(6) must be minimum (most negative) and g’ decreases (becomes more negative) as x moves 3 to 6

Since the graph of g is symmetric with respect to x=3,

Therefore, graph of g’ from 0 to 6 is