help? area? points? integral?

Question

Answer 1

(2.a)

Step 1:

The function is  on the interval .

Number of rectangles are .

The sum of all circumscribed rectangle is upper sum.

, where .

Width .

Find upper Sum.

Right end points : .

Area of higher sum is 

Step 2:

Apply summation formula  .

Apply summation formula  .

Apply summation formula .

Here .

The area of higher sum is .

Solution :

The area of the region is 2.37 sq-units.

Answer 2

(2.b)

Step 1:

The function is  on the interval is .

Number of rectangles are .

The sum of all inscribed rectangle is lower sum.

, where .

Width .

Find lower Sum.

Left end points : .

Area of lower sum is .

Apply summation formula  .

Apply summation formula  .

Apply summation formula .

Here .

The area of lower sum is .

Solution :

The area of the region is 10.37 sq-units.

Answer 3

(2.c)

Step 1:

The function is  on the interval is .

Number of rectangles are .

Using mid point theorem:

The area is .

Consider .

Where , and .

Width .

.

Mid points :

Substitute i  values from 1 to 4.

Step 2:

Area is

The area of the region is 6.815 sq-units.

Solution :

The area of the region is 6.815 sq-units.

Answer 4

(1.a)

Step 1:

The function is  on interval .

Number of rectangles are .

The sum of all circumscribed rectangle is upper sum.

, where  is the right end point.

Width .

Right end point:

.

Step 2:

Find the upper sum.

Area of upper sum is .

Apply summation formulae: , , and

.

In this case .

sq units.

Area of upper sum is sq units.

Solution:

Area of region is sq units.

Answer 5

(1.b)

Step 1:

The function is  on interval .

Number of rectangles are .

The sum of all inscribed rectangle is lower sum.

, where  is left end point.

Width .

Left end point:

.

Step 2:

Find lower Sum.

Lower sum is .

Apply summation formulae: , , and

.

In this case .

sq units.

Solution:

The area of the region is sq units.

Answer 6

(1.c)

Step 1:

The function is  on interval .

Number of rectangles are .

Mid point theorem:

The area is .

Consider .

Where and .

Substitute in .

Width .

Substitute in .

.

.

Step 2:

Find values.

.

Substitute in .

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Step 3:

Using mid point theorem:

Area =

sq-units.

Solution:

Area of the region is sq units.