# Determine the area of the region bounded by y=4x+3, y=6-x-2x^2, x=-4, and x=2.

## Expert Answers

To determine the bounded region, graph the given equations. See the attached image for the graph I've created.

Base on the graph, there are three bounded regions.

To get the area of each bounded region, draw a vertical strip on each region. The vertical strips each have a width of dx. And the height of each vertical strip is `y_(upper)-y_(lower)` .

For the bounded region at the left, its height is:

`h=y_(upper) - y_(lower)`

`h = 4x +3 - (6 -x - 2x^2)`

`h = 2x^2+ 5x - 3`

So the expression that represents the area of this vertical strip is:

`A = height * width`

` ` `A = (2x^2+5x- 3)dx`

The bounded region at the left starts from x = -4 and ends at x=-3. So, the integral needed to get the area of this bounded region is:

`A = int_-4^-3 (2x^2+5x - 3)dx `

Evaluating this integral, the result will be:

`A = (2x^3)/3 + (5x^2)/2-3x|_-4^-3 `

`A = [(2*(-3)^3)/3 + (5*(-3)^2)/2-3(-3)] - [(2*(-4)^3)/3+(5*(-4)^2)/2-3(-4)]`

`A=25/6`

Hence, the area of the bounded region at the left is 25/6 square units.

For the bounded region at the middle, the height of the vertical strip is:

`h = y_(upper) - y_(lower)`

`h = 6 - x - 2x^2-(4x + 3)`

`h = -2x^2-5x + 3`

So the expression that represents the area of this vertical strip is:

`A = height * width`

`A = (-2x^2-5x + 3)dx`

The bounded region at the middle starts from x=-3 and ends at x = 1/2. So, the integral needed to compute its area is:

`A = int_-3^(1/2) (-2x^2-5x + 3)dx `

Evaluating this integral, the result will be:

`A = (-2x^3)/3 - (5x^2)/2+3x |_-3^(1/2`

`A= [(-2(1/2)^3)/3 - (5(1/2)^2)/2 + 3(1/2)] - [(-5(-3)^3)/3-(5(-3)^2)/2+3(-3)]`

`A=343/24`

Hence, the area of the bounded region in the middle is 343/24 square units.

For the bounded region at the right, the height of the vertical strip is:

`h=y_(upper) - y_(lower)`

`h=4x + 3 - (6-x-2x^2)`

`h=2x^2+5x - 3`

So, the area of the vertical strip is:

`A = height * width`

`A = (2x^2+5x - 3)dx`

The bounded region at the right starts from x = 1/2 and ends at x = 2. So, the integral needed to compute its area is:

`A=int_(1/2)^2 (2x^2+5x-3)dx`

Evaluating this integral, we will get:

`A = (2x^3)/3 + (5x^2)/2 - 3x |_(1/2)^2`

`A = [(2(2)^3)/3 + (5(2)^2)/2 - 3(2)] - [(2(1/2)^3)/3+(5(1/2)^2)/2-3(1/2)]`

`A=81/8`

Hence, the are of the bounded region at the right is 81/8 square units.

From here, add the area of the three bounded regions.

`A=25/6 + 343/24 + 81/8`

`A=343/12`

Therefore, the total area of the bounded region of the given equations is `343/12` square units.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

• 30,000+ book summaries
• 20% study tools discount
• Ad-free content
• PDF downloads
• 300,000+ answers
• 5-star customer support