# Find the area of the region enclosed by these graphs and the vertical lines x = 0 and x = 3

f(x)=-x^2+2x+4 g(x)=-x+4

## picture:

http://www.wolframalpha.com/input/?i=plot+y%3D+-x%5E2%2B2x%2B4+%2C+y+%3D-x%2B4

looks like they intersect at (0,4) and((3,1)

(the points satisfy both equations)

area = ∫( -x^2 + 2x + 4 - (-x+4)) dx from 0 to 3

= ∫( -x^2 + 3x ) dx from 0 to 3

= [ -x^3/3 + 3x^2/2 ] from 0 to 3

= (-9 + 27/2

= 9/2

## To find the area of the region enclosed by the graphs of f(x) = -x^2 + 2x + 4 and g(x) = -x + 4, and the vertical lines x = 0 and x = 3, we need to calculate the definite integral of the difference between the two functions over the given interval.

Step 1: Find the points of intersection

To find the points of intersection, we set f(x) equal to g(x):

-x^2 + 2x + 4 = -x + 4

Simplifying:

-x^2 + 3x = 0

Factoring out an x:

x(-x + 3) = 0

This equation has two solutions: x = 0 and x = 3.

Step 2: Determine the integrand

We need to find the difference between the two functions, f(x) and g(x):

h(x) = f(x) - g(x)

h(x) = (-x^2 + 2x + 4) - (-x + 4)

Simplifying:

h(x) = -x^2 + 3x

Step 3: Calculate the definite integral

To find the area, we calculate the definite integral of h(x) from x = 0 to x = 3:

Area = ∫[0 to 3] -x^2 + 3x dx

Step 4: Evaluate the integral

Using standard integration rules, we find:

Area = [(-1/3)x^3 + (3/2)x^2] evaluated from 0 to 3

Area = [(-1/3)(3)^3 + (3/2)(3)^2] - [(-1/3)(0)^3 + (3/2)(0)^2]

Area = (-9/3 + 9/2) - (0 + 0)

Area = (-3 + 9/2)

Area = -3/2 + 9/2

Area = 6/2

Area = 3

Therefore, the area of the region enclosed by the given graphs and the vertical lines x = 0 and x = 3 is 3 square units.

## To find the area of the region enclosed by the graphs of the given functions and the vertical lines x = 0 and x = 3, you can use definite integration.

Step 1: First, let's find the points of intersection between the two functions.

Setting the two functions equal to each other:

-f(x) = g(x)

-x^2 + 2x + 4 = -x + 4

Rearranging the equation:

-x^2 + 2x + x = 0

Combining like terms:

-x^2 + 3x = 0

Factoring out x:

x(-x + 3) = 0

Setting each factor equal to zero:

x = 0 or -x + 3 = 0

Solving the second equation for x:

-x = -3

x = 3

So the points of intersection between the two functions are x = 0 and x = 3.

Step 2: Next, determine the limits of integration for finding the area. The region we are interested in is bounded by the vertical lines x = 0 and x = 3, so the limits of integration will be from x = 0 to x = 3.

Step 3: Now, we need to set up the integral to find the area using the formula for the area between two curves:

Area = ∫(g(x) - f(x)) dx

Since the upper function is g(x) and the lower function is f(x), the integral becomes:

Area = ∫(g(x) - f(x)) dx from x = 0 to x = 3

Plugging in the functions:

Area = ∫((-x + 4) - (-x^2 + 2x + 4)) dx from x = 0 to x = 3

Simplifying the expression inside the integral:

Area = ∫(x^2 - 3x) dx from x = 0 to x = 3

Step 4: Integrate the expression with respect to x:

∫(x^2 - 3x) dx = (1/3)x^3 - (3/2)x^2

Step 5: Evaluate the integral by substituting the limits of integration:

Area = [(1/3)(3)^3 - (3/2)(3)^2] - [(1/3)(0)^3 - (3/2)(0)^2]

Simplifying the expressions:

Area = [27/3 - 27/2] - [0 - 0]

Area = [9 - 13.5]

Area = -4.5

The area of the region enclosed by the graphs of the functions f(x) = -x^2 + 2x + 4 and g(x) = -x + 4, and the vertical lines x = 0 and x = 3 is -4.5 square units.