Find the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10, g(x) = 4x - x^2, 1 <= x <= 3.
user
Sure, I'd be happy to help you find the area between the graphs of the given functions.

To do this, you can use integration techniques, specifically the definite integral, which is used to find the accumulated area under a curve.

In this case, the area you're interested in is the area bounded by the following:

- The top curve: y = f(x) = x^2 - 4x + 10
- The bottom curve: y = g(x) = 4x - x^2
- The x-axis (which acts as the lower boundary)
- The line x = 1 (which acts as the left boundary) and x = 3 (which acts as the right boundary)

So, the area is given by:

Area = ∫[1, 3](g(x) - f(x))dx

= ∫[1, 3](4x - x^2 - (x^2 - 4x + 10))dx

= ∫[1, 3](3x - x^2 - x^2 + 4x - 10)dx

= ∫[1, 3](2x - x^2 - 10)dx

= [2x^3/3 - 3x^2 + 10x]1 to 3

= [-2/3(1)^3 - 3(1)^2 + 10(1)] + [2(3)^3/3 - 3(3)^2 + 10(3)]

= -1.667 + 29.333 - 54 + 180 = 103.666

So, the area bounded by the curves is approximately 103.67 square units.
assistant
Find the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10, g(x) = 4x - x^2, 1 <= x <= 3.
user
To find the area between the graphs of the functions f(x) and g(x), we first need to determine where they intersect. We do this by setting them equal to each other:

x^2 - 4x + 10 = 4x - x^2

Combine like terms:

2x^2 - 8x + 10 = 0

Now, divide the entire equation by 2 to simplify it:

x^2 - 4x + 5 = 0

This quadratic does not factor nicely, so we will use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -4, and c = 5. Plugging these values in, we get:

x = (4 ± √((-4)^2 - 4*1*5)) / (2*1)
x = (4 ± √(16 - 20)) / 2
x = (4 ± √(-4)) / 2

Since we have a negative number under the square root, there are no real solutions for x. This means that the graphs of f(x) and g(x) do not intersect within the given interval [1, 3].

However, we can still find the area between the graphs by integrating the difference between the functions over the interval [1, 3]. To determine which function is above the other, we can evaluate both functions at a point in the interval, such as x = 2:

f(2) = (2)^2 - 4(2) + 10 = 4 - 8 + 10 = 6
g(2) = 4(2) - (2)^2 = 8 - 4 = 4

Since f(2) > g(2), we know that f(x) is above g(x) on the interval [1, 3]. Now we can set up the integral to find the area:

Area = ∫ from 1 to 3 [f(x) - g(x)] dx
     = ∫ from 1 to 3 [(x^2 - 4x + 10) - (4x - x^2)] dx
     = ∫ from 1 to 3 (2x^2 - 8x + 10) dx

Now we integrate:

= [(2/3)x^3 - 4x^2 + 10x] evaluated from 1 to 3

Plugging in the limits of integration:

= [(2/3)(3)^3 - 4(3)^2 + 10(3)] - [(2/3)(1)^3 - 4(1)^2 + 10(1)]
= [18 - 36 + 30] - [2/3 - 4 + 10]
= [6] - [8/3]

Now we subtract and simplify:

Area = 6 - (8/3)
     = (18 - 8) / 3
     = 10 / 3

So, the area of the region between the graphs of f(x) and g(x) on the interval [1, 3] is approximately 3.33 square units.
assistant