RaySix

Post

Finding Area Between Two Curves

Created by @nathanedwards

at November 4th 2023, 8:49:36 pm.

AP_Calculus_AB-Questions

#calculus

#area

#integration

Question:

Consider the curves defined by the functions f(x) = x^2 - 4x + 3 and g(x) = 4 - x. Find the area of the region bounded by the curves f(x) and g(x).

Answer:

To find the area between two curves, we need to determine the x-values at which the curves intersect and integrate the difference of the functions over the interval.

First, let's find the x-values where the curves f(x) and g(x) intersect by setting the two functions equal to each other:

x^2 - 4x + 3 = 4 - x

Rearranging the equation:

x^2 - 3x - 1 = 0

Factoring the quadratic equation:

(x - 1)(x - 3) = 0

Setting each factor equal to zero:

x - 1 = 0 => x = 1 x - 3 = 0 => x = 3

We have found that the curves f(x) and g(x) intersect at x = 1 and x = 3.

Next, we need to determine the upper and lower functions in the interval of interest, [1, 3]. Looking at the functions f(x) = x^2 - 4x + 3 and g(x) = 4 - x, we can see that g(x) is the upper function and f(x) is the lower function over the interval [1, 3].

The area between the curves can be calculated using the following integral:

A = \int_{a}^{b} (g(x) - f(x)) \, dx

Where a = 1 and b = 3.

Substituting the functions into the integral:

A = \int_{1}^{3} [(4 - x) - (x^2 - 4x + 3)] \, dx

Simplifying the integrand:

A = \int_{1}^{3} (-x^2 + 8x - 1) \, dx

Using the power rule for integration:

A = \left[-\frac{x^3}{3} + 4x^2 - x\right]_{1}^{3}

Evaluating the definite integral:

A = \left[-\frac{3^3}{3} + 4(3^2) - 3\right] - \left[-\frac{1^3}{3} + 4(1^2) - 1\right]

A = \left[-9 + 36 - 3\right] - \left[-\frac{1}{3} + 4 - 1\right]

A = 24 - (-\frac{1}{3} + 3)

A = 24 - (2\frac{2}{3})

A = 24 - \frac{8}{3}

A = \frac{64}{3}

Therefore, the area between the curves f(x) and g(x) over the interval [1, 3] is equal to $\frac{64}{3}$ square units.