RaySix

Post

Intersections and Enclosed Area of Curves

Created by @nathanedwards

at November 1st 2023, 6:58:42 pm.

AP_Calculus_AB-Questions

#calculus

#area

#intersection

AP Calculus AB Exam Question:

Consider the curves defined by the equations:

Curve A: $y = x^2 - 4x + 3$
Curve B: $y = x^3 - 3x^2 + 2x$

(a) Find the $x$ -coordinates of the points where the curves A and B intersect.

(b) Find the area of the region enclosed between curves A and B.

Answer:

(a) To find the $x$ -coordinates of the points where curves A and B intersect, we need to set their respective equations equal to each other:

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

Rearranging the equation, we get:

$x^3 - 4x^2 + 5x - 3 = 0$

Unfortunately, this cubic equation cannot be solved easily by factoring. Therefore, we will employ numerical methods to approximate the roots. Using a graphing calculator or any numerical solver tool, we find that the roots of the equation are approximately:

$x_1 \approx 0.865$ , $x_2 \approx 1.318$ , and $x_3 \approx 3.817$ .

Therefore, the $x$ -coordinates of the points of intersection between curves A and B are $x = 0.865$ , $x = 1.318$ , and $x = 3.817$ .

(b) To find the area of the region enclosed between the curves A and B, we need to determine which curve is on top and which is on the bottom over the interval where they intersect. By comparing the curves' equations, we can see that curve B is always greater than curve A. However, we still need the exact interval on the $x$ -axis where the two curves intersect.

To find the interval, we need to determine the bounds for integration. We already have the $x$ -coordinates of the points of intersection, but we also need to find the $y$ -coordinates for those points. By substituting these $x$ -values into either of the original equations, we can find the corresponding $y$ values.

For $x = 0.865$ : $y = (0.865)^2 - 4(0.865) + 3 = 0.456$

For $x = 1.318$ : $y = (1.318)^2 - 4(1.318) + 3 = -0.142$

For $x = 3.817$ : $y = (3.817)^2 - 4(3.817) + 3 = 1.138$

Now, we can set up the definite integral to find the area:

\text{{Area}} = \int_{0.865}^{1.318} (x^3 - 3x^2 + 2x) - (x^2 - 4x + 3) \, dx

Simplifying the integrand, we obtain:

\text{{Area}} = \int_{0.865}^{1.318} x^3 - 3x^2 + 2x - x^2 + 4x - 3 \, dx

Combining like terms and integrating, we have:

\text{{Area}} = \int_{0.865}^{1.318} x^3 - 4x^2 + 6x - 3 \, dx

Integrating term by term, we get:

\text{{Area}} = \left[\frac{1}{4}x^4 - \frac{4}{3}x^3 + 3x^2 - 3x \right]_{0.865}^{1.318}

Evaluating the definite integral, we obtain:

\text{{Area}} = \left(\frac{1}{4}(1.318)^4 - \frac{4}{3}(1.318)^3 + 3(1.318)^2 - 3(1.318) \right) - \left(\frac{1}{4}(0.865)^4 - \frac{4}{3}(0.865)^3 + 3(0.865)^2 - 3(0.865)\right)

\text{{Area}} \approx 1.828

Therefore, the area of the region enclosed between curves A and B is approximately 1.828 square units.