Find the area enclosed by the given curves.
-$$y = 2 x - x ^ { 2 } , y = 2 x - 4$$
A) $\frac { 34 } { 3 }$
B) $\frac { 32 } { 3 }$
C) $\frac { 31 } { 3 }$
D) $\frac { 37 } { 3 }$

Question

Quizplus · Accepted Answer

To find the area enclosed by the curves $y = 2x - x^2$ and $y = 2x - 4$, we first need to find the points of intersection by setting the equations equal to each other:$$2x - x^2 = 2x - 4$$Simplifying this, we get:$$x^2 = 4$$$$x = \pm 2$$So, the points of intersection are at $x = -2$ and $x = 2$.The area enclosed by the curves can be found by integrating the difference between the two functions from $x = -2$ to $x = 2$:$$\int_{-2}^{2} [(2x - x^2) - (2x - 4)] dx$$$$= \int_{-2}^{2} (-x^2 + 4) dx$$$$= \left[ -\frac{1}{3}x^3 + 4x \right]_{-2}^{2}$$$$= \left( -\frac{1}{3}(2)^3 + 4(2) \right) - \left( -\frac{1}{3}(-2)^3 + 4(-2) \right)$$$$= \left( -\frac{8}{3} + 8 \right) - \left( \frac{8}{3} - 8 \right)$$$$= -\frac{8}{3} + 8 + \frac{8}{3} - 8$$$$= \frac{16}{3}$$However, since we need to find the total area enclosed, and the integral gives us the net area (considering the direction of the curve), we actually find that the calculation above does not directly match any of the options provided, indicating a mistake in the calculation process. Let's correct the calculation:The correct integral setup for the area between the curves from $x = -2$ to $x = 2$ is:$$A = \int_{-2}^{2} [(2x - x^2) - (2x - 4)] dx$$$$= \int_{-2}^{2} (4 - x^2) dx$$$$= \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2}$$$$= \left( 4(2) - \frac{(2)^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right)$$$$= \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right)$$$$= 8 - \frac{8}{3} + 8 - \frac{8}{3}$$$$= 16 - \frac{16}{3}$$$$= \frac{48}{3} - \frac{16}{3}$$$$= \frac{32}{3}$$Thus, the correct answer is $\frac{32}{3}$, which corresponds to choice B.

Find the Area Enclosed by the Given Curves y=2x−x2,y=2x−4y = 2 x - x ^ { 2 } , y = 2 x - 4y=2x−x2,y=2x−4

Find the Area Enclosed by the Given Curves $y = 2 x - x ^ { 2 } , y = 2 x - 4$