How do you find the area bounded by #y=6x-x^2# and #y=x^2-2x#?

Question

Christopher Agresta · Accepted Answer

Use #A = int_a^b(y_1(x)-y_2(x))dx# where #y_1(x) >= y_2(x)#

Find the x coordinates of endpoints of the area.

#6x - x^2 = x^2- 2x#

#0 = 2x^2-8x#

#x = 0 and x = 4#

This means that:

#a = 0 and b = 4#

Evaluate both at 2 and observe which is greater:

#y = 6(2)-(2)^2 = 8#

#y = 2^2 - 2(2) = 0#

The first one is greater so we subtract the second from the first in the integral:

#int_0^4(6x-x^2) - (x^2 - 2x)dx =#

#int_0^4(8x-2x^2)dx =#

#{:4x^2-2/3x^3|_0^4 =#

#4(4)^2-2/3(4)^3- 4(0)^2 + 2/3(0)^2 = 64/3#

Adam Adkins · Answer

undefined To find the area bounded by the two curves $ y = 6x - x^2 $ and $ y = x^2 - 2x $, you need to determine the points of intersection between the two curves and then integrate the difference between the curves over the interval of intersection.

First, set the two equations equal to each other to find the points of intersection:

$$ 6x - x^2 = x^2 - 2x $$

Combining like terms:

$$ 2x^2 - 4x = 0 $$

Factoring out 2x:

$$ 2x(x - 2) = 0 $$

This yields two potential solutions:

$$ x = 0 $$ and $$ x = 2 $$

Next, you need to determine which curve is above the other in the interval [0, 2]. To do this, subtract the equations:

$$ (6x - x^2) - (x^2 - 2x) $$

Simplify:

$$ 6x - x^2 - x^2 + 2x $$
$$ = 4x - 2x^2 $$

Now, integrate this expression over the interval [0, 2]:

$$ \int_{0}^{2} (4x - 2x^2) \, dx $$

Integrating term by term:

$$ \int_{0}^{2} 4x \, dx - \int_{0}^{2} 2x^2 \, dx $$
$$ = 2x^2 \Big|_0^2 - \frac{2}{3}x^3 \Big|_0^2 $$

Evaluate at the limits:

$$ = 2(2^2) - 2(0^2) - \frac{2}{3}(2^3) + \frac{2}{3}(0^3) $$
$$ = 8 - \frac{16}{3} $$
$$ = \frac{24}{3} - \frac{16}{3} $$
$$ = \frac{8}{3} $$

So, the area bounded by the two curves $ y = 6x - x^2 $ and $ y = x^2 - 2x $ over the interval [0, 2] is $ \frac{8}{3} $ square units.