What is the area under the curve of #x - x^2# in the positive #(x,y)# region?

Question

Cameron Alexander · Accepted Answer

When you define an integral like this, you simply subtract the equation that is higher from the equation that is lower. We just have to find the intersections with the x-axis. Solving for the intersections, #x = x^2#
#x = 1, 0# Thus, #int_0^1 (x-x^2) - 0dx# #= {:[x^2/2 - x^3/3]|:}_(0)^(1)# #= [1^2/2 - 1^3/3] - [0]# #= [3/6 - 2/6]# #= color(blue)(1/6)# graph{(y - x + x^2)(y)sqrt(0.25 - (x - 0.5)^2) <= 0 [-2, 2, -2, 2]}

Evelyn Agard · Answer

undefined To find the area under the curve of $ x - x^2 $ in the positive $ (x, y) $ region, we need to integrate the function $ x - x^2 $ over the appropriate interval.

The function $ x - x^2 $ intersects the x-axis at $ x = 0 $ and $ x = 1 $, forming a parabolic curve in the positive $ x $-region.

To find the area under the curve in this region, we integrate the function from $ x = 0 $ to $ x = 1 $:

$$ \int_{0}^{1} (x - x^2) \, dx $$

Integrating term by term:

$$ \int_{0}^{1} x \, dx - \int_{0}^{1} x^2 \, dx $$

Using the power rule of integration:

$$ \left[ \frac{x^2}{2} \right]_{0}^{1} - \left[ \frac{x^3}{3} \right]_{0}^{1} $$

$$ \left( \frac{1^2}{2} \right) - \left( \frac{1^3}{3} \right) - \left( \frac{0^2}{2} \right) + \left( \frac{0^3}{3} \right) $$

$$ \left( \frac{1}{2} \right) - \left( \frac{1}{3} \right) - 0 + 0 $$

$$ \frac{1}{2} - \frac{1}{3} $$

$$ \frac{3}{6} - \frac{2}{6} $$

$$ \frac{1}{6} $$

Therefore, the area under the curve of $ x - x^2 $ in the positive $ (x, y) $ region is $ \frac{1}{6} $ square units.