The problem asks us to find the area enclosed by the curves $y = -x^2 + 1$ and $y = x^2 - 2x + 1$. The handwritten expression suggests the process of subtracting the equations.

1. Problem Description

The problem asks us to find the area enclosed by the curves $y = -x^2 + 1$ and $y = x^2 - 2x + 1$. The handwritten expression suggests the process of subtracting the equations.

2. Solution Steps

First, we need to find the points of intersection between the two curves. To do this, we set the two equations equal to each other:

$-x^2 + 1 = x^2 - 2x + 1$

Now we solve for $x$:

$0 = 2x^2 - 2x$

$0 = 2x(x - 1)$

So, the points of intersection occur at $x = 0$ and $x = 1$.

Next, we determine which function is on top. We can test a value between $0$ and $1$, such as $x = 0.5$.

For $y = -x^2 + 1$, when $x = 0.5$, $y = -(0.5)^2 + 1 = -0.25 + 1 = 0.75$.

For $y = x^2 - 2x + 1$, when $x = 0.5$, $y = (0.5)^2 - 2(0.5) + 1 = 0.25 - 1 + 1 = 0.25$.

Since $0.75 > 0.25$, the function $y = -x^2 + 1$ is above $y = x^2 - 2x + 1$ in the interval $[0, 1]$.

Now we can set up the integral to find the area $S$:

$S = \int_{0}^{1} [(-x^2 + 1) - (x^2 - 2x + 1)] dx$

$S = \int_{0}^{1} (-x^2 + 1 - x^2 + 2x - 1) dx$

$S = \int_{0}^{1} (-2x^2 + 2x) dx$

Now we evaluate the integral:

$S = [-\frac{2}{3}x^3 + x^2]_{0}^{1}$

$S = (-\frac{2}{3}(1)^3 + (1)^2) - (-\frac{2}{3}(0)^3 + (0)^2)$

$S = -\frac{2}{3} + 1 - 0$

$S = \frac{1}{3}$

3. Final Answer

The area enclosed by the curves is $\frac{1}{3}$.