Math Snap

STEP 1

Assumptions
1. The function $f(x) = x^2 - 3x + 3$ represents a parabola.
2. The function $g(x) = x$ represents a straight line.
3. We need to find the points of intersection of these two functions.
4. We need to calculate the area of the region between the two curves over the interval defined by their points of intersection.

STEP 2

To find the points of intersection, set $f(x) = g(x)$.
$$x^2 - 3x + 3 = x$$

STEP 3

Rearrange the equation to form a quadratic equation.
$$x^2 - 3x + 3 - x = 0$$ $$x^2 - 4x + 3 = 0$$

STEP 4

Solve the quadratic equation $x^2 - 4x + 3 = 0$ using the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = 1$, $b = -4$, and $c = 3$.

STEP 5

Calculate the discriminant $b^2 - 4ac$.
$$(-4)^2 - 4 \times 1 \times 3 = 16 - 12 = 4$$

STEP 6

Substitute the values into the quadratic formula.
$$x = \frac{-(-4) \pm \sqrt{4}}{2 \times 1}$$ $$x = \frac{4 \pm 2}{2}$$

STEP 7

Calculate the two possible solutions for $x$.
$$x_1 = \frac{4 + 2}{2} = 3$$ $$x_2 = \frac{4 - 2}{2} = 1$$

STEP 8

The points of intersection are when $x = 1$ and $x = 3$. Find the corresponding $y$-values using $g(x) = x$.
For $x = 1$, $y = 1$.
For $x = 3$, $y = 3$.
Thus, the points of intersection are $(1, 1)$ and $(3, 3)$.

STEP 9

To find the area between the curves, integrate the difference $f(x) - g(x)$ from $x = 1$ to $x = 3$.
$$\text{Area} = \int_{1}^{3} (f(x) - g(x)) \, dx$$ $$\text{Area} = \int_{1}^{3} (x^2 - 3x + 3 - x) \, dx$$ $$\text{Area} = \int_{1}^{3} (x^2 - 4x + 3) \, dx$$

STEP 10

Integrate the function $x^2 - 4x + 3$.
$$\int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x + C$$

STEP 11

Evaluate the definite integral from $x = 1$ to $x = 3$.
$$\left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{1}^{3}$$

STEP 12

Calculate the value at $x = 3$.
$$\frac{3^3}{3} - 2(3)^2 + 3(3) = \frac{27}{3} - 18 + 9 = 9 - 18 + 9 = 0$$

STEP 13

Calculate the value at $x = 1$.
$$\frac{1^3}{3} - 2(1)^2 + 3(1) = \frac{1}{3} - 2 + 3 = \frac{1}{3} + 1 = \frac{4}{3}$$

STEP 14

Subtract the result at $x = 1$ from the result at $x = 3$.
$$0 - \frac{4}{3} = -\frac{4}{3}$$

Since the area cannot be negative, take the absolute value.
$$\text{Area} = \frac{4}{3}$$ The area of the region between the two curves is $\frac{4}{3}$ square units.