Math Snap

STEP 1

What is this asking?
We need to find the area trapped between two curves, a parabola and a line!
Watch out!
Make sure to find where the curves intersect; that's super important for setting up the area calculation.
Also, remember which function is on top!

STEP 2

1. Find Intersection Points
2. Set Up the Integral
3. Evaluate the Integral

STEP 3

Let's find where these two curves meet!
We do that by setting $f(x)$ equal to $g(x)$: $x^2 - x + 5 = x + 8$.

STEP 4

Now, let's rearrange this equation to solve for $x$.
Subtract $x$ and $8$ from both sides: $x^2 - 2x - 3 = 0$.

STEP 5

This is a quadratic equation, and it factors nicely: $(x-3)(x+1) = 0$.
So, our intersection points are at $x = \mathbf{3}$ and $x = \mathbf{-1}$.
These are the boundaries of our region!

STEP 6

To find the area between the curves, we need to integrate the difference of the functions from $x = -1$ to $x = 3$.
But which function comes first?

STEP 7

Let's test a point between $-1$ and $3$, like $x = 0$. $f(0) = 5$ and $g(0) = 8$.
Since $g(0)$ is greater than $f(0)$, we know that $g(x)$ is above $f(x)$ in our region.

STEP 8

So, our integral is:
$$\int_{-1}^{3} (g(x) - f(x)) \, dx = \int_{-1}^{3} ((x+8) - (x^2 - x + 5)) \, dx$$

STEP 9

Let's simplify the integrand:
$$\int_{-1}^{3} (-x^2 + 2x + 3) \, dx$$ This represents the area between the curves!

STEP 10

Now, let's integrate term by term:
$$\int_{-1}^{3} (-x^2 + 2x + 3) \, dx = \left[ -\frac{1}{3}x^3 + x^2 + 3x \right]_{-1}^{3}$$

STEP 11

Evaluate at the upper limit, $x = 3$:
$$-\frac{1}{3}(3)^3 + (3)^2 + 3(3) = -9 + 9 + 9 = \mathbf{9}$$

STEP 12

Evaluate at the lower limit, $x = -1$:
$$-\frac{1}{3}(-1)^3 + (-1)^2 + 3(-1) = \frac{1}{3} + 1 - 3 = -\frac{\mathbf{5}}{\mathbf{3}}$$

STEP 13

Subtract the lower limit value from the upper limit value:
$$9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{\mathbf{32}}{\mathbf{3}}$$

The area of the region enclosed by the curves is $\frac{32}{3}$ square units!