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Math  /  Calculus

Question19. Find the area of the given region enclosed by the curves of f(x)=x2−x+5f(x)=x^{2}-x+5 and g(x)=x+8g(x)=x+8.

Studdy Solution

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)f(x) equal to g(x)g(x): x2−x+5=x+8x^2 - x + 5 = x + 8.

STEP 4

Now, let's **rearrange** this equation to solve for xx.
Subtract xx and 88 from both sides: x2−2x−3=0x^2 - 2x - 3 = 0.

STEP 5

This is a **quadratic equation**, and it **factors** nicely: (x−3)(x+1)=0(x-3)(x+1) = 0.
So, our intersection points are at x=3x = \mathbf{3} and x=−1x = \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=−1x = -1 to x=3x = 3.
But which function comes first?

STEP 7

Let's **test** a point between −1-1 and 33, like x=0x = 0. f(0)=5f(0) = 5 and g(0)=8g(0) = 8.
Since g(0)g(0) is greater than f(0)f(0), we know that g(x)g(x) is **above** f(x)f(x) in our region.

STEP 8

So, our **integral** is: ∫−13(g(x)−f(x)) dx=∫−13((x+8)−(x2−x+5)) dx \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: ∫−13(−x2+2x+3) dx \int_{-1}^{3} (-x^2 + 2x + 3) \, dx This represents the area between the curves!

STEP 10

Now, let's **integrate** term by term: ∫−13(−x2+2x+3) dx=[−13x3+x2+3x]−13 \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=3x = 3: −13(3)3+(3)2+3(3)=−9+9+9=9 -\frac{1}{3}(3)^3 + (3)^2 + 3(3) = -9 + 9 + 9 = \mathbf{9}

STEP 12

**Evaluate** at the lower limit, x=−1x = -1: −13(−1)3+(−1)2+3(−1)=13+1−3=−53 -\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−(−53)=9+53=273+53=323 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{\mathbf{32}}{\mathbf{3}}

STEP 14

The area of the region enclosed by the curves is 323\frac{32}{3} square units!

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