Current objective

Find the area of a region bounded above by two different functions

Question
Determine the area, in square units, bounded above by  
f
(
x
)
=
x
2
+
10
x
+
25
  and  
g
(
x
)
=
−
2
x
−
2
  and bounded below by the  
x
 -axis over the interval  
[
−
5
,
−
1
]
 .
Give an exact fraction, if necessary, for your answer and do not include units.

To find the area of the region bounded by the functions \( f(x) = x^2 + 10x + 25 \) and \( g(x) = -2x - 2 \), and the \( x \)-axis over the interval \([-5, -1]\), we need to break the problem down into steps.

### Step 1: Set up the integral for the area

We will compute the area as the difference between the upper curve and the lower curve over the given interval. The area bounded between two functions \( f(x) \) and \( g(x) \) is given by:

\[
A = \int_{a}^{b} (f(x) - g(x)) \, dx
\]

where \( f(x) \) is the function on top, \( g(x) \) is the function below, and \( [a, b] \) is the interval. Here:

- \( f(x) = x^2 + 10x + 25 \)
- \( g(x) = -2x - 2 \)
- Interval: \([-5, -1]\)

Thus, the area \( A \) can be written as:

\[
A = \int_{-5}^{-1} \left( (x^2 + 10x + 25) - (-2x - 2) \right) \, dx
\]

Simplifying the expression inside the integral:

\[
A = \int_{-5}^{-1} \left( x^2 + 10x + 25 + 2x + 2 \right) \, dx
\]
\[
A = \int_{-5}^{-1} \left( x^2 + 12x + 27 \right) \, dx
\]

### Step 2: Compute the integral

Now, we integrate each term:

\[
A = \int_{-5}^{-1} x^2 \, dx + \int_{-5}^{-1} 12x \, dx + \int_{-5}^{-1} 27 \, dx
\]

- The integral of \( x^2 \) is \( \frac{x^3}{3} \)
- The integral of \( 12x \) is \( 6x^2 \)
- The integral of \( 27 \) is \( 27x \)

Thus, we have:

\[
A = \left[ \frac{x^3}{3} + 6x^2 + 27x \right]_{-5}^{-1}
\]

### Step 3: Evaluate the definite integral

Now, substitute \( x = -1 \) and \( x = -5 \) into the expression:

For \( x = -1 \):

\[
\frac{(-1)^3}{3} + 6(-1)^2 + 27(-1) = \frac{-1}{3} + 6 - 27 = \frac{-1}{3} - 21 = \frac{-64}{3}
\]

For \( x = -5 \):

\[
\frac{(-5)^3}{3} + 6(-5)^2 + 27(-5) = \frac{-125}{3} + 6(25) - 135 = \frac{-125}{3} + 150 - 135 = \frac{-125}{3} + 15 = \frac{-80}{3}
\]

Now subtract the two results:

\[
A = \left( \frac{-64}{3} \right) - \left( \frac{-80}{3} \right) = \frac{-64 + 80}{3} = \frac{16}{3}
\]

### Final Answer:
The area bounded by the curves and the \( x \)-axis is \( \frac{16}{3} \) square units.

Would you like further details on this solution or have any questions?

Here are 5 related questions:
1. How do you determine which function is the upper and which is the lower curve?
2. Can you find the area between two functions if the curves intersect within the interval?
3. What if the region is unbounded? How does that affect the area calculation?
4. How do you set up an integral if one of the bounds is infinite?
5. How do we handle regions where one of the curves crosses the x-axis?

**Tip**: Always check whether the curves cross each other within the interval before setting up the integral to avoid errors.

Determine the area, in square units, bounded above by f(x) = x^2 + 10x + 25 and g(x) = -2x - 2, and bounded below by the x-axis over the interval [-5, -1]. Give an exact fraction, if necessary, for your answer and do not include units.

Find the exact area bounded between the quadratic function f(x) = x^2 + 10x + 25, the linear function g(x) = -2x - 2, and the x-axis over the interval [-5, -1]. This detailed solution uses integral calculus, showing step-by-step integration to calculate the area as 16/3 square units.

Math Problem Statement

Question Determine the area, in square units, bounded above by
f ( x )

x 2 + 10 x + 25 and
g ( x )

Solution

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