Question content area top
Part 1
Find the area between the curves.
xequalsnegative 2​, xequals1​, yequals11​x, yequalsx squared minus 12
Question content area bottom
Part 1
The area between the curves is 
  
enter your response here.
​(Simplify your​ answer.)

To find the area between the curves \( y = 11x \) and \( y = x^2 - 12 \) over the interval \( x = -2 \) to \( x = 1 \), we proceed as follows:

### Step 1: Set up the integral
The area between two curves, \( y = f(x) \) and \( y = g(x) \), from \( x = a \) to \( x = b \) is given by:
\[
\text{Area} = \int_{a}^{b} |f(x) - g(x)| \, dx
\]

In this case:
- \( f(x) = 11x \)
- \( g(x) = x^2 - 12 \)
- \( a = -2 \)
- \( b = 1 \)

### Step 2: Determine which function is on top
To find the top function over the interval, evaluate both functions at a few points or set them equal to each other to see where they intersect:
\[
11x = x^2 - 12
\]
Rearrange to form a quadratic equation:
\[
x^2 - 11x - 12 = 0
\]
Solve this equation using the quadratic formula:
\[
x = \frac{11 \pm \sqrt{121 + 48}}{2} = \frac{11 \pm \sqrt{169}}{2} = \frac{11 \pm 13}{2}
\]
This gives:
\[
x = 12 \quad \text{and} \quad x = -1
\]
Since we're only concerned with \( x = -2 \) to \( x = 1 \), observe that \( y = 11x \) is above \( y = x^2 - 12 \) on this interval.

### Step 3: Set up the integral
The area between the curves from \( x = -2 \) to \( x = 1 \) is:
\[
\text{Area} = \int_{-2}^{1} \left(11x - (x^2 - 12)\right) \, dx
\]
Simplify the integrand:
\[
\text{Area} = \int_{-2}^{1} (11x - x^2 + 12) \, dx
\]

### Step 4: Integrate
Now, integrate term-by-term:
\[
\text{Area} = \int_{-2}^{1} (11x - x^2 + 12) \, dx = \left[ \frac{11x^2}{2} - \frac{x^3}{3} + 12x \right]_{-2}^{1}
\]

### Step 5: Evaluate the integral
Substitute the limits of integration:

1. **At \( x = 1 \):**
   \[
   \frac{11 \cdot 1^2}{2} - \frac{1^3}{3} + 12 \cdot 1 = \frac{11}{2} - \frac{1}{3} + 12 = \frac{33 + 22 - 2}{6} = \36

Find the area between the curves x = -2, x = 1, y = 11x, and y = x^2 - 12.

Learn how to find the area between the curves y = 11x and y = x^2 - 12 over the interval from x = -2 to x = 1. This calculus problem involves setting up and evaluating a definite integral, applying the quadratic formula, and understanding the Fundamental Theorem of Calculus. Suitable for students in Grades 11-12.

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