To find the area of the shaded region between the curves $f(x) = x^4 - 8x^3 + 24x^2$ and $g(x) = 5x + 38$ over the interval $[-1, 2]$, follow these steps:

1. Find the Points of Intersection

The curves intersect at the given points $(-1, 33)$ and $(2, 48)$. This means that the area we want to compute is the region bounded by $f(x)$ and $g(x)$ from $x = -1$ to $x = 2$.

2. Set Up the Integral for the Area

The area between two curves, $g(x)$ and $f(x)$, over an interval $[a, b]$ is given by:

$\text{Area} = \int_{a}^{b} \left[ g(x) - f(x) \right] \, dx$

where $g(x)$ is the upper curve and $f(x)$ is the lower curve. Here:

• $g(x) = 5x + 38$
• $f(x) = x^4 - 8x^3 + 24x^2$
• Interval: $a = -1$ and $b = 2$

3. Compute $g(x) - f(x)$

Let's find the difference between $g(x)$ and $f(x)$:

$g(x) - f(x) = (5x + 38) - (x^4 - 8x^3 + 24x^2) = -x^4 + 8x^3 - 24x^2 + 5x + 38$

4. Set Up the Integral

Now, set up the definite integral from $-1$ to $2$:

$\text{Area} = \int_{-1}^{2} \left( -x^4 + 8x^3 - 24x^2 + 5x + 38 \right) \, dx$

5. Evaluate the Integral

To find the area, we integrate each term separately:

$\int_{-1}^{2} \left( -x^4 + 8x^3 - 24x^2 + 5x + 38 \right) \, dx$

Let's compute this step-by-step:

1. Integral of $-x^4$: $\int -x^4 \, dx = -\frac{x^5}{5}$

2. Integral of $8x^3$: $\int 8x^3 \, dx = 2x^4$

3. Integral of $-24x^2$: $\int -24x^2 \, dx = -8x^3$

4. Integral of $5x$: $\int 5x \, dx = \frac{5x^2}{2}$

5. Integral of $38$: $\int 38 \, dx = 38x$

Now, substitute the limits $[-1, 2]$:

$\left[ -\frac{x^5}{5} + 2x^4 - 8x^3 + \frac{5x^2}{2} + 38x \right]_{-1}^{2}$

6. Compute the Definite Integral

Evaluate at $x = 2$:

$-\frac{2^5}{5} + 2 \times 2^4 - 8 \times 2^3 + \frac{5 \times 2^2}{2} + 38 \times 2$

$= -\frac{32}{5} + 32 - 64 + 10 + 76 = \frac{150}{5} = 30$

Evaluate at $x = -1$:

$-\frac{(-1)^5}{5} + 2 \times (-1)^4 - 8 \times (-1)^3 + \frac{5 \times (-1)^2}{2} + 38 \times (-1)$

$= \frac{1}{5} + 2 + 8 + \frac{5}{2} - 38 = -\frac{121}{5} = -24.2$

Now compute the difference:

$30 - (-24.2) = 54.2$

Thus, the area of the shaded region is:

$\boxed{54.2} \text{ square units.}$

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Do you want more details or have any questions? Here are some related problems you might find useful:

1. What is the area between two curves over a different interval?
2. How to find points of intersection between curves analytically?
3. What happens if $f(x)$ and $g(x)$ swap their positions on the graph?
4. How does changing coefficients in $f(x)$ and $g(x)$ alter the area?
5. What is the geometric interpretation of integrals for areas?

Tip: Always confirm the order of $f(x)$ and $g(x)$ on the graph to correctly identify the upper and lower functions.