To solve this problem, we need to:

1. Sketch the region bounded by the graphs of the functions $f(x) = x^5 + 5$ and $g(x) = x + 5$.
2. Find the area of the region between these curves.

Step 1: Analyze the functions

• $f(x) = x^5 + 5$ is a polynomial function with a degree of 5. It's a curve that will have a steep upward and downward slope as $x$ moves away from zero.
• $g(x) = x + 5$ is a straight line with slope 1 and a y-intercept of 5.

Step 2: Find the intersection points

We find the points where $f(x) = g(x)$, which gives us: $x^5 + 5 = x + 5$ Simplifying this: $x^5 = x$ Factoring: $x(x^4 - 1) = 0$ $x(x^2 - 1)(x^2 + 1) = 0$ This gives: $x = 0, \quad x = 1, \quad x = -1$ So the curves intersect at $x = -1, 0,$ and $1$.

Step 3: Set up the area integral

The area between the two curves is given by the integral of the difference $g(x) - f(x)$ between the intersection points $x = -1$ and $x = 1$: $\text{Area} = \int_{-1}^{1} \left[ (x + 5) - (x^5 + 5) \right] \, dx$ Simplifying the integrand: $\text{Area} = \int_{-1}^{1} \left[ x - x^5 \right] \, dx$

Step 4: Compute the integral

We split the integral: $\text{Area} = \int_{-1}^{1} x \, dx - \int_{-1}^{1} x^5 \, dx$ Both integrals are over symmetric limits, and $x^5$ and $x$ are odd functions. The integral of any odd function over a symmetric interval $[-a, a]$ is zero. Therefore: $\int_{-1}^{1} x \, dx = 0 \quad \text{and} \quad \int_{-1}^{1} x^5 \, dx = 0$ Thus, the total area is: $\text{Area} = 0 - 0 = 0$ There is no enclosed area between the two curves from $x = -1$ to $x = 1$, as they cancel each other out symmetrically.

Conclusion:

The area between the curves is zero.

Would you like a detailed breakdown of any specific step or a visualization of the region?

5 Related Questions:

1. How do you calculate the area when the curves are not symmetric?
2. How do you approach areas where the curves cross multiple times?
3. What does it mean if the area calculation gives a negative result?
4. How would you compute the area if $f(x)$ was a different higher-degree polynomial?
5. How does symmetry help simplify definite integrals?

Tip:

Whenever you're dealing with odd functions over symmetric intervals, always check if the integral evaluates to zero to save time!