To find the area of the shaded region under the curve given by $y = -x^2 + 16$ between $x = 4$ and $x = 5$, we need to set up an integral for the function over this interval.

Step-by-Step Solution:

1. Identify the Function and Interval:

   • The function given is $y = -x^2 + 16$.
   • The shaded area is between $x = 4$ and $x = 5$.

2. Set up the Integral:

   • The area $A$ under the curve from $x = 4$ to $x = 5$ can be found using the definite integral: $A = \int_{4}^{5} (-x^2 + 16) \, dx$

3. Calculate the Integral:

   • Find the antiderivative of $-x^2 + 16$: $\int (-x^2 + 16) \, dx = -\frac{x^3}{3} + 16x$
   • Now, evaluate this from $x = 4$ to $x = 5$: [ A = \left[-\frac{5^3}{3} + 16 \cdot 5\right] - \left[-\frac{4^3}{3} + 16 \cdot 4\right] ]

4. Simplify Each Part:

   • For $x = 5$: $-\frac{5^3}{3} + 16 \cdot 5 = -\frac{125}{3} + 80 = \frac{-125 + 240}{3} = \frac{115}{3}$
   • For $x = 4$: $-\frac{4^3}{3} + 16 \cdot 4 = -\frac{64}{3} + 64 = \frac{-64 + 192}{3} = \frac{128}{3}$

5. Calculate the Difference:

   • Now, subtract the two values: $A = \frac{115}{3} - \frac{128}{3} = -\frac{13}{3}$
   • Since we are looking for the area, we take the absolute value: $A = \frac{13}{3} \approx 4.33$

Answer:

The area of the shaded region is $\frac{13}{3}$ or approximately $4.33$ square units.

Would you like further details on the integration steps, or have any questions?

Related Questions:

1. How would the area change if the interval were from $x = 3$ to $x = 5$?
2. What is the maximum height of the curve $y = -x^2 + 16$?
3. How would you set up an integral to find the area from $x = 0$ to $x = 4$?
4. What is the point of intersection of the curve $y = -x^2 + 16$ with the x-axis?
5. How does the shape of the graph $y = -x^2 + 16$ indicate it is a parabola opening downwards?

Tip:

When finding areas under curves, remember to take the absolute value if the function lies below the x-axis in the specified interval.