Questions: Use the definite integral to find the area between the x-axis and f(x) over the indicated interval. f(x)=-x^2+9 ;[0,5] A. 5/9 B. 98/3 C. 10/3 D. 10/9

Solution Steps

We start with the function \( f(x) = -x^2 + 9 \). This is a downward-opening parabola.

To find the area between the curve and the \( x \)-axis over the interval \([0, 5]\), we set up the definite integral: \[ \text{Area} = \int_{0}^{5} (-x^2 + 9) \, dx \]

Calculating the integral, we find: \[ \text{Area} = \left[ -\frac{x^3}{3} + 9x \right]_{0}^{5} \] Evaluating this from 0 to 5 gives: \[ \text{Area} = \left( -\frac{5^3}{3} + 9 \cdot 5 \right) - \left( -\frac{0^3}{3} + 9 \cdot 0 \right) = \left( -\frac{125}{3} + 45 \right) = \left( -\frac{125}{3} + \frac{135}{3} \right) = \frac{10}{3} \]

Final Answer