Write the definite integral for the area of the region bounded by the grpahs of y=9-x^2 and y=0
the curve crosses the x-axis at x = -3 and x=3
The integral is
Int(9-x^2) dx [-3,3]
in suitable notation.
To find the definite integral for the area of the region bounded by the graphs of y = 9 - x^2 and y = 0, you can use the following steps:
1. Determine the x-values where the two graphs intersect. In this case, we need to find the x-values at which y = 9 - x^2 is equal to y = 0. Since y = 0, we set 9 - x^2 = 0 and solve for x. This equation simplifies to x^2 = 9, so x could be either 3 or -3.
2. Determine the limits of integration. Since we are interested in the region bounded by the graphs, we need to integrate with respect to x over the interval between the x-values where the functions intersect. In this case, the limits of integration are from x = -3 to x = 3.
3. Set up the integral. The integral represents the area under the curve y = 9 - x^2 between the limits of integration. Since we are finding the area below the curve, the integral has a value of the function multiplied by -1. Thus, the integral expression for the area is:
∫[from -3 to 3] (-(9 - x^2)) dx
4. Evaluate the integral. Integrating -(9 - x^2) with respect to x over the interval from -3 to 3 will give you the area of the region bounded by the two graphs.