Find the area bounded by the x=axis below. f(x) above, and the given pair of vertical lines. f(x)= 9-x² , x=0, x=3

Answer 1

integral of 9-x^2 = 9x -x^3/3

9(3) - (3)^3/3 = 27 - 9 = 18

You might try graphing the equation

x^2 is a standard upward opening parabola

-x^2 is flipping the graph upside down, rotating it around the x axis, so it's a downward opening parabola with vertex at the origin = maximum point

now shift if up by 9. It's x intercepts are 3 and -3 and the new vertex is (0,9)

draw a straight line from the vertex to the x intercept (3,0). You have a triangle with area (3x9)/2 = 27/2=13.5

that's a lower bound for the area you want which is 18. You know you need an area larger than 13.5

It's just a useful check to see if your answer is in the right range.

Answer 2

integral ( 9-x^2) dx , lower limit=0 and upper =3

=9x-x^3/3=9(3)-(3)^3/3=18