Find the area of the region bounded by the graphs of the given functions x=3y²−9 x=0 y=0 y=1

To find the area of the region bounded by the given functions, we need to first find the points of intersection.

Setting the two given functions equal to each other:

3y^2 - 9 = 0

y^2 = 3

y = \pm \sqrt{3}

So, the points of intersection are at (0,0) and at \left(\pm \sqrt{3}, \frac{1}{2}\right) .

Since we are interested in the area between x=3y^2-9 and x=0 , we need to integrate the difference of their equations with respect to y from y=0 to y=1 .

A = \int_{0}^{1} \left(3y^2-9-0\right) \,dy

A = \int_{0}^{1} (3y^2-9) \, dy

A = \left[y^3-9y\right]_{0}^{1}

A = \left(1^3-9(1)\right) - \left(0^3-9(0)\right)

A = 1-9 = -8

Therefore, the area of the region bounded by the graphs of the given functions is \boxed{8 sq. units} .