Find the area of the region enclosed by

y=6x2 and y=x2+15

To find the area of the region enclosed by the two curves, we need to determine the points of intersection between the curves and integrate the difference between the two curves over that interval.

First, let's find the points of intersection by setting the two equations equal to each other:

6x^2 = x^2 + 15

Subtracting x^2 from both sides:

5x^2 = 15

Divide both sides by 5:

x^2 = 3

Taking the square root of both sides:

x = ±√3

Now we can set up the integral to calculate the area. Since the region is enclosed, we can integrate the difference between the curves over the interval (-√3, √3).

The formula for the area between two curves is:

A = ∫(f(x) - g(x)) dx

where f(x) and g(x) are the equations of the curves.

In this case, f(x) = 6x^2 and g(x) = x^2 + 15.

So, the integral becomes:

A = ∫(6x^2 - (x^2 + 15)) dx

A = ∫(5x^2 - 15) dx

To find the antiderivative of 5x^2 - 15, we integrate term by term:

A = (5/3)x^3 - 15x + C

Now, substitute the limits of integration:

A = (5/3)(√3)^3 - 15(√3) - [(5/3)(-√3)^3 - 15(-√3)]

Simplifying further:

A = (5/3)(3√3) - 15√3 - (5/3)(-3√3) + 15√3

A = 5√3 - 15√3 + 5√3 - 15√3

Combining like terms:

A = -20√3

Therefore, the area of the region enclosed by the curves y = 6x^2 and y = x^2 + 15 is -20√3 square units.