find the area under the region bounded by the curves y=x^2-3 and y=2x.
To find the area under the region bounded by the curves y=x^2-3 and y=2x, we need to determine the points of intersection between these two curves.
First, let's set y equal to each other: x^2-3 = 2x.
Rearranging the equation, we get: x^2 - 2x - 3 = 0.
Now, we can solve this quadratic equation to find the x-coordinates of the intersection points. We can do this by factoring, completing the square, or using the quadratic formula.
Factoring the quadratic equation, we get: (x-3)(x+1) = 0.
So, the solutions are x = 3 and x = -1.
Now that we have the x-values of the intersection points, we can find the y-values by substituting these x-values into the equations of the curves.
For y=x^2-3:
When x = 3, y = (3)^2 - 3 = 6.
When x = -1, y = (-1)^2 - 3 = -2.
For y=2x:
When x = 3, y = 2(3) = 6.
When x = -1, y = 2(-1) = -2.
Thus, we see that the two curves intersect at (3, 6) and (-1, -2).
To find the area, we integrate the difference of the two functions within the range of x = -1 to x = 3.
The area A can be calculated by integrating the expression [y = (2x) - (x^2-3)] with respect to x.
A = ∫[2x - (x^2-3)] dx, evaluated from x = -1 to x = 3.
Evaluating the integral, we get:
A = ∫[2x - (x^2-3)] dx
= ∫[2x - x^2 + 3] dx
= [x^2 - (x^3/3) + 3x] evaluated from x = -1 to x = 3.
Substituting the limits of integration:
A = [(3)^2 - ((3)^3/3) + 3(3)] - [(-1)^2 - ((-1)^3/3) + 3(-1)]
= [9 - 9 + 9] - [1 - (-1/3) - 3]
= 18 - (1 - 1/3 - 3)
= 18 - 1 + 1/3 + 3
= 21 + 1/3
= 21.33 (approximately).
Therefore, the area under the region bounded by the curves y=x^2-3 and y=2x is approximately 21.33 square units.