Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
y=5x^2,y=x^2+3
integrate with respect to x, so you don't have horizontal strips that change boundary curves. Then, since the region is symmetric about x=0, the area is
2∫[0,√3/2] (x^2+3)-5x^2 dx
each vertical strip has a height equal to the distance between the curves.
Cannot see sketches. Cannot copy and paste diagrams here.
To sketch and find the area of the region enclosed by the given curves, let's start by graphing the two curves: y = 5x^2 and y = x^2 + 3.
By observing the equations, we notice that they intersect at two points. To find these points, we can set the equations equal to each other:
5x^2 = x^2 + 3
Subtracting x^2 from both sides:
4x^2 = 3
Dividing both sides by 4:
x^2 = 3/4
Taking the square root of both sides:
x = ±√(3/4)
So the two x-coordinates where the curves intersect are x = √(3/4) and x = -√(3/4).
Now we can proceed with sketching the curves on a graph:
1. Plot the vertex of the parabola y = x^2 + 3, which is (0, 3).
2. Plot a few additional points on the parabola by substituting different x-values into the equation and calculating the corresponding y-values. For example, if x = 1, then y = 1^2 + 3 = 4.
3. Plot the vertex of the parabola y = 5x^2, which is also (0, 0).
4. Plot a few additional points on the parabola by substituting different x-values into the equation and calculating the corresponding y-values. For example, if x = 1, then y = 5(1)^2 = 5.
5. Connect the points for each parabola to sketch the curves.
Now that we have sketched the curves, we can determine the region enclosed by them. Based on the graph, we can observe that the curve y = 5x^2 is above the curve y = x^2 + 3. The region enclosed is bounded by these two curves.
To find the area of the region, we need to integrate the curve that is below the other curve with respect to either x or y. Since the curve y = x^2 + 3 is below the curve y = 5x^2 in our graph, we will integrate with respect to y.
The limits of integration will be the y-values where the two curves intersect. We found earlier that they intersect at x = √(3/4) and x = -√(3/4). Plugging these x-values into y = 5x^2 will give us the limits of integration.
For the region enclosed by the curves, the limits of integration with respect to y are:
√(3/4) ≤ x ≤ -√(3/4)
Now we can set up the integral to find the area of the region. Since we are integrating with respect to y, the integral is:
A = ∫[√(3/4), -√(3/4)] (y - (x^2 + 3)) dx
Evaluating this integral will give us the desired area of the region enclosed by the given curves.