Find the area between y=x^2 and y=1 using Calculus.
The two graphs intersect at (-1,1) and (1,1).
Consider the area as a collection of thin rectangles of width dx and height 1-x^2. Then, taking advantage of the symmetry of the region, the area is
2∫[0,1] 1-x^2 dx
Now evaluate the integral to get your answer.
find the two points where the line (y = 1) intersects the parabola (y = x^2)
... this is the range for the integration
drawing a sketch is helpful
... you're finding the area of the "nose" of the parabola
think of dividing the area into vertical strips
... the length of the strips is ... 1 - x^2
the width of the strips is ... dx
the integral is ... ∫ 1-x^2 dx
... the integration is from ... x=-1 to x=1
area is ... [1 - (1^3 / 3)] - {-1 - [(-1)^3 / 3]}
To find the area between two curves using calculus, you can use definite integration. Here's the step-by-step process to solve this problem:
Step 1: Identify the points of intersection.
To find the area between the curves y = x^2 and y = 1, you need to first determine the points where these two curves intersect. Set the two equations equal to each other and solve for x:
x^2 = 1.
This equation has two solutions: x = 1 and x = -1.
Step 2: Determine the limits of integration.
Since you want to find the area between the curves, you need to determine the interval over which you're integrating. In this case, the interval is from -1 to 1 since those are the x-values where the two curves intersect.
Step 3: Set up the definite integral.
Next, you need to set up the definite integral that represents the area between the curves. Since we are finding the area between the curves and the upper curve is y = 1, the integral will be:
∫[from -1 to 1] (1 - x^2) dx.
Step 4: Evaluate the definite integral.
To evaluate the integral, you can use the power rule for integration:
∫(1 - x^2) dx = x - (1/3)x^3 + C.
Now, substitute the limits of integration and evaluate the integral:
[(1 - (1/3)(1)^3) - (1 - (1/3)(-1)^3)].
Simplifying the expression further will give you the area between the curves y = x^2 and y = 1.