Asked by Sinachi
Find the area enclosed between the curve y=x^3 and the straight line y=x
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To find the area enclosed between the curve y=x^3 and the straight line y=x, we need to determine the points of intersection between the two curves and then integrate the difference between the two functions over that interval.
First, let's find the points of intersection.
Setting y = x for both curves, we have:
x = x^3
This simplifies to:
0 = x^3 - x
Factoring out an x, we have:
0 = x(x^2 - 1)
So, we have three possible solutions:
x = 0, x = 1, and x = -1
Now, we need to determine the interval over which we'll integrate. Since the curve y = x^3 is below the line y = x for x < -1 and above the line y = x for x > 1, we only need to consider the interval from x = -1 to x = 1.
So, the area enclosed between the curve y = x^3 and the line y = x is given by the integral:
∫[from -1 to 1] (x - x^3) dx
Integrating this expression gives:
∫[from -1 to 1] (x - x^3) dx = [x^2/2 - x^4/4] [from -1 to 1]
= (1/2 - 1/4) - (1/2 - 1/4)
= 1/4 - 1/4
= 0
Therefore, the area enclosed between the curve y = x^3 and the line y = x is 0 units squared.
First, let's find the points of intersection.
Setting y = x for both curves, we have:
x = x^3
This simplifies to:
0 = x^3 - x
Factoring out an x, we have:
0 = x(x^2 - 1)
So, we have three possible solutions:
x = 0, x = 1, and x = -1
Now, we need to determine the interval over which we'll integrate. Since the curve y = x^3 is below the line y = x for x < -1 and above the line y = x for x > 1, we only need to consider the interval from x = -1 to x = 1.
So, the area enclosed between the curve y = x^3 and the line y = x is given by the integral:
∫[from -1 to 1] (x - x^3) dx
Integrating this expression gives:
∫[from -1 to 1] (x - x^3) dx = [x^2/2 - x^4/4] [from -1 to 1]
= (1/2 - 1/4) - (1/2 - 1/4)
= 1/4 - 1/4
= 0
Therefore, the area enclosed between the curve y = x^3 and the line y = x is 0 units squared.
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