To find the area enclosed by the graphs of y = x and y = x³, we first need to determine the points where the two curves intersect. This can be done by setting the equations equal to each other:
x = x³
Rearranging this, we get:
x³ – x = 0
Factoring out x, we have:
x(x² – 1) = 0
This gives us:
x = 0 or x² – 1 = 0,
which leads to x = ±1. Therefore, the curves intersect at the points:
(-1, -1), (0, 0), and (1, 1).
Next, we need to find the area between these curves from x = -1 to x = 1. The area can be computed using the integral of the top function minus the bottom function over the interval. Here, from x = -1 to x = 1, the curve y = x is above the curve y = x³.
The area (A) is given by:
A = ∫-11 (x – x³) dx
Now, we calculate the integral:
A = ∫-11 (x – x³) dx = ∫-11 x dx – ∫-11 x³ dx
The first integral:
∫ x dx = (1/2)x²
Evaluating from -1 to 1:
=(1/2)(1)² – (1/2)(-1)² = (1/2) – (1/2) = 0
The second integral:
∫ x³ dx = (1/4)x⁴
Evaluating from -1 to 1:
=(1/4)(1)⁴ – (1/4)(-1)⁴ = (1/4) – (1/4) = 0
So, putting these together:
A = 2 * ((1/2) – (1/4)) = 2 * (1/4) = 1
Thus, the area enclosed by the graphs of y = x and y = x³ is 1 square unit.