Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer.
Between y = −x and y = −x^3 for x in [−1, 1]
both odd functions through origin
same area both sides of y axis
2 times integral 0 to 1 of xdx - x^3dx
= 2(1^2/2 - 1^4/4) = 2(1/2-1/4) = 2(1/4)
=1/2
To find the area of the indicated region between the curves y = -x and y = -x^3 for x in [-1, 1], we can use the definite integral.
Step 1: Graph the curves
First, let's graph the curves y = -x and y = -x^3 to visualize the region.
The curve y = -x is a straight line that passes through the origin with a slope of -1.
The curve y = -x^3 is a downward-opening cubic curve symmetric about the y-axis. It passes through the origin and approaches negative infinity as x approaches negative infinity and positive infinity as x approaches positive infinity.
By graphing these curves, we can see that they intersect at two points: (-1, 1) and (1, -1).
Step 2: Determine the bounds of integration
The region between the curves is bounded by the x-values -1 and 1.
Step 3: Set up the integral
To find the area, we need to calculate the definite integral of the difference between the curves.
∫[a, b] (f(x) - g(x)) dx
In this case, f(x) represents the upper curve (y = -x) and g(x) represents the lower curve (y = -x^3).
So, the integral becomes:
A = ∫[-1,1] (-x - (-x^3)) dx
Step 4: Evaluate the integral
Using technology or antiderivative techniques, we can evaluate the integral:
A = ∫[-1,1] (-x + x^3) dx
Integrating, we get:
A = [(-x^2/2) + (x^4/4)] from -1 to 1
Substituting the limits of integration:
A = [(-(1)^2/2) + ((1)^4/4)] - [(-(-1)^2/2) + ((-1)^4/4)]
Simplifying,
A = [(-1/2) + (1/4)] - [(-1/2) + (1/4)]
A = 1/4
Therefore, the area of the region between the curves y = -x and y = -x^3 for x in [-1, 1] is 1/4 square units.