Find the area bounded by the curves y = x^3 and y = 4x.
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Step-by-Step Solution
Step 1: Find the points of intersection
To find the points where the two curves intersect, we set their equations equal to each other. This gives us an equation in terms of x$. We then factor the equation to find the values of x where the curves meet.
Step 2: Determine the intervals and upper/lower curves
From the factored equation x(x-2)(x+2) = 0$, we find the x-coordinates of the intersection points. These points divide the area into intervals. We need to determine which curve is above the other in each interval.
Step 3: Set up the integral for the area
For the interval [-2, 0]$, we test a point, say x = -1$. y = 4(-1) = -4 and y = (-1)^3 = -1$. Since -1 > -4$, y = x^3 is the upper curve. For the interval [0, 2]$, we test x = 1$. y = 4(1) = 4 and y = (1)^3 = 1$. Since 4 > 1$, y = 4x is the upper curve. Wait, I made a mistake in the explanation. Let's re-evaluate. For x = -1$, y = 4x = -4 and y = x^3 = -1$. So x^3 is above 4x$. For x = 1$, y = 4x = 4 and y = x^3 = 1$. So 4x is above x^3$. The problem statement is symmetric, so the area from -2 to 0 where x^3 is above 4x will be equal to the area from 0 to 2 where 4x is above x^3$. Therefore, the total area is the sum of the absolute differences of the integrals.
Step 4: Corrected integral setup
Based on the analysis of the upper and lower curves, the integral for the area from x = -2 to x = 0 is \int_{-2}^{0} (x^3 - 4x) dx$, and the integral for the area from x = 0 to x = 2 is \int_{0}^{2} (4x - x^3) dx$. We sum these two integrals to get the total bounded area.
Step 5: Evaluate the integrals
We find the antiderivative for each integrand. Then, we evaluate the definite integrals using the Fundamental Theorem of Calculus, substituting the limits of integration.
Step 6: Calculate the total area
We substitute the limits of integration into the antiderivatives and perform the arithmetic. The first integral evaluates to 0 - (4 - 8) = 4$. The second integral evaluates to (8 - 4) - 0 = 4$. Summing these values gives the total bounded area.