Problem 56 - Applied Calculus

Question

Find the area bounded by the given curves. \(y=x^{3}\) and \(y=4 x\)

Step-by-Step Solution

Answer

The area bounded by the curves is 16 square units.

Step 1: Find the Intersection Points

To determine the area bounded by the curves, we first need to find their points of intersection. Set the equations equal to each other: \[x^3 = 4x\]Rearrange and factor the equation:\[x(x^2 - 4) = 0\]This gives the solutions: \(x = 0, x^2 = 4\), where \(x = \pm 2\). So, the curves intersect at \((0, 0)\), \((2, 8)\), and \((-2, -8)\).

Step 2: Setup the Integral for Area Calculation

We can solve for the area by integrating the difference between the two functions from \(-2\) to \(2\). The area \(A\) is given by the integral:\[A = \int_{-2}^{2} (4x - x^3) \, dx\] This represents the area between the curves, from the leftmost to the rightmost intersection points, where \(y = 4x\) is the top function.

Step 3: Integrate the Function

Evaluate the integral:\[\int_{-2}^{2} (4x - x^3) \, dx = \int_{-2}^{2} 4x \, dx - \int_{-2}^{2} x^3 \, dx\]Compute each integral separately:\[\int 4x \, dx = 2x^2 + C \text{ and } \int x^3 \, dx = \frac{x^4}{4} + C\]Thus,\[\int_{-2}^{2} 4x \, dx = \left[2x^2\right]_{-2}^{2} = 8 + 8 = 16\]\[\int_{-2}^{2} x^3 \, dx = \left[\frac{x^4}{4}\right]_{-2}^{2} = \left(\frac{16}{4}\right) - \left(\frac{16}{4}\right) = 0\]

Step 4: Compute the Area

Using the results from above, find the area between the two curves:\[A = 16 - 0 = 16\]This means the area bounded by the curves is 16 square units.

Key Concepts

Area Between Curves

Finding the area between curves is like figuring out the space covered between two distinct paths or shapes on a graph. Consider two curves: one might be above the other across a particular interval. To find this area, we subtract the lower curve from the upper one.

• Upper Curve: Typically the curve with higher values in the given interval.
• Lower Curve: The one with lower values in the same interval.

In our exercise, the curves are described by the equations:

• \( y = x^3 \)
• \( y = 4x \)

These represent a cubic curve and a linear line intersecting each other. To define the boundaries for integration, we seek the intersection points. Knowing these points is crucial because they mark where we measure our area.

Integrals

Integrals are the building blocks for finding areas under curves or between curves, in calculus. In simple terms, integration is the process of summing an infinite number of infinitesimally small areas beneath a curve to find the total area.
For the area between functions, we use a definite integral which has designated starting and stopping points, often the intersection points where the curves meet. For our exercise:

• The integral is set up from \( -2 \ \text{to} \ 2 \), based on the intersection point.
• We evaluate the difference \( \int (4x - x^3) \, dx \), with \( 4x \) as the top function.

Each term within the integral covers different parts of the area between curves. When we compute the integral, we break it up into separate pieces and find the difference, giving us the area of interest.

Intersection Points

Finding intersection points between curves allows us to establish where two curves meet, and therefore, where the limits of integration are.

• To find these mathematically, set the equations equal: \( x^3 = 4x \), for our problem.
• Next, solve for \( x \): Factor or solve this equation \( x(x^2 - 4) = 0 \).
• Your solutions might include zero or equal roots for the given parabola or line.

Our intersection results in three points: \( (0, 0), (2, 8), (-2, -8) \). These points determine the egde of our region, illustrating the shift between the two curves. By knowing exactly where these curves meet, we can effectively apply the process of integration to find precise area measurements.