Problem 62 - Calculus and its applications

Question

Find the area of the region bounded by the given graphs. $$ y=x^{2}, y=x^{3}, x=-1 $$

Step-by-Step Solution

Answer

The area of the region is \(\frac{1}{6}\).

Step 1: Understand the region

The problem involves two curves and a vertical line. The curves are the parabola \(y = x^2\) and the cubic curve \(y = x^3\). We also have the line \(x = -1\). First, we should understand the region bounded by these graphs. The region's vertical limits are between \(x = -1\) and the point where \(y = x^2\) and \(y = x^3\) intersect.

Step 2: Find intersection points

To find where \(y = x^2\) and \(y = x^3\) intersect, we set them equal: \(x^2 = x^3\). Solving for \(x\), we factor: \(x^2(x - 1) = 0\). Thus, \(x = 0\) and \(x = 1\). So, the region is bounded between \(x = -1\), 0, and \(x = 1\).

Step 3: Determine the curves' positions

On different intervals, determine which curve is above the other. Between \(x = 0\) and \(x = 1\), the parabola \(y = x^2\) is above \(y = x^3\). Between \(x = -1\) and \(x = 0\), we need to check the positions by plugging any value from the interval or observing the nature of functions (cubic vs. quadratic growth and asymmetry at negative values).

Step 4: Set up integrals between intersection points

Set up the integrals for computing the area. For \(-1 \leq x \leq 0\), \(y = x^2\) is above \(y = x^3\). Similarly, for \(0 \leq x \leq 1\), \(y = x^2\) remains above. Therefore, the integral is \[ \int_{-1}^{0} (x^2 - x^3) \, dx + \int_{0}^{1} (x^2 - x^3) \, dx \].

Step 5: Calculate the integrals

Compute \( \int (x^2 - x^3) \, dx = \int (x^2) \, dx - \int (x^3) \, dx = \frac{x^3}{3} - \frac{x^4}{4} + C \). Evaluate this from \(-1\) to \(0\) and from \(0\) to \(1\).

Step 6: Solve each integral separately

For \([-1, 0]\), calculate \(\left[ \frac{0^3}{3} - \frac{0^4}{4} \right] - \left[ \frac{(-1)^3}{3} - \frac{(-1)^4}{4} \right] = 0 + \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \). For \([0, 1]\), calculate \(\left[ \frac{1^3}{3} - \frac{1^4}{4} \right] - \left[ \frac{0^3}{3} - \frac{0^4}{4} \right] = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \).

Step 7: Add the areas

Sum the areas obtained from both calculated integrals: \(\frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}\). This gives the total bounded area between the curves \(y = x^2\), \(y = x^3\), and \(x = -1\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, primarily used to compute the area under curves between two points. When solving problems involving definite integrals, one essentially sums up the values of a function over an interval, which can be particularly useful for finding areas in the plane.
In the context of our exercise, the definite integrals help us calculate the area of the region between two curves and a vertical line. To set up a definite integral, choose the lower and upper limits based on the interval of interest—in our case, these were the x-values of the intersection points and the given boundary.
The function to be integrated is determined by the difference between the upper and lower curves:

• If the upper curve is given by a function \( f(x) \) and the lower curve by \( g(x) \), then we integrate \( f(x) - g(x) \).
• The area is computed as \( \int_{a}^{b} (f(x) - g(x)) \, dx \), where \( a \) and \( b \) are the limits of integration.

This setup captures the vertical distances between the curves across the interval, and summing them yields the total area.

Area Between Curves

Finding the area between curves involves understanding how the curves differ across an interval and applying integration. This is done by calculating the vertical distance between two functions that describe the curves.
In our problem, the curves described by \( y = x^2 \) (a parabola) and \( y = x^3 \) (a cubic curve) differ depending on the value of \( x \).
To solve for this area, first, identify which function is above the other over the interval.

• From
  [-1, 0] and
  [0, 1], \( x^2 \) is above \( x^3 \).
• Calculate the integral \( \int_{-1}^{1} (x^2 - x^3) \, dx \). Since this involves two segments, sum up the results from the integrals of each interval.

This approach is key to determining the enclosed area's exact size.

Intersection Points

Intersection points are crucial when dealing with areas between curves because they often determine the limits of integration. These are the points where curves meet or cross.
In the exercise, finding these points is a necessary step for accurately determining the region of interest. To find where two functions, like \( y = x^2 \) and \( y = x^3 \), intersect, set the functions equal to one another: \( x^2 = x^3 \).
Solving this equation involves simple algebra:

• Factor it as \( x^2(x - 1) = 0 \), giving solutions \( x = 0 \) and \( x = 1 \).
• The solutions represent the intersection points, forming the boundaries for the integration intervals \([-1, 0]\) and \([0, 1]\).

Understanding this process is fundamental in calculus, as it helps accurately define the segments over which to integrate when calculating areas.