Problem 15 - Calculus and its applications

Question

Find the area of the region bounded by the graphs of the given equations. $$ y=x, y=x^{3}, x=0, x=1 $$

Step-by-Step Solution

Answer

The area is \( \frac{1}{4} \).

Step 1: Identify Intersection Points

The first step in setting up the integral is to determine where the curves intersect. We need to find the points of intersection of the graphs of the equations \( y=x \) and \( y=x^3 \). Equating the two equations gives \( x = x^3 \). This simplifies to \( x^3 - x = 0 \), or \( x(x^2 - 1) = 0 \), which factors to \( x(x-1)(x+1) = 0 \). The intersection points are \( x = 0 \) and \( x = 1 \).

Step 2: Set Up the Integral

The area between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral \[\int_{0}^{1} (x - x^3) \, dx\]Here, \( x^3 \) is subtracted from \( x \) because \( y = x \) is above \( y = x^3 \) between \( x = 0 \) and \( x = 1 \).

Step 3: Integrate the Function

Compute the integral \[\int_{0}^{1} (x - x^3) \, dx = \int_{0}^{1} x \, dx - \int_{0}^{1} x^3 \, dx\]Calculate each part separately:- \( \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2} \)- \( \int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4} - 0 = \frac{1}{4} \)

Step 4: Evaluate the Area

Subtract the results of the two definite integrals:\[\text{Area} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\]Thus, the area of the region bounded by the graphs of the given equations is \( \frac{1}{4} \).

Key Concepts

Area Between Curves

Finding the area between two curves is a fundamental application of integral calculus. It involves determining how much space is enclosed within the boundaries formed by these curves. In this problem, we have two functions: \( y = x \) and \( y = x^3 \). To find the area between these curves:

• Identify which function is above the other within the desired interval. This is crucial as it determines what to subtract from what.
• For \( x = 0 \) to \( x = 1 \), it is evident that \( y = x \) lies above \( y = x^3 \). Therefore, determine the area by integrating the difference of these functions: \((x - x^3)\).
• Evaluate this integral over the interval \([0, 1]\) to get the total area between the curves.

This approach captures the full enclosed region, leading to the calculation of the total area between these two distinct curves.

Definite Integral

A definite integral is a tool used to calculate the accumulation of quantities, such as areas under curves. In this context, it represents the net area under a curve over a specific interval. Here's how it works:

• The notation \( \int_{a}^{b} f(x) \, dx \) specifies the integration of the function \( f(x) \) from \( x = a \) to \( x = b \).
• For our specific problem, the definite integral \( \int_{0}^{1} (x - x^3) \, dx \) gives the area between the curves from \( x = 0 \) to \( x = 1 \).
• This involves breaking down the integral into separable functions: \( \int_{0}^{1} x \, dx \) and \( \int_{0}^{1} x^3 \, dx \), and then evaluating each separately.
• The resultant areas are then subtracted because one function represents the upper boundary, while the other the lower.

The definite integral succinctly aids in summing these values to arrive at the correct area.

Intersection Points

Intersection points serve as boundaries where two or more curves cross or touch. Detecting these points is crucial as they often define the limits for our integrals.

• To find these points, set the functions equal: \( y = x \) and \( y = x^3 \). Equating them leads to \( x = x^3 \), simplifying to \( x(x^2 - 1) = 0 \).
• This further factors into \( x(x-1)(x+1) = 0 \), giving intersection points at \( x = 0 \), \( x = 1 \), and \( x = -1 \). However, only \( x = 0 \) and \( x = 1 \) fall within our desired interval.
• These points, 0 and 1, become the bounds for integration as they mark where the graphs intersect and change dominance.

Understanding and identifying intersection points is vital as they anchor the start and end points for our integrations, crucial for calculating enclosed areas.