Problem 26 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The area is \(\frac{1}{12}\).

Step 1: Identify the Points of Intersection

Before finding the area, we determine where the curves intersect. To find the points of intersection, set the equations equal to each other: \(x^2 = x^3\). This simplifies to \(x^3 - x^2 = 0\) or \(x^2(x-1) = 0\). The solutions are \(x = 0\) and \(x = 1\). Thus, the graphs intersect at \((0,0)\) and \((1,1)\).

Step 2: Set Up the Definite Integral for Area

The area between the curves from \(x = 0\) to \(x = 1\) can be found by integrating the difference of the top function and the bottom function. Here, \(y = x^2\) is always greater than \(y = x^3\) within our interval. Hence, the area is described by the integral \(A = \int_{0}^{1} (x^2 - x^3) \, dx\).

Step 3: Evaluate the Integral

Calculate the integral \(\int (x^2 - x^3) \, dx\). The antiderivative of \(x^2\) is \(\frac{x^3}{3}\) and of \(x^3\) is \(\frac{x^4}{4}\). Thus, the integral becomes \(\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1}\).

Step 4: Substitute and Solve

Substitute the limits into the evaluated antiderivative: \([\frac{1^3}{3} - \frac{1^4}{4}] - [\frac{0^3}{3} - \frac{0^4}{4}]\). This simplifies to \(\frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}\).

Step 5: Conclusion

The area of the region bounded by the graphs of \(y = x^2\) and \(y = x^3\) between \(x = 0\) and \(x = 1\) is \(\frac{1}{12}\).

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with the concept of integrals. It is primarily concerned with the process of finding the antiderivative, or the inverse operation of differentiation. In simple terms, when you perform integration, you are essentially trying to find a quantity when you know its rate of change.

When calculating the total area under or above a curve, integral calculus is a powerful tool. This allows us to accumulate infinite tiny quantities across a specific interval. Think of it as adding up infinitely small slices to find the overall volume or area.

• Integrals can be either indefinite or definite.
• Indefinite integrals do not have set boundaries and result in a general form plus a constant of integration, often represented by "C".
• Definite integrals have specific limits and provide an exact value, often used for finding areas under curves.

Understanding integrals is essential for solving problems involving areas, volumes, and other applications requiring the summation of changing quantities.

Definite Integrals

Definite integrals are a specific type of integrals where the limits of integration are defined. These limits indicate the start and finish points on the x-axis, for which you need to find the area under the curve.

When computing a definite integral, you not only find the antiderivative but also evaluate it at the upper and lower limits. This gives you a precise numerical value, which often represents area or accumulated quantity.

• The notation is typically expressed as \ \( \int_{a}^{b} f(x) \, dx \ \) \ , where \(a\) and \(b\) are the limits of integration.
• To solve, you calculate the antiderivative of the function \ \( f(x) \ \), then substitute \(b\) and \(a\) to find the difference \ \( F(b) - F(a) \ \).
• The result quantifies the total accumulation of \( f(x) \) between \( a \) and \( b \).

This approach is fundamental in problems dealing with finding areas between curves, as seen in our example exercise.

Area Between Curves

To find the area between two curves, you use definite integrals. This is crucial in calculus when you need to determine the space enclosed by the graphs of two functions.

Start by identifying the region where the two curves overlap or intersect, which allows you to set the correct limits for integration.

• First, find the points of intersection by setting the equations equal to each other.
• Determine which function is on top (larger y-values) over the interval and which is on the bottom (smaller y-values).
• Formulate the integral as \ \( \int_{a}^{b} [f(x) - g(x)] \, dx \ \), where \( f(x) \) is the upper function and \( g(x) \) is the lower.

By evaluating this integral, you obtain the net area between the two curves over the specified interval. This technique was used in the original problem to find the area bounded by \( y = x^2 \) and \( y = x^3 \), resulting in an area of \ \( \frac{1}{12} \ \).