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Chapter 4: Problem 32

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

Short Answer

Expert verified
The area of the region is \( \frac{1}{12} \).

Step by step solution

01

- Identify the Points of Intersection

First, find where the curves intersect by setting the equations equal to each other: \[ x^2 = x^3 \] Solve for x: \[ x^2 - x^3 = 0 \] \[ x^2 (1 - x) = 0 \] This gives the points of intersection: \[ x = 0 \] and \[ x = 1 \]. These points are at (0, 0) and (1, 1).
02

- Set Up the Integral to Find the Area

The area between two curves can be found using the integral: \[ \text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx \] In this case, \( f(x) \) is the upper function (\( y = x^2 \)) and \( g(x) \) is the lower function (\( y = x^3 \)) from \( x = 0 \) to \( x = 1 \).
03

- Evaluate the Integrals

Evaluate the two integrals: \[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} - 0 = \frac{1}{3} \] \[ \int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4} - 0 = \frac{1}{4} \]
04

- Subtract the Integrals to Find the Area

Now subtract the results of the integrals: \[ \text{Area} = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

integration
Integration is a fundamental concept in calculus that helps us calculate the area under a curve.
It is the reverse process of differentiation, and it accumulates the

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