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Chapter 4: Problem 4

Find the area under the given curve over the indicated interval. $$ y=x^{2} ; \quad[0,3] $$

Short Answer

Expert verified
The area under the curve \( y = x^2 \) from 0 to 3 is 9.

Step by step solution

01

Understand the problem

We need to find the area under the curve of the function \( y = x^2 \) over the interval \([0, 3]\). This is essentially finding the definite integral of the function from 0 to 3.
02

Write the integral expression

The definite integral to find the area under the curve from \( x=0 \) to \( x=3 \) is written as: \[\int_{0}^{3} x^2 \, dx\]
03

Apply the power rule for integration

The power rule for integration states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. For \( x^2 \), \( n = 2 \). Thus, the integral is \[\int x^2 \, dx = \frac{x^{3}}{3} + C\]
04

Evaluate the definite integral

Apply the limits of integration (from 0 to 3) to the antiderivative: \[\left. \frac{x^{3}}{3} \right|_{0}^{3} = \frac{3^{3}}{3} - \frac{0^{3}}{3} \]
05

Simplify the evaluation

Simplify the expression: \[\frac{3^{3}}{3} = \frac{27}{3} = 9\]So, the area under the curve from 0 to 3 is 9.

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Key Concepts

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

Area under Curve
Finding the area under a curve involves calculating the definite integral of a function over a specified interval. In this exercise, we are considering the function \( y = x^2 \) over the interval \([0, 3]\). This means we want to find the total area enclosed between the curve, the x-axis, and the vertical lines \(x = 0\) and \(x = 3\). The definite integral gives us this area by summing up an infinite number of infinitesimal rectangles under the curve. The definite integral
  • summarizes areas added together within the limits of integration,
  • can represent exact areas under the curve when calculated correctly.
Understanding that calculating the area under the curve involves integration is key to solving these types of problems.
Power Rule for Integration
The power rule for integration is a fundamental tool for finding antiderivatives, which allows us to solve integrals for polynomial functions like our given \( y = x^2 \). According to the power rule:
  • For any function \( x^n \), the integral is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
  • "\(C\)" represents the constant of integration,
  • This rule only applies when \( n eq -1 \).
Applying the rule to \( x^2 \), we integrate to obtain \( \frac{x^{3}}{3} + C \). This simplifies determining antiderivatives for polynomial functions, a necessary step in evaluating definite integrals.
Antiderivative
The antiderivative, sometimes called the "indefinite integral," of a function is a function whose derivative is the original function. In simpler terms, finding the antiderivative is like reversing the process of differentiation. For a function \( x^2 \), its antiderivative based on the power rule is \( \frac{x^{3}}{3} + C \). To solve a definite integral, we need this antiderivative to evaluate the integral over a particular interval. Here, the function \( f(x) = \frac{x^3}{3} \) acts as the antiderivative of \( y = x^2 \), and crucially helps us compute the exact area under the curve between \( x = 0 \) and \( x = 3 \).
Evaluation of Integral
The evaluation of a definite integral involves applying the antiderivative to the upper and lower limits of the interval. In our problem for the integral \( \int_{0}^{3} x^2 \, dx \), the steps are:
  • First, substitute \( x = 3 \) into the antiderivative \( \frac{x^3}{3} \), giving \( \frac{27}{3} = 9 \).
  • Then substitute \( x = 0 \), which results in \( \frac{0^3}{3} = 0 \).
Subtract these results
  • upper value \(9 - 0\) from the lower value,
  • leading to an area of 9 between \( x = 0 \) to \( x = 3 \).
This subtraction gives the total area under the curve within your given limits, clarifying the role of definite integrals in determining precise area measurements.

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