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Magazine Chapter 7: Problem 21
Evaluate the integral. $$ \int_{0}^{1} 5 x^{3} \sqrt{1-x^{2}} d x $$
Short Answer
Expert verified
The integral evaluates to 0.
Step by step solution
01
Identify the Integral Type and Consider a Substitution
The integral has the function \( x^3 \sqrt{1-x^2} \), which suggests a trigonometric substitution might simplify it. Recognize the form \( \sqrt{1-x^2} \), which is reminiscent of \( \sin^2(\theta) \) since \( \sin^2(\theta) = 1 - \cos^2(\theta) \). Consider the substitution \( x = \sin(\theta) \), so \( dx = \cos(\theta) d\theta \).
02
Transform the Limits of Integration
Convert the limits of integration from terms of \( x \) to terms of \( \theta \). When \( x = 0 \), \( \theta = \arcsin(0) = 0 \); when \( x = 1 \), \( \theta = \arcsin(1) = \pi/2 \). So, the new limits for \( \theta \) are from \( 0 \) to \( \pi/2 \).
03
Substitute and Simplify the Integral
Substitute \( x = \sin(\theta) \) into the integral. The integral becomes: \[ \int_{0}^{\pi/2} 5 (\sin(\theta))^3 \sqrt{1-\sin^2(\theta)} \cos(\theta) \,d\theta \]. Simplify using \( \sqrt{1-\sin^2(\theta)} = \cos(\theta) \): \[ \int_{0}^{\pi/2} 5 \sin^3(\theta) \cos^2(\theta) \,d\theta \].
04
Simplify the Expression
Rewrite \( \sin^3(\theta) \cos^2(\theta) \) as \( \sin^3(\theta) \cos(\theta) \cos(\theta) \). It simplifies to \( \sin^3(\theta) \cos^2(\theta) = \frac{1}{4} \sin(2\theta) (1-\cos(2\theta)) \) using trigonometric identities.
05
Apply Trigonometric Identities and Integrate
Using \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), rewrite the integral: \[ \int_{0}^{\pi/2} 5 \frac{1}{4} \sin(2\theta) (1-\cos(2\theta)) \,d\theta \]. The integral now is \[ \frac{5}{4} \int (\sin(2\theta) - \sin(2\theta)\cos(2\theta)) \,d\theta \].
06
Solve the Integral
Integrate each term separately. The antiderivative of \( \sin(2\theta) \) is \(-\frac{1}{2}\cos(2\theta)\), and using a known result \( \int \sin(a\theta)\cos(a\theta)d\theta=\frac{1}{4a}\sin^2(a\theta)\), for \( \sin(2\theta)\cos(2\theta)\) it simplifies and solves easily. Apply the limits \( \theta=0 \) to \( \theta=\pi/2 \) to find the final numeric value.
07
Calculate the Definite Integral
Upon calculating, the terms involving \( \sin(2\theta) \) and \( \sin(2\theta)\cos(2\theta) \) from \( 0 \) to \( \pi/2 \) evaluate to 0 because of trigonometric periodicity, finalizing the computation as 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus is a core part of mathematics that deals with the concept of integration. It is essentially the reverse process of differentiation. While differentiation focuses on finding the rate of change, integration helps us find the total accumulation of quantities or the area under a curve. In this exercise, integration is used with a definite integral denoting the area under the curve defined from specific bounds, here from 0 to 1. The integral of a function is represented by the symbol \( \int \) and is followed by the function we want to integrate, with respect to a variable, typically denoted as \( dx \), meaning in terms of \( x \). This is essential when working with continuous data, allowing us to perform calculations over a specified interval.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the included variables. They serve as crucial tools for simplifying complex trigonometric expressions. In our problem, the substitution \( x = \sin(\theta) \) utilizes the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) to transform the original integral into a form that is easier to evaluate.
It's important to remember some key identities:
It's important to remember some key identities:
- \( \sin^2(\theta) = 1 - \cos^2(\theta) \)
- \( \cos^2(\theta) = 1 - \sin^2(\theta) \)
- \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)
Definite Integral
A definite integral is different from an indefinite integral because it computes the accumulation of quantity over an interval, not just the antiderivative. With limits of integration, like in this exercise from 0 to 1, we determine the net area under the curve of the given function over the specified interval.
In practice, once we've transformed our integral through substitution and simplification, we apply the limits of integration. Here, by changing the function to involve \( \theta \), our limits also changed from the original bounds (0 to 1) to equivalent ones in terms of \( \theta \), which are 0 to \( \pi/2 \) following the substitution \( x = \sin(\theta) \). Evaluating the integral from these limits simplifies it into numerical values.
The ability to calculate definite integrals is crucial in fields that require precise computations over certain intervals, like physics for determining positions or economics for area-based value calculations.
In practice, once we've transformed our integral through substitution and simplification, we apply the limits of integration. Here, by changing the function to involve \( \theta \), our limits also changed from the original bounds (0 to 1) to equivalent ones in terms of \( \theta \), which are 0 to \( \pi/2 \) following the substitution \( x = \sin(\theta) \). Evaluating the integral from these limits simplifies it into numerical values.
The ability to calculate definite integrals is crucial in fields that require precise computations over certain intervals, like physics for determining positions or economics for area-based value calculations.
