/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Verify by direct computation tha... [FREE SOLUTION] | 91Ó°ÊÓ

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Verify by direct computation that $$ \int_{1}^{3} x^{2} d x=-\int_{3}^{1} x^{2} d x $$

Short Answer

Expert verified
We find the definite integrals \(\int_{1}^{3} x^2 dx\) and \(-\int_{3}^{1} x^2 dx\). By evaluating the antiderivatives and comparing the two values, we find that both are equal to \(9-\frac{1}{3}\). Thus, we can verify that \(\int_{1}^{3} x^{2} dx=-\int_{3}^{1} x^{2} dx\).

Step by step solution

01

Compute the first integral

Integrate the function \(x^2\) with respect to \(x\) from 1 to 3. To find the integral, we use the power rule of integration: \[\int x^n dx = \frac{1}{n + 1}x^{n+1}\] In this case, we have \(n=2\), so the antiderivative of \(x^2\) is: \[\frac{1}{2+1}x^{2+1} = \frac{1}{3}x^3\] Now, find the definite integral by evaluating the antiderivative at the bounds 1 and 3: \(=\frac{1}{3}\cdot3^3 - \frac{1}{3}\cdot1^3\) \(=9-\frac{1}{3}\) #Step 2: Compute the second integral#
02

Compute the second integral

Integrate the function \(x^2\) with respect to \(x\) from 3 to 1. Using the same antiderivative as in step 1, we now evaluate the antiderivative at the bounds 3 and 1 but in a reversed order: \(-\biggl[\frac{1}{3}\cdot1^3 - \frac{1}{3}\cdot3^3\biggr]\) \(=-\biggl[\frac{1}{3}-9\biggr]\) #Step 3: Check for equality #
03

Compare the two integrals

In step 1, we found the value of the first integral: \(9-\frac{1}{3}\). In step 2, we found the value of the second integral: \(-\biggl[\frac{1}{3}-9\biggr]\). Now, we compare the two values: \[\int_{1}^{3} x^2 dx = 9-\frac{1}{3}\] \[-\int_{3}^{1} x^2 dx = -\biggl[\frac{1}{3}-9\biggr] = 9-\frac{1}{3}\] Since both values are equal, we can verify that: \[\int_{1}^{3} x^{2} dx=-\int_{3}^{1} x^{2} dx\]

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Key Concepts

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

Power Rule of Integration
The power rule of integration is a basic technique used to find the antiderivative of a function in the form of \( x^n \). To apply this rule, simply increase the exponent by one and divide by the new exponent. More formally, this is expressed as
\[ \int x^n \, dx = \frac{1}{n + 1}x^{n+1} + C \]
where \( C \) is the constant of integration. When dealing with definite integrals, the constant of integration becomes irrelevant as it will cancel out when evaluating the bounds. For example, the antiderivative of \( x^2 \) is \( \frac{1}{3}x^3 + C \), following the power rule. This straightforward approach simplifies the process of integrating polynomials and is essential for anyone learning calculus.
Antiderivative
An antiderivative of a function \( f(x) \) is a differentiable function \( F(x) \) whose derivative is \( f(x) \). In other words, the antiderivative is the inverse operation to taking a derivative, and hence it is sometimes referred to as the indefinite integral. Antiderivatives are not unique; they include a constant of integration \( C \) because derivatives of constants are zero. For example, the antiderivative of \( x^2 \) is given by \( \frac{1}{3}x^3 + C \), where \( C \) can be any real number. In the context of definite integrals, like in our original exercise, the difference of antiderivatives evaluated at the upper and lower bounds gives you the exact area under the curve between those points.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a critical link between differentiation and integration. It states that if \( F(x) \) is an antiderivative of \( f(x) \) on an interval \( [a, b] \), then
\[ \int_a^b f(x) \, dx = F(b) - F(a) \]
This incredible result provides a way to precisely calculate the area under the curve of the function \( f(x) \) by merely evaluating its antiderivative at the bounds of the interval. The original exercise is a clear application of this theorem—by obtaining the antiderivative of the function being integrated and then subtracting its value at \( a \) from the value at \( b \), we can determine the definite integral.
Integration Bounds
Integration bounds, or limits of integration, are the values that define the specific section of the curve you are interested in when calculating a definite integral. They are represented by the lower limit \( a \) and the upper limit \( b \) in the notation \( \int_a^b \). In the exercise, the bounds are \( 1 \) and \( 3 \), signifying the stretch of the curve of \( x^2 \) from \( x=1 \) to \( x=3 \) where the area under the curve is to be found. Changing the order of these bounds, as shown in the exercise, will change the sign of the integral's result because the direction of the area calculation is essentially reversed. The bounds are crucial as they transform the antiderivative into specific, calculable values.

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