/*! 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 50 Evaluate the definite integral. ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 50

Evaluate the definite integral. $$ \int_{1}^{3} x^{2} \ln x d x $$

Short Answer

Expert verified
The value of the definite integral \( \int_{1}^{3} x^{2} \ln x dx \) is given by \( [\frac{1}{3} x^{3} (\ln x - 1) + \frac{2}{9} x^{3}]_{1}^{3} \), which when calculated will yield the final numerical answer.

Step by step solution

01

Using the formula for integration by parts.

We choose \( u = x^{2} \) and \( dv = \ln x dx \). Hence, \( du = 2x dx \) and \( v = \int \ln x dx = \int 1 \cdot \ln x dx = x \ln x - x \). The formula for integration by parts gives a new integral: \( \int u dv = uv - \int v du = [x^{2} (x \ln x - x)] - \int (x \ln x - x) 2x dx = x^{3} \ln x - x^{3} - 2 \int x^{2} \ln x dx + 2 \int x^{2} dx \).
02

Solving for the integral.

The integral \( \int x^{2} \ln x dx \) appears on both sides of the equation. Therefore, let's add \( 2 \int x^{2} \ln x dx \) to both sides to get \( 3 \int x^{2} \ln x dx = x^{3} \ln x - x^{3} + 2 \int x^{2} dx \). Hence, \( \int x^{2} \ln x dx = \frac{1}{3} (x^{3} \ln x - x^{3} + 2 \int x^{2} dx) \). The integration of \( x^{2} \) is straightforward. We get \( \int x^{2} dx = \frac{1}{3} x^{3} \). Substituting this into the equation gives \( \int x^{2} \ln x dx = \frac{1}{3} (x^{3} \ln x - x^{3} + 2(\frac{1}{3} x^{3})) \). This simplifies to \( \int x^{2} \ln x dx = \frac{1}{3} x^{3} (\ln x - 1) + \frac{2}{9} x^{3} \).
03

Evaluating the definite integral.

The definite integral of \( x^{2} \ln x \) from \( x = 1 \) to \( x = 3 \) is given by \( \int_{1}^{3} x^{2} \ln x dx = [\frac{1}{3} x^{3} (\ln x - 1) + \frac{2}{9} x^{3}]_{1}^{3} \). After evaluating this expression at \( x = 3 \) and at \( x = 1 \) and subtracting, you will obtain the final answer.

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Most popular questions from this chapter

Use the Trapezoidal Rule and Simpson's Rule to approximate the value of the definite integral for the indicated value of \(n\). Compare these results with the exact value of the definite integral. Round your answers to four decimal places. $$ \int_{0}^{2} x^{3} d x, n=8 $$

Use the error formulas to find \(n\) such that the error in the approximation of the definite integral is less than \(0.0001\) using (a) the Trapezoidal Rule and (b) Simpson's Rule. $$ \int_{0}^{1} x^{3} d x $$

Consumer Trends The rate of change \(S\) in the number of subscribers to a newly introduced magazine is modeled by \(d S / d t=1000 t^{2} e^{-t}, 0 \leq t \leq 6\), where \(t\) is the time in years. Use Simpson's Rule with \(n=12\) to estimate the total increase in the number of subscribers during the first 6 years.

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{2} \frac{x}{\sqrt{4-x^{2}}} d x $$

Median Age The table shows the median ages of the U.S. resident population for the years 1997 through \(2005 .\) (Source: U.S. Census Bureau) \begin{tabular}{|l|c|c|c|c|c|} \hline Year & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline Median age & \(34.7\) & \(34.9\) & \(35.2\) & \(35.3\) & \(35.6\) \\ \hline \end{tabular} \begin{tabular}{|l|c|c|c|c|} \hline Year & 2002 & 2003 & 2004 & 2005 \\ \hline Median age & \(35.7\) & \(35.9\) & \(36.0\) & \(36.2\) \\ \hline \end{tabular} (a) Use Simpson's Rule to estimate the average age over the time period. (b) A model for the data is \(A=31.5+1.21 \sqrt{t}\), \(7 \leq t \leq 15\), where \(A\) is the median age and \(t\) is the year, with \(t=7\) corresponding to 1997 . Use integration to find the average age over the time period. (c) Compare the results of parts (a) and (b).
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