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Chapter 12: Problem 49

Evaluate the definite integral. $$ \int_{1}^{4} x \ln x d x $$

Short Answer

Expert verified
The definite integral of \(x \ln x\) from 1 to 4 is \(4 \ln(4) + 3.75\).

Step by step solution

01

Identify \(u\) and \(dv\)

Let \(u\) be \(ln(x)\) and \( dv\) be \(x dx\). These parts are selected this way to make the integral simpler, as the derivative of \(ln(x)\) is easier to manage.
02

Calculate \(du\) and \(v\)

Now compute \(du\) as the derivative of \(u\), and \(v\) as the antiderivative of \(dv\). We obtain that \(du = \dfrac{1} {x} dx\) and \(v = \dfrac{x^2} {2}\).
03

Apply Integration by Parts

Substitute \(u\), \(v\), \(dv\) and \(du\) into the integration by parts formula \(\int u dv = u \cdot v - \int v du\). This simplifies to \(\int_{1}^{4} x \ln x dx = [\ln(x) \cdot \dfrac{x^2} {2}]_{1}^{4} - \int_{1}^{4} \dfrac{x^2} {2} \cdot \dfrac{1} {x} dx\). Simplifying the second integral, obtain \(\int_{1}^{4} \dfrac{x} {2}dx\).
04

Simplify Integral and Evaluate

Perform integration of the simpler integral, and substitute the limits of integration. The result is: \(\int_{1}^{4} \dfrac{x} {2} dx = [\dfrac{x^2} {4}]_{1}^{4} = 4-0.25 = 3.75\). After that, substitute limits of integration into the part of the integration by parts formula: \([\ln(x) \cdot \dfrac{x^2} {2}]_{1}^{4} = [4 \ln(4) -1 \ln(1)] = 4 \ln(4)\). The final result will be the sum of these two components.
05

Obtain Final Result

The final step is to add the results obtained in the previous step. So, \(\int_{1}^{4} x \ln x dx\) = \(4 \ln(4) + 3.75\).

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Most popular questions from this chapter

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. $$ \int_{0}^{2} \frac{x}{\sqrt{4-x^{2}}} d x $$

Population Growth \(\ln\) Exercises 57 and 58, use a graphing utility to graph the growth function. Use the table of integrals to find the average value of the growth function over the interval, where \(N\) is the size of a population and \(t\) is the time in days. $$ N=\frac{375}{1+e^{4.20-0.25 t}}, \quad[21,28] $$

Use a spreadsheet to complete the table for the specified values of \(a\) and \(n\) to demonstrate that \(\lim _{x \rightarrow \infty} x^{n} e^{-a x}=0, \quad a>0, n>0\) \begin{tabular}{|l|l|l|l|l|} \hline\(x\) & 1 & 10 & 25 & 50 \\ \hline\(x^{n} e^{-a x}\) & & & & \\ \hline \end{tabular} $$ a=\frac{1}{2}, n=5 $$

Use a program similar to the Simpson's Rule program on page 906 to approximate the integral. Use \(n=100\). $$ \int_{2}^{5} 10 x e^{-x} d x $$

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. $$ \int_{0}^{1}\left(\frac{x^{2}}{2}+1\right) d x, n=4 $$
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