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Chapter 5: Q 71. (page 429)

Solve the definite integral.
${鈭�}_{1}^{2} \ln x d x$

Short Answer

Expert verified

The solution is $2 \ln 2 - 1$ .

Step by step solution

01

Step 1. Given information.

The given integral is ${鈭�}_{1}^{2} \ln x d x$ .

02

Step 2. Solve the integral.

$\begin{array}{rcl} {鈭�}_{1}^{2} \ln x d x & = & {[\ln x 鈭� d x - 鈭� \{\frac{d}{d x} (\ln x) 鈭� d x\} d x]}_{1}^{2} \\ = & {[\ln x 路 x - 鈭� \frac{1}{x} 路 x d x]}_{1}^{2} \\ = & {[\ln x 路 x - 鈭� d x]}_{1}^{2} \\ = & {[\ln x 路 x - x]}_{1}^{2} \\ = & [2 \ln 2 - 2] - [1 \ln 1 - 1] \\ = & 2 \ln 2 - 2 + 1 \\ = & 2 \ln 2 - 1 \end{array}$

03

Step 3. Simplified answer.

Hence the required solution is $2 \ln 2 - 1$ .

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Most popular questions from this chapter

Consider the integral $鈭� \sin x \cos x d x$ .

(a) Solve this integral by using u-substitution with $u = \sin x$ and $d u = \cos x d x$ .

(b) Solve the integral another way, using u-substitution with $u = \cos x$ and $d u = 鈭� \sin x d x$ .

(c) How must your two answers be related? Use algebra to prove this relationship.

Complete the square for each quadratic in Exercises 28鈥�33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$2 x^{2} 鈭� 4 x + 1$

Solve given definite integral.

${鈭�}_{1}^{2} 鈥� \frac{1}{x \sqrt{9 鈭� x^{2}}} d x$

Explain why it makes sense to try the trigonometric substitution $x = \sec 鈦� u$ if an integrand involves the expression $\sqrt{x^{2} 鈭� 1}$

Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u 鈭� a^{2}}$ is $鈭� a \tan u$ if $x < 鈭� a$ and is $a \tan u$ if $x > a$ . Your explanation should include a discussion of domains and absolute values.

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