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Chapter 5: Q 73. (page 429)

Solve the definite integral.
${鈭�}_{- 1}^{1} \frac{x}{e^{x}} d x$

Short Answer

Expert verified

The solution is $- \frac{2}{e}$ .

Step by step solution

01

Step 1. Given information.

The given integral is ${鈭�}_{- 1}^{1} \frac{x}{e^{x}} d x$ .

02

Step 2. Solve the integral.

$\begin{array}{rcl} {鈭�}_{- 1}^{1} \frac{x}{e^{x}} d x & = & {鈭�}_{- 1}^{1} x e^{- x} d x \\ = & {[x 鈭� e^{- x} d x - 鈭� \{\frac{d}{d x} (x) 鈭� e^{- x} d x\} d x]}_{- 1}^{1} \\ = & {[- x e^{- x} - 鈭� - e^{- x} d x]}_{- 1}^{1} \\ = & {[- x e^{- x} + 鈭� e^{- x} d x]}_{- 1}^{1} \\ = & {[- x e^{- x} - e^{- x}]}_{- 1}^{1} \\ = & [- 1 e^{- 1} - e^{- 1}] - [- (- 1) e^{- (- 1)} - e^{- (- 1)}] \\ = & [- 2 e^{- 1}] - [e - e] \\ = & - \frac{2}{e} - 0 \\ = & - \frac{2}{e} \end{array}$

03

Step 3. Simplified answer.

Hence, the required solution is $\begin{array}{rcl} - \frac{2}{e} \end{array}$ .

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Most popular questions from this chapter

Solve the integral $鈭� x^{3} \sqrt{x^{2} - 1} d x$ three ways:

(a) with the substitution $u = x^{2} - 1,$ followed by back substitution;

(b) with integration by parts, choosing localid="1648814744993" $u = x^{2} and d v = x \sqrt{x^{2} - 1} d x;$

(c) with the trigonometric substitution x = sec u.

Solve the integral

$鈭� \sqrt{4 - x} \sqrt{x - 2} d x$

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

Solve the integral: $鈭� \ln (x^{3}) d x$ .

Write $\sin (2 \cos^{- 1} x)$ as an algebraic function.

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