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Chapter 5: Q 88. (page 431)

Use the product rule to derive the formula for integration by parts in theorem 5.7.

Short Answer

Expert verified

The solution is $鈭� u v d x = u 鈭� v d x - 鈭� [\frac{d}{d x} (u) 鈭� v d x] d x$ .

Step by step solution

01

Step 1. Given information.

The product rule of differentiation.

$\frac{d}{d x} (u v) = u \frac{d}{d x} (v) + v \frac{d}{d x} (u)$

02

Step 2. Derivation of the formula for integration by parts.

If $y = u v$ then, $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ .

Rearranging this rule:

$u \frac{d v}{d x} = \frac{d (u v)}{d x} - v \frac{d u}{d x}$

Now integrate both sides:

$鈭� u \frac{d v}{d x} d x = 鈭� \frac{d (u v)}{d x} d x - 鈭� v \frac{d u}{d x} d x$

The first term on the right simplifies since we are simply integrating.

$鈭� u \frac{d v}{d x} d x = u v - 鈭� v \frac{d u}{d x} d x$

03

Step 3. Simplified answer.

This is the formula known as integration by parts.

$鈭� u \frac{d v}{d x} d x = u v - 鈭� v \frac{d u}{d x} d x$

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Most popular questions from this chapter

Solve given definite integral.

${鈭�}_{3}^{5} 鈥� \sqrt{x^{2} 鈭� 9} d x$

For each function u(x) in Exercises 9鈥�12, write the differential du in terms of the differential dx.

$u (x) = x^{2} + x + 1$

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

$鈭� e^{u} d u$

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

$鈭� \frac{1}{\sqrt{u}} d u$

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.

See all solutions

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